SOLUTIONS MANUAL for Abstract Algebra Structures and Applications, 1e Stephen Lovet

1 | Set Theory 1.1 – Sets and Functions Exercise: 1 Section 1.1 Question: Let U = {n ∈ N | n ≤ 10} and consider the subsets A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 4, 5}, and C = {1, 2, 5, 7, 8}. Calculate the following operations. a) A ∩ B b) (B ∪ C) − A c) (A ∩ B) ∩ (A ∩ C) ∩ (B ∩ C) d) ((A − B) − C) ∩ (A − (B − C)) Solution: We apply the definitions of set operations: a) A ∩ B = {1, 3, 5} b) (B ∪ C) − A = {1, 2, 3, 4, 5, 7, 8} − {1, 3, 5, 7, 9} = {2, 4, 8} c) A ∩ B∩A ∩ C∩B ∩ C = {1, 3, 5}∩{1, 5, 7}∩{1, 2, 5} = {2, 4, 6, 7, 8, 9, 10}∩{2, 3, 4, 6, 8, 9, 10}∩{4, 6, 8, 9, 10} d) ((A − B) − C) ∩ (A − (B − C)) = ({7, 9} − C) ∩ (A − {3, 4}) = {9} ∩ {1, 5, 7, 9} = {9} Exercise: 2 Section 1.1 Question: Let U = {a, b, c, d, e, f, g} and consider the subsets A = {a, b, d}, B = {b, c, e}, and C = {c, d, f }. Calculate the following operations. a) C ∩ (A ∪ B) b) (A ∪ C) − B c) (A ∪ B ∪ C) − (A ∩ B ∩ C) d) (A − B) ∪ (B − C) Solution: We apply the definitions of set operations: a) C ∩ (A ∪ B) = C ∩ {a, b, c, d, e} = {c, d} b) (A ∪ C) − B = {a, b, c, d, f } − B = {a, d, f } c) (A ∪ B ∪ C) − (A ∩ B ∩ C) = {a, b, c, d, e, f } − ∅ = {a, b, c, d, e, f } d) (A − B) ∪ (B − C) = {a, d} ∪ {b, e} = {a, b, d, e} Exercise: 3 Section 1.1 Question: As subsets of the reals, describe the differences between the sets {3, 5}, [3, 5] and (3, 5). Solution: The set {3, 5} contains the integers 3 and 5. The closed interval [3, 5] contains all real numbers between 3 and 5 including 3 and 5, while the open interval (3, 5) contains all real numbers between 3 and 5 not including 3 and 5. Exercise: 4 Section 1.1 Question: Prove that the following are true for all sets A and B. a) A ∩ B ⊆ A. b) A ⊆ A ∪ B. Solution: We use the definitions of subsets and the intersection and union of sets. a) Let x ∈ A ∩ B. Then x ∈ A and x ∈ B =⇒ x ∈ A, so A ∩ B ⊆ A. b) Let x ∈ A. We know that A ∪ B = {y | y ∈ A or y ∈ B}, so x ∈ A =⇒ x ∈ A ∪ B. Hence A ⊆ A ∪ B. Exercise: 5 Section 1.1 Question: Let A and B be subsets of a set S. a) Prove that A ⊆ B if and only if P(A) ⊆ P(B) b) Prove that P(A ∩ B) = P(A) ∩ P(B). c) Show that P(A ∪ B) = P(A) ∪ P(B) if and only if A ⊆ B or B ⊆ A. Solution: 1

2

CHAPTER 1. SET THEORY a) (=⇒): Suppose A ⊆ B. Then, ∀a ∈ A, a ∈ B. Since P(B) contains all the possible subsets of B, all the possible subsets of A must be in P(B) because A ⊆ B. Therefore, P(A) ⊆ P(B). (⇐=): Suppose P(A) ⊆ P(B). Then ∀{a} ∈ P(A), {a} ∈ P(B). Therefore, there must exist a subset C of P(B) that contains every {a} from P(A). The subset C leads to the conclusion that every a ∈ A must also be in B. Therefore, A ⊆ B. b) By definition, P(A ∩ B) = {{t1 , t2 , ..., tn } | ti ∈ A, ti ∈ B}. This implies {ti } ∈ P(A) and {ti } ∈ P(B). Therefore, by definition of intersection, P(A ∩ B) = P(A) ∩ P(B). c) (=⇒): Suppose there are two sets A and B such that neither A ⊆ B nor B ⊆ A. Let a ∈ A − B and b ∈ B − A. Then the set {a, b} is in P(A ∪ B) but not in P(A) or in P(B). Therefore by the contrapositive, P(A ∪ B) = P(A) ∪ P(B) if A ⊆ B or B ⊆ A. (⇐=): Suppose A ⊆ B. Then, A ∪ B = B so P(A ∪ B) = P(B). Now suppose B ⊆ A. Then A ∪ B = A so P(A ∪ B) = P(A). Either way, P(A ∪ B) = P(A) ∪ P(B).

Exercise: 6 Section 1.1 Question: Give the list description of P({1, 2, 3, 4}). Solution: Using the definition of a power set, P({1, 2, 3, 4}) ={∅, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}}.

Exercise: 7 Section 1.1 Question: Give the list description of {{a1 , a2 , . . . , ak } ∈ P({1, 2, 3, 4, 5}) a1 + a2 + · · · + ak = 8}. Solution: We need to find all the subsets of {1, 2, 3, 4, 5} whose elements add to a total of 8. Recall that no subset has repeated elements so {4, 4, } does not make sense. The set is {{1, 2, 5}, {1, 3, 4}, {3, 5}} .

Exercise: 8 Section 1.1 Question: Let A, B, and C be subsets of a set S. a) Prove that (A − B) − C = (A − C) − (B − C). b) Find and prove a similar formula for A − (B − C). Solution: S

S

B

A

B

C

A

C

a) In the first Venn diagram, the lighter shade represents (A − B), and the darker shade, which overlaps some of (A − B), represents (A − B) − C. In the second Venn diagram, the lighter shade represents (A − C), while the darker shade represents (A − C) − (B − C). We observe from the diagrams that the darker regions are equal.

1.1. SETS AND FUNCTIONS

3 S

S

B

A

B

C

A

C

b) In the Venn diagram above, the lighter shade represents B −C, and the darker shade represents A−(B −C). In the second diagram, the lighter region represents A − B, and the darker region represents A − C, which overlaps some of A − B. Thus, (A − B) ∪ (A − C) = A − (B − C). Exercise: 9 Section 1.1 Question: Let A, B, and C be subsets of a set S. a) Prove that A4B = ∅ if and only if A = B. b) Prove that A ∩ (B4C) = (A ∩ B)4(A ∩ C). Solution: Let A, B, and C be subsets of a set S. a) Suppose that A4B = ∅. Then by definition of the symmetric difference (A − B) ∪ (B − A) = ∅. If the union of two sets is the empty set, then each of the two sets must be empty. Hence we deduce that A − B = ∅ and B − A = ∅. Now for and two sets U and T , the identity U − T = ∅ is equivalent to U ⊆ T . Hence we deduce that A ⊆ B and B ⊆ A. Consequently, A = B. The argument of the opposite direction is identical. Suppose that A = B. Then A ⊆ B and B ⊆ A. Thus A − B = ∅ and B − A = ∅. We deduce that A4B = (A − B) ∪ (B − A) = ∅. b) There are a variety of ways to prove the identity A ∩ (B4C) = (A ∩ B)4(A ∩ C). We could use a well designed Venn diagram. We could also use a membership table which lists all possibilities of an element whether it is in or not in one √ of the given three sets. Here is a membership table for both side of the equality. In this table, we put an in a column to indicate membership and nothing to indicate non-membership. √ Hence if there is a in the A and C column and nothing in the B column, that refers to the situations of an element in A, not in B and in C. A √ √ √ √

B √ √

C √ √

√ √

(B4C)

A ∩ (B4C)

√ √

√ √

(A ∩ B) √ √

(A ∩ C) √ √

(A ∩ B)4(A ∩ C) √ √

√ √

√ √

Since the A ∩ (B4C) and column and the (A ∩ B)4(A ∩ C) of this membership table are the same, then the sets are equal. Exercise: 10 Section 1.1 Question: Let S be a set and let {Ai }i∈I be a collection of subsets of S. Prove the following. [ \ a) Ai . Ai = i∈I

b)

\ i∈I

i∈I

Ai =

[

Ai .

i∈I

Solution: Let S be a set and let {Ai }i∈I be a collection of subsets of S.

4

CHAPTER 1. SET THEORY a) We will prove the equation by proving set inclusion in both directions. [ [ (=⇒) Let a ∈ Ai . Then a ∈ / Ai or a ∈ / Ai for every i ∈ I. And this implies that a ∈ Ai for every i∈I

i∈I

i ∈ I. So then a ∈

\

[

Ai . And so

i∈I

\

(⇐=) Let a ∈

\

Ai ⊆

i∈I

Ai .

i∈I

Ai . Then a ∈ Ai for every i ∈ I. Which implies that a ∈ / Ai for eversy i ∈ I. And so

i∈I

a∈ /

[ i∈I

And so

[

Ai . Which implies a ∈

Ai . So

i∈I

[

Ai =

i∈I

\

\

Ai ⊆

i∈I

[

Ai .

i∈I

Ai .

i∈I

b) We will prove this equation by proving set inclusion in both directions. \ \ (=⇒) Let a ∈ Ai . So a ∈ / Ai . So a ∈ / Ai for at least one i ∈ I. So a ∈ Ai for at least one i ∈ I. So i∈I

a∈

[

i∈I

Ai . Which implies that

i∈I

\

Ai ⊆

i∈I

(⇐=) Let a ∈

[

[

Ai .

i∈I

Ai . Then a ∈ Ai for at least one i ∈ I. This implies that a ∈ / Ai for at least one i ∈ I.

i∈I

It follows that a ∈ /

\

Ai . Which implies that a ∈

i∈I

i∈I

So

\ i∈I

Ai =

[

\

Ai . So

[

Ai ⊆

i∈I

\

Ai .

i∈I

Ai .

i∈I

Exercise: 11 Section 1.1 Question: Let P be the parabola in the plane whose equation is y = x2S . Let {Aq }q∈P be the collection of subsets of R2 where Aq is the tangent line to P at q. Determine with proof q∈P Aq . Solution: Sketching a picture of the standard parabola, we see that all the tangent lines are in some sense beneath the parabola. Also, taking the (infinite, uncountable) union of all the tangent lines to the parabola, we might guess that we would get all points (a, b) in the plane such that b ≤ a2 . We need to prove this hypothesis. Label the coordinates of a point q ∈ P as q = (x0 , x20 ). From calculus, the tangent line to P at q has the equation y = x20 + 2x0 (x − x0 ) =⇒ y = 2x0 x − x20 . Now suppose that some point (a, b) in the plane is on a tangent line. Then, for some x0 , we have b = 2x0 a − x20 . Since a, b are given and x0 is unknown, this is an equation in x0 . The quadratic formula gives for x0 x0 =

2a ±

√

p 4a2 − 4b = a ± a2 − b. 2

In particular, we note that there exists an x0 if and only if a2 − b ≥ 0, confirming the hypothesis that b ≤ a2 .

Exercise: 12 Section 1.1 3 3 Question: Let {AT k }k∈N be the collection of subsets of R such that Ak = {(x, y, z) ∈ R | z ≥ ky}. Determine S both k∈N Ak and k∈N Ak . Solution: Each subset Ak represents all points greater than the plane that is bound on the line z = ky. In the diagram below, imagine the x-axis is coming out of the page, and let k1 , k2 , and k3 represent the bounds of the subsets A1 , A2 , and A3 respectively. We note that the larger k gets, S the more steep the plane becomes. However, the condition does not hold true for z < 0 when y is 0. Therefore, k∈N Ak = {(x, y, z) | z ≥ y}∪{(x, y, z) | x < 0}, T and k∈N Ak = {(x, y, z) | z ≥ y} ∩ {(x, y, z) | x ≤ 0}.

1.1. SETS AND FUNCTIONS

5 z

k3 k2

k1

y

Exercise: 13 Section 1.1 Question: In geometry of the plane, a subset S of the plane is called convex if for all p, q ∈ S the line segment pq connecting p and q is a subset of S. Prove that the intersection of two convex sets is a convex set. Solution: Let S and R both be convex sets. Consider S ∩ R. Case 1: If S ∩ R is empty, than it is trivially a convex set. Case 2: If it is non-empty, for any p, q ∈ S ∩ R, consider the line segment pq. Since p, q ∈ S and S is convex, we know that pq ⊆ S. By the same reasoning, since p, q ∈ R and R is convex, this implies that pq ⊆ R. Since pq is a subset of both S and R, we have pq ⊆ S ∩ R. So S ∩ R is convex. This proves that S ∩ R is convex. Exercise: 14 Section 1.1 Question: Inclusion-Exclusion Principle. Let A, B, and C be finite subsets of a set S. a) Prove that |A ∪ B| = |A| + |B| − |A ∩ B|. b) (*) Prove that |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. Solution: a) Let |A| = n and |B| = m. Now A ∪ B is all of the elements that are either in A or B. If we say that |A ∪ B| = |A| + |B| = n + m, for any element c that exists in both A and B, or equivalently A ∩ B, we account for that element twice, once in |A| and the second in |B|. So we must subtract 1 for the |A ∩ B| elements we account for twice. So |A ∪ B| = |A| + |B| − 1(|A ∩ B|) = |A| + |B| − |A ∩ B|. b) Let D = A∪B and consider |D∪C|. By applying the first result of this exercise, |D∪C| = |D|+|C|−|D∩C|. And if we put A ∪ B back in for D and apply the first result again we get |A ∪ B| + |C| − |(A ∪ B) ∩ C| = |A| + |B| − |A ∩ B| + |C| − |(A ∪ B) ∩ C|. Now we will consider |(A ∪ B) ∩ C|. (A ∪ B) ∩ C contains all the elements that are in (A or B) and C. If A and C have n elements in common, or equivalently |A ∩ C| = n, and likewise |B ∩C| = m, then |(A∪B)∩C| is certainly at most m+n. However, we should not double count whatever elements are in both A∩C and B ∩C, or A∩B ∩C. So we must subtract out that amount, and we arrive at |(A ∪ B) ∩ C| = |A ∩ C| + |B ∩ C| − |A ∩ B ∩ C|. Plugging this result back into our original equation we get |A| + |B| − |A ∩ B| + |C| − |(A ∪ B) ∩ C| = |A| + |B| + |C| − |A ∩ B| − (|A ∩ C| + |B ∩ C| − |A ∩ B ∩ C|) = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|. Exercise: 15 Section 1.1 Question: Let U be a set and A, B ⊆ U . a) Show by any means that A ∩ B = A ∪ B. b) Show by any means that A ∪ B = A ∩ B. Solution:

6

CHAPTER 1. SET THEORY a) Observe the Venn diagrams below: U

A

B

U

A

B

In the above diagrams, the left diagram represents A ∩ B, and the right diagram represents A ∪ B. In the A ∪ B diagram, represents A, represents B, and represents where they overlap. A ∪ B is represented by the union of these shaded regions. We can observe the shaded regions from both diagrams describes the same set, thus A ∩ B = A ∪ B. b) Observe the Venn diagrams below: U

A

B

U

A

B

In the above diagrams, the left diagram represents A ∪ B, and the right diagram represents A ∩ B. In the A ∩ B diagram, represents A, represents B, and represents A ∩ B. Notice, the shaded region describes the same set that is shaded in the A ∪ B diagram, hence A ∪ B = A ∩ B.

Exercise: 16 Section 1.1 Question: (*) Let n be a positive integer. Describe an algorithm (a finite list of well-defined instructions to accomplish a task) to list all the subsets of {1, 2, 3, . . . , n}. Solution: Let S = {{∅}} and k = 1. For every set in S, add a new set to S representing the union of {k} and the set in S. Once this process is completed union {k} to S. Increase k by 1. Repeat this process until k = n. In other words, while k < n, S = S ∪ {s | t ∪ {k}, ∀t ∈ S} ∪ {k}.

Exercise: 17 Section 1.1 Question: For each of these real-valued functions determine the largest possible domain D as a subset of R and then prove whether f : D → R is an injection, surjection, both, or neither. a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 1 Solution: We decide if the function is an injection, surjection, both, or neither. a) Let f (x) = −3x + 4. This function can be defined over the whole domain R. Suppose that f (x1 ) = f (x2 ). Then −3x1 + 4 = −3x2 + 4. This implies that −3x1 = −3x2 so x1 = x2 . This shows that f (x) is injective. To prove surjectivity, we attempt to solve for x in the expression y = f (x) for an arbitrary y. We get x = 4−y 3 . Since there is a solution in x for any y, then f is surjective. (f is bijective.) b) Let f (x) = −3x2 + 7. The largest possible domain of definition is R. Note that f (−1) = 4 = f (1). This shows that f is not injective. To p test for surjectivity, we attempt to solve for x in y = f (x). The equation y = −3x2 + 7 leads to x = ± (7 − y)/3. However, if y > 7 there is no solution for x. Hence, since the codomain of f is R, f is not surjective. (f is neither.) c) Let f (x) = (x+1)/(x+2). This function is defined on R−{−2}. This is the domain. To test for injectivity,

1.1. SETS AND FUNCTIONS

7

suppose that f (x1 ) = f (x2 ). Then we have x1 + 1 x1 + 1 = =⇒ (x1 + 1)(x2 + 2) = (x1 + 2)(x2 + 1) x1 + 2 x1 + 2 =⇒ x1 x2 + 2x1 + x2 + 2 = x1 x2 + x1 + 2x2 + 2 =⇒ 2x1 + x2 = x1 + 2x2 =⇒ x1 = x2 . This shows that f is injective. To test for surjectivity, we attempt to solve y = f (x) for x given arbitrary y. We have 1 − 2y x+1 =⇒ yx + 2y = x + 1 =⇒ xy − x = 1 − 2y =⇒ x = . y= x+2 y−1 We see that there is no solution for x if y = 1. Hence f (x) is not surjective. d) Consider the function f (x) = x5 + 1. This is defined over all R so this is the largest possible domain in R. To check for injectivity, consider the equality f (x1 ) = f (x2 ). This gives x51 + 1 = x52 + 1 =⇒ x51 − x52 = 0 =⇒ (x1 − x2 )(x41 + x31 x2 + x21 x22 + x1 x32 + x42 ) = 0. Obviously the equation holds if x1 = x2 . Now we look for solutions of the second term. Note that if x2 = 0, then the quartic equation implies that x1 = 0. But then x1 = x2 , which we already know to be a possibility. Assuming that x1 6= x2 , after division by x42 the quartic term implies

x1 x2

4

+

x1 x2

3

+

x1 x2

2

+

x1 x2

+1=0

A graph of the function g(x) = x4 + x3 + x2 + x + 1 shows that g(x) has no solutions. Thus, the only solution to √ f (x1 ) = f (x2 ) is x1 = x2 . Thus f is injective. For surjectivity, we see that y = f (x) implies that x = 5 y − 1, which is defined for all y. Hence f is surjective. (f is bijective.) Exercise: 18 Section 1.1 Question: Given an explicit example of a function f : Z → Z that is a) bijective; b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective. Solution: Recall that bxc takes x and returns the nearest integer less than or equal to x. The solutions presented here are not the only options! a) f (x) = −x. b) f (x) = bx/2c. c) f (x) = x ∗ 2. d) f (x) = x2 . Exercise: 19 Section 1.1 Question: Given an explicit example of a function f : N → N that is a) bijective; b) surjective but not injective; c) injective but not surjective; d) neither injective nor surjective. Solution: Recall that bxc takes x and returns the nearest integer less than or equal to x. The answers presented here are not the only options! a) f (x) = x. b) f (x) = bx/2c. c) f (x) = 2 ∗ x.

8

CHAPTER 1. SET THEORY d) f (x) = 2 ∗ (bx/2c).

Exercise: 20 Section 1.1 Question: Let f : A → B and g : B → C be functions. Prove that if f and g are bijective, then g ◦ f is bijective and (g ◦ f )−1 = f −1 ◦ g −1 . Solution: Assume g ◦ f is not injective. Then there would exist a1 , a2 ∈ A such that a1 6= a2 and g ◦ f (a1 ) = g ◦ f (a2 ). Let b1 = f (a1 ) and b2 = f (a2 ). We know that b1 6= b2 because f is bijective. Using substitution, we notice that g(b1 ) = g(b2 ), but b1 6= b2 . This creates a contradiction since g is bijective. Thus, g ◦ f is injective. Now assume that g ◦ f is not surjective. Then there would exist at least one c ∈ C such that ∀a ∈ A, g(f (a)) 6= c. By substituting we observe ∀b ∈ B, g(b) 6= c. However, since g is bijective, this creates a contradiction. Therefore g ◦ f is surjective. Hence, g ◦ f is bijective. Let f (a) = b and g(b) = c. Then g(f (a)) = c so (g ◦ f )−1 (c) = a. Therefore, (g ◦ f )−1 (c) = a = f −1 (b) = f −1 (g −1 (c)) so (g ◦ f )−1 = f −1 ◦ g −1 . Exercise: 21 Section 1.1 Question: Suppose that f and g are functions and that f ◦ g is injective. a) Prove that g is injective. b) Does it also follow that f is injective? Justify your answer (with a proof or counter-example). Solution: a) For the sake of contradiction, we assume g is not injective. Then for some a, b that exist in the domain of g, g(a) = g(b). However, then f (g(a)) = f (g(b)) as well. This contradicts the injectivity of f ◦ g. Therefore g must be injective. b) No it does not. Consider the following functions f, g : Z −→ Z: ( 2x if x ≥ 0 g(x) = 2|x| + 1 if x < 0. f (x) = x2 . Now, g sends non-negative numbers to a unique, non-negative, even number and negative numbers to a unique, positive, odd number. However, f is injective on non-negative numbers so that f ◦ g is injective, but is not injective on all of Z. (Note: What is necessary is that f be injective on the range of g (in our case, the nonnegative numbers) but not necessarily the codomain of g(in our case, all of Z)) Exercise: 22 Section 1.1 Question: Suppose that f and g are functions and that f ◦ g is surjective. a) Prove that f is surjective. b) Does it also follow that g is surjective? Justify your answer (with a proof or counter-example). Solution: a) Consider any a in the codomain of f . Since a is also in the codomain of f ◦ g and f ◦ g is a surjective function, there must exist some b in the domain of f ◦ g so that a = (f ◦ g)(b). Set c = g(b). Then (f ◦ g)(b) = f (g(b)) = f (c) = a. So every element of the codomain is hit by f and f is surjective. b) No it does not. Consider the functions f, g : N −→ N where f (x) = bx/2c and g(x) = 2x. Then for any y ∈ N, (f ◦ g)(y) = f (g(y)) = f (2 ∗ y) = b(2 ∗ y)/2c = byc = y. So f ◦ g is surjective, but g is certainly not since no odd, positive numbers will be hit. Exercise: 23 Section 1.1 Question: Restate the definition of (a) injective and (b) surjective as applied to a function f : A → B in terms of properties of the sets f −1 ({b}). Solution:

1.1. SETS AND FUNCTIONS

9

a) For injectivity, |f −1 ({b})| ≤ 1. b) For surjectivity, |f −1 {b})| ≥ 1. Exercise: 24 Section 1.1 Question: For the following functions f , find the pre-image (or fiber) f −1 (T ) of the given set T in the codomain. √ a) f : R → R with f (x) = sin x and T = { 3/2}. b) f : R → R with g(x) = x2 − 2 and T = [1, 2]. c) f : R → R with h(x) = x3 − 2x and T = [−1, 0]. Solution: Note that the pre-image of f −1 (T ) represents all of the domain values that map to the elements of T. √ a) f −1 (T ) = sin−1 ( 23 ) = { π3 , 2π 3 } b) f −1 (T ) = g −1 ([1, 2]) √ x2 − 2 = 1 ⇒ x = ± 3 x2 − 2 = 2 ⇒ x = ±2 f (x) x2 − 2

2 1

x −2

−1

1

2

−1 −2 √ √ Therefore, the pre-image of g −1 ([1, 2]) is {[−2, − 3], [ 3, 2]}. c) f −1 (T ) = h−1 ([−1, 0]) Notice that 1 is a zero of the polynomial x3 − 2x + 1. This allows us to find the x-values that satisfy x3 − 2x = −1. x3 − 2x + 1 = (x − 1)(x2 + x − 1) √ −1 ± 5 x = 1, 2 √ x3 − 2x = 0 ⇒ x = 0, ± 2 f (x) x3 − 2x

2 1 −2

−1

1

2

x

−1 −2 √ √ √ √ Therefore, the pre-image of h−1 ([−1, 0]) is {[ −1−2 5 , − 2], [0, −1+2 5 ], [1, 2]}.

10

CHAPTER 1. SET THEORY

Exercise: 25 Section 1.1 Question: Let f : A −→ B be a function from the set A to the set B. Let S and T be subsets of the domain A. 1. Show that f (S ∪ T ) = f (S) ∪ f (T ). 2. Show that f (S ∩ T ) ⊆ f (S) ∩ f (T ). 3. Find an example of a function f : A → B and subsets S and T in A such that f (S ∩ T ) 6= f (S) ∩ f (T ). Solution: a) We prove first that f (S ∪ T ) ⊆ f (S) ∪ f (T ). Suppose that y ∈ f (S ∪ T ). Then there exists x ∈ S ∪ T such that f (x) = y. We have x ∈ S or x ∈ T , so y = f (x) ∈ f (S) or y = f (x) ∈ f (T ). Hence y ∈ f (S) ∪ f (T ), so we deduce that f (S ∪ T ) ⊆ f (S) ∪ f (T ). Conversely, suppose that y ∈ f (S) ∪ f (T ). This y ∈ f (S) or y ∈ f (T ). We deduce that there exists x ∈ S with f (x) = y or there exists x0 ∈ T with f (x0 ) = y. Thus, there exists an x ∈ S ∪ T such that x = f (y) and we deduce that f (S) ∪ f (T ) ⊆ f (S ∪ T ). With these two set inclusions, we conclude that f (S ∪ T ) = f (S) ∪ f (T ). b) Let y ∈ f (S ∩ T ). By definition, there exists x ∈ S ∩ T such that y = f (x). Then x ∈ S and x ∈ T so y = f (x) ∈ f (S) and y = f (x) ∈ f (T ). Thus y ∈ f (S) ∩ f (T ). We conclude that f (S ∩ T ) ⊆ f (S) ∩ f (T ). c) Consider the function f : R → R with f (x) = x2 . Setting S = [−2, −1] and T = [1, 2], we find that f (S ∩ T ) = f (∅) = ∅ f (S) ∩ f (T ) = [1, 4] ∩ [1, 4] = [1, 4] Obviously, these are not equal sets. Note that if we attempted to prove that f (S) ∩ f (T ) ⊆ f (S ∩ T ), the reasoning would go as follows. Let y ∈ f (S) ∩ f (T ). Then y ∈ f (S) and y ∈ f (T ). Then there exists x ∈ S with f (x) = y and there exists x0 ∈ T with f (x0 ) = y. However, since x does not have to be equal to x0 , there does not have to exist an element x00 ∈ S ∩ T such that f (x00 ) = y. Exercise: 26 Section 1.1 Question: Let f : A −→ B be a function from the set A to the set B. Let V and W be subsets of the codomain B. Show the following. a) f −1 (V ∪ W ) = f −1 (V ) ∪ f −1 (W ). b) f −1 (V ∩ W ) = f −1 (V ) ∩ f −1 (W ). Solution: a) We will prove the equality by showing set inclusion in both directions. (=⇒) Consider any element a ∈ f −1 (V ∪ W ). Then f (a) ∈ V ∪ W so that f (a) exists in V or W . Then a ∈ f −1 (V ) or f −1 (W ). Which implies a ∈ f −1 (V ) ∪ f −1 (W ). This shows that f −1 (V ∪ W ) ⊆ f −1 (V ) ∪ f −1 (W ). (⇐=) Consider any element b ∈ f −1 (V ) ∪ f −1 (W ). So b exists in at least one of f −1 (V ) or f −1 (W ). Then f (b) exists in V or W . Which implies f (b) ∈ V ∪ W . So then b ∈ f −1 (V ∪ W ). This shows that f −1 (V ) ∪ f −1 (W ) ⊆ f −1 (V ∪ W ). This proves the equality f −1 (V ∪ W ) = f −1 (V ) ∪ f −1 (W ). b) We will prove the equality by showing set inclusion in both directions. (=⇒) Consider any element a ∈ f −1 (V ∩ W ). So f (a) exists in both V and W . Then a exists in both f −1 (V ) and f −1 (W ). Which implies that a ∈ f −1 (V ) ∩ f −1 (W ). This shows that f −1 (V ∩ W ) ⊆ f −1 (V ) ∩ f −1 (W ). (⇐=) Consider any element a ∈ f −1 (V ) ∩ f −1 (W ). So a exists in both f −1 (V ) and f −1 (W ). Then f (a) exists in both V and W . Then, by definition, f (a) ∈ V ∩ W . Which implies that a ∈ f −1 (V ∩ W ). This shows that f −1 (V ) ∩ f −1 (W ) ⊆ f −1 (V ∩ W ). This proves the equality f −1 (V ∩ W ) = f −1 (V ) ∩ f −1 (W ).

1.1. SETS AND FUNCTIONS

11

Exercise: 27 Section 1.1 Question: Let S and T be finite sets with |S| = |T |. Prove that a function f : S → T is injective if and only if it is surjective. Solution: (=⇒): Suppose f : S → T is injective. Then ∀s1 , s2 ∈ S such that f (s1 ) = f (s2 ) implies s1 = s2 . Therefore, every s ∈ S maps to a unique t ∈ T . Since |S| = |T |, there does not exist t ∈ T such that t is not associated with a unique s ∈ S. Thus f is surjective. (⇐=): Suppose f : S → T is surjective. This means that there are at least |T | elements in S that map to T . Assume there exists f (s1 ) = f (s2 ) such that s1 = s2 . Then f implies that |S| elements are mapped to at most |S| − 1 elements in T . Thus there exists at least one t1 ∈ T where there does not exist s ∈ S such that f (s) = t1 . This creates a contradiction because f is surjective. Therefore f is injective.

Exercise: 28 Section 1.1 Question: Let S be a set. For each subset A ⊆ S we define the characteristic function of A as the function χA : S → {0, 1} such that ( 1 if s ∈ A χA (s) = 0 if s ∈ / A. Prove the following. a) The association A 7→ χA is a bijection between P(S) and the set of functions from S to {0, 1}. b) χA∩B (s) = χA (s) · χB (s) for all s ∈ S. c) χA∪B (s) = χA (s) + χB (s) − χA (s) · χB (s) for all s ∈ S. d) χA (s) = 1 − χA (s) for all s ∈ S. e) χA−B (s) = χA (s)(1 − χB (s)) for all s ∈ S. Solution: a) First, notice that each distinct characteristic function represents a unique subset of S. From proposition 1.1.11 we know that there are 2n unique characteristic functions of S. We can write these subsets as n-tuples where each entry contains either a 1 or a 0. We can also picture any function from S to {0, 1} as a n-tuple (f (s1 ), f (s2 ), ..., f (sn )) where |S| = n. Note that each n-tuple will represent a distinct function from S to {0, 1}. In comparing n-tuples, we observe that every function from S to {0, 1} represents a unique characteristic function of S implying that the association is injective. Since each entry has two options, and there are n entries, there are 2n distinct functions from S to {0, 1}. Thus the cardinalities of the functions of S to {0, 1} and P(S) are equal. Using the result in exercise 1.1.27, the association must be surjective as well. Hence, the association from A 7→ χA is a bijection between P(S) and the set of functions from S to {0, 1}. b) Suppose s ∈ A ∩ B. Then, s ∈ A and s ∈ B. This implies χA (s) = 1 and χB (s) = 1. From here we deduce that χA∩B (s) = 1 = χA (s) · χB (s). Now suppose s ∈ / A ∩ B. Then s 6∈ A or s 6∈ B, so χA (s) = 0 or χB (s) = 0. Either way, χA∩B (s) = 0 = χA (s) · χB (s). Therefore, χA∩B (s) = χA (s) · χB (s) for all s ∈ S. c) Suppose s ∈ A ∪ B. Then s ∈ A or s ∈ B. If s ∈ A ∩ B or if s ∈ B ∩ A then we deduce that χA (s) + χB (s) works for both cases. If s ∈ A ∩ B, then we need to account for counting an element twice so we subtract χA∩B (s) which we know to be χA (s) · χB (s) from above. Hence χA∪B (s) = χA (s) + χB (s) − χA (s) · χB (s) for all s ∈ S. d) Suppose s ∈ A. Then s 6∈ A, so χA (s) = 0 and χA (s) = 1. If s ∈ A then s 6∈ A and we get the opposite result. Notice to reverse the results of χA (s) we subtract it from 1. Therefore χA (s) = 1 − χA (s). e) Note that A − B = A ∩ B and χB (s) = 1 − χB (s). Thus χA−B (s) = χA (s)(1 − χB (s)) for all s ∈ S.

Exercise: 29 Section 1.1 Question: Provide all the details in the proof of Proposition 1.1.23. Solution: We start with f (n) = 2n + 21 where f : Z → N. Plugging in some points such as −2, −1, 0, 1, 2

12

CHAPTER 1. SET THEORY

we observe the following: f (−2) = 3 f (−1) = 1 f (0) = 0 f (1) = 2 f (2) = 4

Noticing the negative integers point to odd natural numbers and the positive integers point to even natural numbers, we can describe any integer n in terms of any natural number m by using the definition of even and odd numbers ( m if m is even n = 2 m+1 − 2 if m is odd. Since every integer n is represented by a different natural number m we can quickly deduce that the association is by definition. Assume that there are two integers a and b such that a 6= b but f (a) = f (b). Then, surjective 2a + 21 = 2b + 12 which implies that 2a + 12 = 2b + 12 ⇒ a = b. This creates a contradiction, hence the association is injective as well. Thus since there exists a bijection between Z and N. Therefore, Z is countable.

Exercise: 30 Section 1.1 Question: Let A, B, and C be sets. Prove that if |A| ≤ |B| and |B| ≤ |C|, then |A| ≤ |C|. Solution: Let A, B, and C be sets such that |A| ≤ |B| and |B| ≤ |C|. Also let a be an any element in A, f : A → B, and g : B → C. There exists a unique element in B such that f (a) = b, ∀a ∈ A. Also, there is a unique c ∈ C such that g(b) = c, ∀b ∈ B. By using substitution and composition we find g(f (a)) = c, ∀a ∈ A. This establishes a injective function f : A → C therefore |A| ≤ |C|. Exercise: 31 Section 1.1 Question: Let A and B be finite sets. Prove that the number of distinct functions A → B is |B||A| . Solution: Let the n-tuple, (f (a1 ), f (a2 ), f (a3 ), ..., f (an )), represent a distinct function f : A → B where |A| = n. Each entry in the n-tuple can be mapped to a different element in B resulting in m possibilities where |B| = m. Since every entry has m options and there are n entries, the total number of possible distinct n-tuples is m1 × m2 × m3 × ... × mn = mn . By substituting we observe that the number of distinct functions is equal to |B||A| .

1.2 – The Cartesian Product; Operations; Relations Exercise: 1 Section 1.2 Question: Let A, B, C be sets. Explain why A × B × C is not the same set as A × (B × C). Solution: The set A × B × C contains 3-tuples with the components (a, b, c), however, the set A × (B × C) contains pairs with the components (a, (b, c)). Exercise: 2 Section 1.2 Question: Let A, B, C, D be sets. Explain why A × (B × C) × D is not the same set as (A × B) × (C × D). Solution: The elements of the set A × (B × C) × D contains 3-tuples with three components, (a, (b, c), d), whereas the elements of the set (A × B) × (C × D) contains pairs with two components, ((a, b), (c, d)). Exercise: 3 Section 1.2 Question: Let A = {1, 2, 3, 4} and B = {a, b}. Write out as a list a) A × B; b) A × A; c) B × B × A. Solution: a) {(1, a), (2, a), (3, a), (4, a), (1, b), (2, b), (3, b), (4, b)} b) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

13

c) {(a, a, 1), (a, a, 2), (a, a, 3), (a, a, 4), (a, b, 1), (a, b, 2), (a, b, 3), (a, b, 4), (b, a, 1), (b, a, 2), (b, a, 3), (b, a, 4), (b, b, 1), (b, b, 2), (b, b, 3), (b, b, 4)} Exercise: 4 Section 1.2 Question: Write in list form {1, 3} × {2, 4} × {3, 5}. Solution: {(1, 2, 3), (1, 2, 5), (1, 4, 3), (1, 4, 5), (3, 2, 3), (3, 2, 5), (3, 4, 3), (3, 4, 5)} Exercise: 5 Section 1.2 Question: Write in list form {1} × {1, 2} × {1, 2, 3}. Solution: The elements in {1} × {1, 2} × {1, 2, 3} are triples. The Cartesian product set is {(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), (1, 2, 2), (1, 2, 3)}.

Exercise: 6 Section 1.2 Question: Justify the statement that A × ∅ = ∅ for all sets A. Solution: By definition of the Cartesian product, A × ∅ = {(a, b)| a ∈ A and b ∈ ∅}. Since ∅ has no elements, A × ∅ has no elements and therefore is equal to ∅. Exercise: 7 Section 1.2 Question: (*) This exercise offers a proof that if A and B are countable sets, then A × B is countable. a) Find a bijection between N∗ and N∗ × N∗ . [Hint: Count through the pairs (x, y) ∈ N∗ × N∗ by successively going through the lines of the form x + y = n for n = 2, 3, 4, . . ..] b) Use the bijection in the previous part to prove that if A and B are countable sets, then A × B is countable. Solution: a) Note that N∗ × N∗ represents all the pairs of natural numbers which can be represented on a graph in the first quadrant. Using the hint, we establish a series of lines following the pattern x + y = n + 1 where n = 1, 2, 3, .... Observe the intersections of the first couple of lines: n = 1 ⇒ (1, 1) n = 2 ⇒ (1, 2), (2, 1) n = 3 ⇒ (1, 3), (2, 2), (3, 1)

As the pattern continues, we quickly observe that as n increases, every element in N∗ × N∗ is mapped to once and only once. This establishes a bijection between N∗ × N∗ and N∗ . b) Suppose A and B are countable sets. Then both A and B have a bijection with N∗ . It follows that there exists a bijection between A × B and N∗ × N∗ . Since we have already established a bijection between N∗ × N∗ and N∗ , there exists a bijection between A × B and N∗ . Therefore, A × B is countable. Exercise: 8 Section 1.2 Question: The operation ∗ on vectors of Rn defined by ~u ∗ ~v = proj~u ~v , i.e., projection of ~v onto ~u. Solution: Associative: Find any counter-example. Let ~a, ~b, and ~c be vectors in R2 such that ~a =< 1, 1 >, ~b =< 2, 1 >, a. Therefore, ∗ is not associative. and ~c =< 3, 1 >. Then (~a ∗ ~b) ∗ ~c = 34 ~a and ~a ∗ (~b ∗ ~c) = 21 10 ~ Commutative: Find any counter-example. Using the same vectors above, observe that ~a ∗~b = 23 ~a and ~b∗~a = 35~b. Since ~a ∗ ~b 6= ~b ∗ ~a, ∗ is not commutative. Identity: Note that in order for ~a ∗~e = ~a, proj~a ~e = ~a which is only possible when ~e = ~a since proj~a ~a = ~a. Thus it is impossible to have a unique identity. Inverse: Since ∗ does not have an identity, it cannot be closed under inverses.

14

CHAPTER 1. SET THEORY

Idempotent: As shown earlier, proj~a ~a = ~a therefore, ∗ is idempotent.

Exercise: 9 Section 1.2 Question: For the operation ? on the open interval [0, 1) described by a ? b = a + b − ba + bc where bxc is the greatest integer less than or equal to x, determine whether it is associative, commutative, has an identity, has inverses or is idempotent. Solution: Associative: Let a, b, c ∈ [0, 1). Then (a ? b) ? c = (a + b − ba + bc) ? c = a + b − ba + bc + c − ba + b − ba + bc + cc since bx + nc = bxc + n for all x ∈ R and all n ∈ Z = a + b + c − ba + bc + ba + bc − ba + b + cc = a + b + c − ba + b + cc. Furthermore a ? (b ? c) = a ? (b + c − bb + cc) = a + b + c − bb + cc − ba + b + c − bb + ccc = a + b + c − bb + cc + bb + cc − ba + b + cc = a + b + c − ba + b + cc. We conclude that (a ? b) ? c = a ? (b ? c). Hence ? is associative. Commutative: For all a, b ∈ [0, 1) we do have a ? b = a + b − ba + bc = b + a − bb + ac = b ? a. So ? is commutative. Identity: The element 0 ∈ [0, 1) serves as the identity. Has Inverses: Let a ∈ [0, 1). If a = 0, then a is its own inverse. If a > 0, then the element b = 1 − a, which is also in [0, 1), satisfies a ? b = a + (1 − a) − ba + 1 − ac = 1 − b1c = 0. Hence, every element in [0, 1) has an inverse. Idempotent: Consider the element 0.5. We have 0.5 ? 0.5 = 0, so 0.5 gives one counter example to idempotence.

Exercise: 10 Section 1.2 Question: The operation 4 on nonnegative integers N defined by n4m = |m − n|. Solution: Associative: Let a = 1, b = 2, and c = 3. Observe a 4 (b 4 c) = ||c−b|−a| = 0 while (a 4 b) 4 c = |c−|b−a|| = 2. Therefore 4 is not associative. Commutative: a 4 b = |b − a| = |a − b| = b 4 a. Therefore 4 is commutative. Identity: a 4 e = |e − a| = |a − e| = e 4 a = a. Hence e = 0. Inverse: Every element is it’s own inverse. a 4 a = |a − a| = 0. Idempotent: a 4 a = 0 hence 4 is not idempotent.

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

15

Exercise: 11 Section 1.2 Question: The operation ~ on points in the plane R2 where A ~ B is the midpoint of A and B. Solution: Associative: Let A = (a1 , a2 ), B = (b1 , b2 ), and C = (c1 , c2 ). A ~ (B ~ C) = 2a1 +b41 +c1 and (A ~ B) ~ C = a2 +b2 +2c2 . Thus, if A 6= C then A ~ (B ~ C) 6= (A ~ B) ~ C. Hence, ~ is not associative. 4 Commutative: The midpoint between A and B is the same as the midpoint between B and A, so ~ is commutative. Identity: Note that A ~ E = A if A = E. Since E is dependent on A, there does not exist an identity for ~. Inverse: Because there is no identity for ~, there cannot be any inverses. 1 a2 +a2 , 2 ) = (a1 , a2 ) = A. Therefore ~ is idempotent. Idempotent: A ~ A = ( a1 +a 2

Exercise: 12 Section 1.2 Question: The operation × + on C defined by a+ ×b = a + b + ab. Solution: Associative: Consider a, b, and c ∈ C. a+ ×(b+ ×c) = a + b + c + bc + ab + ac + abc = a + b + ab + c + ac + bc + abc = (a+ ×b)+ ×c Therefore × + is associative. Commutative: a+ ×b = a + b + ab = b + a + ba = b+ ×a. Therefore × + is commutative. Identity: Consider any q ∈ C. q+ ×0 = 0+ ×q = q so 0 is the identity. Inverse: Let b be the inverse of any a ∈ C. Then a+ ×b = 0 which implies a + b(1 + a) = 0. We can quickly −a deduce b = 1+a . From this observation we find that there exists an inverse for all values in C except −1. Thus, × + is not closed under inverses. Idempotent: Let a = 1. Then a+ ×a = 1 + 1 + 1 = 3 which is not equal to a. Hence, × + is not idempotent.

Exercise: 13 Section 1.2 Question: The operation 4 on P(S), where S is any set. Solution: Associative: S

S

B

A

B

C

A

C

Note in the above diagrams, the darker shading represents the operation within the parenthesis. We observe through the Venn diagrams that a 4 (b 4 c) = (a 4 b) 4 c. Thus, 4 is associative. Commutative:

16

CHAPTER 1. SET THEORY S

A

B

Observing the Venn diagram, A 4 B = B 4 A therefore 4 is commutative. Identity: A 4 ∅ = ∅ 4 A = A hence, ∅ is the identity of A. Inverse: Each set is it’s own inverse as A 4 A = ∅. Idempotent: A 4 A = ∅ implying 4 is not idempotent. Exercise: 14 Section 1.2 Question: The cross product on R3 Solution: Associative: Find any counter-example. Let ~a =< 1, 2, 3 >, ~b =< 1, 1, 1 >, and ~c =< 1, 2, 1 >. Then (~a × ~b) × ~c =< 4, 0, −4 > and ~a × (~b × ~c) =< 2, −4, 2 >. Since (~a × ~b) × ~c 6= ~a × (~b × ~c), the cross product is not associative over R3 . Commutative: For any two vectors in R3 , ~a × ~b = −~b × ~a. Hence the cross product is not commutative. Identity: The cross product is perpendicular to the plane created by the two vectors being multiplied by definition, therefore, there cannot exist a vector that satisfies ~a × ~e = ~a. Thus the cross product does not have an identity. Inverse: Because the cross product does not have an identity, it cannot be closed under inverses. Idempotent: Note, ~b × ~b =< 0, 0, 0 > which does not equal ~b in every case therefore the cross product is not idempotent. Exercise: 15 Section 1.2 Question: For the power operator a∧ b = ab on the set N∗ of positive integers determine whether it is associative, commutative, has an identity, has inverses or is idempotent. Solution: Associative: The following give a counter example 2∧ (3∧ 4) = 2∧ 81 = 281

while

(2∧ 3)∧ 4 = (23 )∧ 4 = 212

Hence ∧ is not associative. Commutative : Since 2∧ 3 = 8 and 3∧ 2 = 9, the operation ∧ is not commutative. Identity : Assume that ∧ has an identity e. Then ae = a for all a ∈ N∗ . Hence e = 1. However, by definition, we must also have ea = 1a = a for all a ∈ N∗ , which leads to a contradiction. Hence ∧ does not have an idenity element. Inverses: The operation cannot have an inverse since it does not have an identity. Idempotent: Since 2∧ 2 = 4 6= 2, then ∧ is not idempotent. Exercise: 16 Section 1.2 Question: The composition operator ◦ on the set F(A, A) of functions from a set A to A (where A is any set). Solution:

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

17

Associative: Let f , g, and h be functions in F(A, A). Then (h ◦ g) ◦ f = h(g(f (a))) = h ◦ (g ◦ f ) where a ∈ A. Therefore, ◦ is associative. Commutative: Consider the function j where j(a) = a1 or in other words, every input gives the same output a1 . Now let k be a function such that k(a1 ) 6= a1 . Then j ◦ k(a) = a1 , but k ◦ j(a) 6= a1 . Therefore, ◦ is not commutative. Identity: Consider the function e where e(a) = a. Let q be any function in F(A, A). Then q ◦ e = q and e ◦ q = q. Therefore, e is the identity. Inverse: Let q be any function in F(A, A). Then q ◦ q −1 = q(q −1 (a)) = a and q −1 ◦ q = q −1 (q(a)) = a where a ∈ A. However, not every function from A to A is bijective, for example, the function j as described above. Therefore, ◦ is not closed under inverses. Idempotent: Let l be a function such that l(a1 ) = a2 and l(a2 ) = a3 . Then l(l(a1 )) = a3 which is not equal to l(a1 ). Thus, ◦ is not idempotent. Exercise: 17 Section 1.2 Question: Prove that for all A, B, C ∈ P(S), A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Solution: Let x be in A ∩ (B ∪ C). Then, x ∈ A and x ∈ (B ∪ C) by the definition of the intersection of sets. So, x ∈ A and x ∈ B, or x ∈ A and x ∈ C. By the definitions of the union and intersection of sets, x ∈ (A ∩ B) ∪ (A ∩ C). Therefore, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). Exercise: 18 Section 1.2 Question: Let S be a set with a binary operation ∗. Assume that (a ∗ b) ∗ a = b for all a, b ∈ S. Prove that a ∗ (b ∗ a) = b for all a, b ∈ S. Solution: Assume (a ∗ b) ∗ a = b for all a, b ∈ S. (a ∗ b) ∗ a = b (a ∗ b) ∗ a ∗ a = (b ∗ a) a ∗ (a ∗ b) ∗ a ∗ a = a ∗ (b ∗ a) (a ∗ b) ∗ a = a ∗ (b ∗ a) b = a ∗ (b ∗ a) Therefore, a ∗ (b ∗ a) = b for all a, b ∈ S. Exercise: 19 Section 1.2 Question: Consider the operations a∧ b = ab and a × b on N∗ . Prove that ∧ is right-distributive over × but not left-distributive over ×. Solution: Let a, b, c ∈ N∗ . For right-distributivity, we observe that (a × b)∧ c = (ab)c = ac bc = (a∧ c) × (b∧ c), so ∧ is right-distributive over ×. In contrast, as a counter example to left-distributivity 2∧ (3 × 4) = 212

while

2∧ 3 × 2∧ 4 = 23 · 24 = 27 .

Exercise: 20 Section 1.2 Question: Let S be a finite set with |S| = n. How many binary operations exist on S? Solution: Let S be a finite set with |S| = n. By proposition 1.2.4 we know that the size of S × S is |S| · |S|. By definiton of a function, we know that for every tuple in S × S there exists an element in S that is mapped

18

CHAPTER 1. SET THEORY

to. Therefore, every element in S × S has |S| possible associations. Since there are |S| · |S| elements, each with |S| possibilities, then there are |S||S|·|S| distinct functions possible from S × S to S. By substitution, there are 2 nn possible binary operators on S. Exercise: 21 Section 1.2 Question: Let S = {1, 2}. How many binary operations on S are associative? Solution: We observe that S × S = {(1, 1), (1, 2), (2, 1), (2, 2)}. Since each element maps to either 1 or 2, there are 24 binary operations on S. We can represent the binary operations as 4-tuples, (s1 , s2 , s3 , s4 ), where (1, 1) = s1 , (1, 2) = s2 , (2, 1) = s3 , and (2, 2) = s4 . Therefore, by checking each of the sixteen binary operations, the following upheld associativity: (1, 1, 1, 1), (1, 2, 2, 1), (2, 2, 2, 2), (2, 1, 1, 2), (1, 1, 1, 2), (1, 1, 2, 2), (1, 2, 1, 2), (1, 2, 2, 2) Hence, there are eight binary operations on S that are associative. Exercise: 22 Section 1.2 Question: Let A and B be finite sets. Find the number of distinct relations from A to B. Solution: Let A and B be finite sets. By definition, a distinct relation is a distinct subset of A × B. Recall, according to Proposition 1.2.4 that |A × B| = |A| · |B|. By Proposition 1.1.11, we know that |P(A × B)| = 2|A|·|B| . Therefore there are 2|A|·|B| distinct relations from A to B. Exercise: 23 Section 1.2 2 Question: Let A be a finite set with n elements. Prove that the number of reflexive relations on A is 2n −n and that the number of symmetric relations on A is 2n(n+1)/2 Solution: Let A be a finite set with n elements. Define S to be the set of all possible reflexive elements in A2 . Note, the smallest reflexive relation, which we will call B, is of size n such that B = {(a, a) | a ∈ A, ∀a}. Thus, we define S = {s | s ∈ A2 − B and B ∪ {s} is a reflexive relation}. We observe that S = A2 − B so |S| = n2 − n. With the union of P(S) and B we find all possible reflexive relations. By proposition 1.1.11 we 2 2 know that the size of P(S) is 2n −n . Therefore, there are 2n −n possible reflexive relations on A. Similar to the reflexive relations, if we can find the set of possible symmetric elements, then its power set will result in the number of symmetric relations on A. n=2

n=3

A×A

n=4

A×A

A×A

Each graph above represents the total number of symmetric elements for a set A of size n. By definition, every pair in the form of (a, a) is a distinct symmetric element. These elements, contained in B as defined before, are represented by the squares shown in the graphs above. Let T be the set of symmetric pairs in A2 − B such that T = {{(a, b), (b, a)} | ∀ a, b ∈ A2 }.These pairs are represented by the circles on the graph. To illustrate, when n = 2, the only symmetric pair is {(a1 , a2 ), (a2 , a1 )} represented by the single dot on the graph. Thus when n = 2, |T | = 1 and |B| = 2. Let X = T ∪ B. X contains all the symmetric elements of A2 . Hence we see when |A| = 2, that there are 3 symmetric elements. We observe that every time n increases, n symmetric Pn elements are added to X. It follows that |X| = i=1 i. By substitution, |X| = n(n+1) . Thus, the cardinality of 2 its power set results in 2

n(n+1) 2

symmetric relations on A.

Exercise: 24 Section 1.2 Question: For any set S, consider the relation G on P(S) defined by A G B to mean that A ∩ B 6= ∅. Solution:

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

19

Reflexive: Choose any subset A. Then, A G A implies A ∩ A = A 6= ∅ which is true for all cases except the empty set. However, ∅ ∈ P(S) so G is not reflexive. Symmetric: Let A and B be subsets of S. Then, A G B ⇒ A ∩ B 6= ∅ ⇒ B ∩ A 6= ∅ ⇒ B G A Therefore, G is symmetric. Antisymmetric: Consider the set S = {1, 2, 3, 4}. Let A = {1, 2} and B = {3, 4}. We observe that both A G B and B G A exist, however it does not imply that they are equal. Hence, G is not antisymmetric. Transitive: Consider the set S = {1, 2, 3, 4}. Let A = {1, 2}, B = {3}, and C = {2, 4}. Notice that while A G B and B G C satisfy the conditions of transitivity, A G C does not exist because A ∩ C = {2}. Hence, G is not transitive.

Exercise: 25 Section 1.2 Question: The relation % on S the set of people defined by p1 % p2 if p1 is taller than or the same height as p2 . Solution: Reflexive: Any person is the same height as himself, therefore % is reflexive. Symmetric: Consider the case where an individual p1 is taller than p2 . Then p1 % p2 but p2 6% p1 . Hence, % is not symmetric. Antisymmetric: Assume (p1 % p2 ) and (p2 % p1 ), but p1 6= p2 . This implies that either p1 is taller than p2 or vice versa. Without loss of generality, suppose p1 is taller than p2 . Then p1 % p2 but p2 6% p1 which is a contradiction. Therefore, if p1 % p2 and p2 % p1 , then p1 = p2 . Thus % is antisymmetric. Transitive: Assume that p1 6% p3 , but p1 % p2 and p2 % p3 where p2 6= p3 . This implies that p2 is taller than p3 and p1 is taller than or equal to p2 . Hence, p1 would have to be taller than p3 which is a contradiction. Therefore, if p1 % p2 and p2 % p3 , then p1 % p3 implying % is transitive. Exercise: 26 Section 1.2 Question: The relation R on Z defined by nRm if n ≥ m2 . Solution: Reflexive: Find any counter-example. Let n = 2. This creates a contradiction since 2 6≥ 4, therefore R is not reflexive. Symmetric: Find any counter-example. Let n = 9 and m = 2. It is clear to see that nRm but the reverse is not true. This creates a contradiction so R is not symmetric. Antisymmetric: Let n and m be integers such that nRm and mRn. This implies that n ≥ m2 and m ≥ n2 . By squaring both sides and substitution, we find that n ≥ n4 which is only true if n = 1. Applying the same method the other direction, we find that m = 1 as well. Therefore, the only case where nRm and mRn is when n = m = 1. Thus, R is antisymmetric. Transitive: Let a, b and c be integers. Assume a 6≥ c2 , but aRb and bRc. Using some substitution, this implies that a ≥ b2 ≥ c2 . This creates a contradiction, therefore R is transitive.

Exercise: 27 Section 1.2 Question: Consider the relation on S = R2 defined by (x1 , y1 ) (x2 , y2 ) to mean x21 + y12 ≤ x22 + y22 . Prove which of the properties reflexivity, symmetry, antisymmetry, and transitivity hold. Solution: Reflexivity: Let (x, y) ∈ R2 . Then x2 + y 2 ≤ x2 + y 2 , so (x, y) (x, y). Hence is reflexive.

20

CHAPTER 1. SET THEORY

Symmetry: Consider the elements (1, 1) and (1, 2). Then 12 + 12 ≤ 12 + 22 so (1, 1) (1, 2). However 12 + 22 6≤ 12 + 12 so (1, 2) 6 (1, 1). Hence is not symmetric. Antisymmetry: Note that (1, 2) (2, 1) and (2, 1) (1, 2) but since (1, 2) 6= (2, 1), then is not antisymmetric. Transitivity: Consider three points (x1 , y1 ), (x2 , y2 ), and (x3 , y3 ). Suppose that (x1 , y1 ) (x2 , y2 ) and (x2 , y2 ) (x3 , y3 ). Then x21 + y12 ≤ x22 + y22 and x22 + y22 ≤ x23 + y32 . Hence x21 + y12 ≤ x22 + y22 so (x1 , y1 ) (x3 , y3 ). Hence is transitive. Exercise: 28 Section 1.2 Question: The relation $ on R defined by a $ b to mean ab = 0. Solution: Reflexive: Consider the case where a ∈ R and a 6= 0. Then a $ a ⇒ a2 = 0. This creates a contradiction as this is not true in every case, thus $ is not reflexive. Symmetric: Suppose a $ b. By definition, ab = 0 implying that either a = 0 or b = 0. In either case, ba = 0. Hence, b $ a therefore $ is symmetric. Antisymmetric: Let a 6= 0 and b = 0. Then a $ b and b $ a, but a 6= b. Hence, $ is not antisymmetric. Transitive: Let a 6= 0, b = 0, and c 6= 0. Then a $ b and b $ c, but a 6$ c since a 6= 0 and c 6= 0. Thus, $ is not transitive.

Exercise: 29 Section 1.2 Question: For any set S, consider the relation

on P(S) defined by A

B to mean that A ∪ B = S.

Solution: Reflexive: Let A ∈ P(S) and A 6= S. Then, A ∪ A = A which is not equal to S, therefore Symmetric: Suppose A, B are in P(S) and A is symmetric.

is not reflexive.

B. Then A ∪ B = B ∪ A = S which implies B

Antisymmetric: Let A = ∅ and B = S. Then, A antisymmetric.

B and B

A, but A 6= B. Therefore,

A. Hence, cannot be

Transitive: Let A = ∅, B = S, and C 6= S such that A C. However, notice that A B and B A C implies A ∪ C = S which creates a contradiction since C 6= S. Thus, is not transitive.

Exercise: 30 Section 1.2 Question: The relation on the set of pairs of points in the plane S = R2 ×R2 defined by (P1 , Q1 ) if the segment [P1 , P2 ] intersects [Q1 , Q2 ].

C, but

(P2 , Q2 )

Solution: Reflexive: Let P1 , P2 be in S. It is not hard to see that (P1 , P1 ) itself. Hence, is reflexive.

(P2 , P2 ) because any line intersects with

Symmetric: Suppose (P1 , Q1 ) (P2 , Q2 ). Then [P1 , P2 ] intersects [Q1 , Q2 ] implying that [Q1 , Q2 ] intersects [P1 , P2 ] and (Q1 , P1 ) (Q2 , P2 ). Thus, is symmetric. Antisymmetric: Let (P1 , Q1 ) (P2 , Q2 ) such that [P1 , P2 ] is perpendicular to [Q1 , Q2 ]. It is not hard to show that (P1 , Q1 ) (P2 , Q2 ) and (Q1 , P1 ) (Q2 , P2 ), but it is impossible for [P1 , P2 ] to be equal to [Q1 , Q2 ]. Therefore, is not antisymmetric. Transitive: Let [P1 , P2 ], [Q1 , Q2 ], and [R1 , R2 ] be lines in S and let [P1 , P2 ] be parallel to [R1 , R2 ] and perpendicular to [Q1 , Q2 ]. Then (P1 , Q1 ) (P2 , Q2 ) and (Q1 , R1 ) (Q2 , R2 ), but it is impossible for (P1 , R1 ) (P2 , R2 ) since they are parallel to each other. Therefore, is not transitive.

1.2. THE CARTESIAN PRODUCT; OPERATIONS; RELATIONS

21

Exercise: 31 Section 1.2 Question: Let S be a set and let R be a relation on S. Prove that if a relation is reflexive, symmetric, and anti-symmetric, then it is the = relation on S. Solution: Let S be a set and let R be a relation on S such that R is reflexive, symmetric, and antisymmetric. By definition, R contains all elements of the form (s, s) ∀s ∈ S. Since R is symmetric, if s R t then t R s where s, t ∈ S. Moreover, R is antisymmetric implying that if there is a symmetric pair (s, t), (t, s) ∈ R, then s = t. Hence, it is impossible for R to contain an element of the form (s, t) where s 6= t. Therefore, R = {(s, s), ∀s ∈ S} which is the = relation.

Exercise: 32 Section 1.2 Question: Let P be the set of people who are living now. Let R be the relation on P defined by aRb if a and b are in the same nuclear family, i.e. if a is a self, child, parent, sibling, or spouse of b. a) Decide whether R is reflexive, symmetric, antisymmetric, or transitive. b) List all the family relations included in R(2) = R ◦ R. c) Give four commonly used family terms for relations in R(3) = R ◦ R ◦ R though not in R(2) . Solution: a) For any person p, p R p since ’self’ is included in the nuclear family, therefore R is reflexive. Assume there exists two people p1 and p2 such that p1 R p2 but p2 is not related to p1 . It is not hard to see that this creates a contradiction as any two people who are in the same nuclear family satisfy the relation R. Therefore R is symmetric. In regards to antisymmetry, consider any p1 R p2 where p1 is the child of p2 . We know that p1 R p2 and p2 R p1 , however p1 and p2 are not the same person. Hence, R is not antisymmetric. Let p1 be the child of p2 and p2 be the sibling of p3 . We can easily observe that p1 R p2 and p2 R p3 , yet p1 is not in the same nuclear family as p3 . Therefore R is not transitive. b) R2 = {self, child, parent, sibling, spouse, grandchild, child-in-law, grandparent, uncle / aunt, niece / nephew, sibling-in-law, parent-in-law} c) {cousin, great-grandchild, great-grandparent, great-uncle / aunt}

Exercise: 33 Section 1.2 Question: We can define the graph of a relation R from R to itself as the subset of R2 {(x, y) ∈ R2 | x R y}. a) Sketch the graph of the relation ≤. b) Sketch the graph of the relation l defined by x l y if |x − y| = 1. c) Provide defining geometric characteristics of a subset of R2 for a relations on R that are i) reflexive; ii) symmetric; iii) transitive; iv) antisymmetric. Solution: y

y=x

x

a)

22

CHAPTER 1. SET THEORY y

y =x+1 y =x−1

x

b) c) i) A reflexive relation must contain the line y = x. ii) A symmetric relation must have a reflective mapping over the line y = x. iii) An antisymmetric relation cannot have a reflective mapping over the line y = x and must include points from the line y = x. iv) In a transitive relation, for any points (x, y) and (y, z) there exists a right triangle that contains the points (x, y), (y, z), and (x, z). Exercise: 34 Section 1.2 Question: Let S = {a, b, c, d, e} and consider the relation R on S described by R = {(a, a), (a, c), (a, d), (b, c), (b, e), (c, b), (c, d), (e, a), (e, b)}. Determine as a list in S × S, the composite relation R ◦ R. Solution: R ◦ R = {(a, a), (a, b), (a, c), (a, d), (b, a), (b, b), (b, d), (c, c), (c, e), (e, a), (e, c), (e, d), (e, e)} Exercise: 35 Section 1.2 Question: Let R be a relation on a set A. Denote by R(n) the n-composite relation of R with itself: n times

R

(n) def

z }| { = R ◦ R ◦ · · · ◦ R.

Prove that the relation R is transitive if and only if R(n) ⊆ R for all n = 1, 2, 3, . . .. Solution: (=⇒): Suppose R is transitive. Assume that R(n) contains an element (x, y) such that (x, y) 6∈ R. Then there exists some z ∈ A such that (x, z) and (z, y). This creates a contradiction by the definition of a transitive relation, thus ∀(x, y) ∈ R(n) , (x, y) ∈ R. Therefore, R(n) ⊆ R. (⇐=): Suppose R(n) ⊆ R. Assume that there exists a R b and b R c such that a is not related to c. By definition of relation composition, a R c exists in R(n) . If a R c exists in R(n) , then it must also exist in R by definition of a subset causing a contradiction. Hence, if a R b and b R c, then a R c. Therefore, R is transitive. Exercise: 36 Section 1.2 Question: Let R be a relation that is reflexive and transitive. Prove that Rn = R for all n ∈ N∗ . Solution: Let R be a relation that is reflexive and transitive. From exercise 1.2.35 we know that Rn ⊆ R. Consider the case when n = 2 and let (r1 , r2 ) be any pair in R. Then r1 R r1 and r1 R r2 , since R is reflexive, therefore r1 R r2 must be in R2 . We can easily observe that the same is true for R3 in that r1 R r1 and r1 R r1 and r1 R r2 , so (r1 , r2 ) must be in R3 . We can continue this process for any n to show that any (r1 , r2 ) in R must also be in Rn . This implies that R ⊆ Rn . Therefore, Rn = R for all n ∈ N∗ .

1.3 – Equivalence Relations Exercise: 1 Section 1.3 Question: Let S = Z × Z and let R be the relation on S defined by (a, b)R(c, d) means that a + d = b + c. Show that R is an equivalence relation. Concisely describe the equivalence classes of R. Solution: Let S = Z × Z and let R be the relation on S defined by (a, b)R(c, d) means that a + d = b + c. For any (a, b) in S it is not hard to see that (a, b) R (a, b) means a + b = a + b therefore R is reflexive. Suppose

1.3. EQUIVALENCE RELATIONS

23

(a, b) R (c, d). Then, a + d = b + c which is equivalent to c + b = d + a. This implies (c, d) R (a, b) therefore R is symmetric. Suppose (a, b) R (c, d) and (c, d) R (e, f ). Then a + d = b + c and c + f = d + e. Using subtraction and substitution we find that a − b = e − f . With some arranging we observe a + f = b + e implying (a, b) R (e, f ). Hence, R is transitive. Therefore, since R is reflexive, symmetric, and transitive it is an equivalence relation. Each distinct equivalence class describes the solutions to a function f : Z −→ Z where f (x) = x + c for c ∈ Z. Exercise: 2 Section 1.3 Question: Let C be the set of people in your abstract algebra class. Describe a “natural” relation satisfying each of the combination of properties listed below. 1. Reflexive and symmetric, but not transitive. 2. Reflexive and transitive, but not symmetric. 3. Symmetric and transitive, but not reflexive. 4. An equivalence relation. Solution: We describe relations for each of the following situations. a) Reflexive and symmetric, but not transitive. R, where a R b if a and b live within one kilometer (or one mile) of each other. b) Reflexive and transitive, but not symmetric. R, where a R b if a earns a final grade that is less than or equal to the final grade that b earns. c) Symmetric and transitive, but not reflexive. The simplest relation that satisfies these conditions is the empty relation. Symmetry and transitivity are satisfied trivially. (The hypothesis is always false so the conditional statement is always true.) Note that if a R b, where a 6= b, then by symmetry b R a and then by transitivity a R a. So for this combinations of properties to hold, there must be an element that is not in relation to any other element. d) An equivalence relation. R, where a R b if a and b entered college the same semester. Exercise: 3 Section 1.3 Question: Let P be the set of living people. For all a, b ∈ P , define the relation a R b if a and b have met. Solution: It is not hard to see that R is both reflexive and symmetric by definition. However, let a, b, and c be people such that a R b and b R c. It does not follow that person a has met person c in every case implying R is not transitive. Therefore, R is not an equivalence relation. Exercise: 4 Section 1.3 Question: Let P be the set of living people. For all a, b ∈ P , define the relation a R b if a and b live in a common town. Solution: Obviously, a R a exists for all people within the town. It quickly follows that if a R b then b R a and therefore is symmetric as well. Suppose a R b and b R c. Then a and b live in a common town, and b and c live in the same town. Hence, a and c live in the same town meaning a R c and R is transitive. Therefore, R is an equivalence relation. Exercise: 5 Section 1.3 Question: Let C be the set of circles in R2 and let R be the relation of concentric on C. Prove or disprove whether the described relation is an equivalence relation. If the relation is not an equivalence relation, determine which properties it lacks. Solution: Two circles are concentric if and only if they have the same center. Reflexivity: If C is a circle in C, then it has the same center as itself. Symmetry: If C1 , C2 ∈ C, then if C1 has the same center as C2 , then C2 has the same center as C1 . So symmetry holds. Transitivity: Let C1 , C2 , C3 ∈ C. Suppose that C1 is concentric with C2 and that C2 is concentric with C3 . Then C1 and C2 have the same center and C2 and C3 have the same center. Hence C1 and C3 have the same center, so C1 is concentric with C3 . Hence concentric is transitive.

24

CHAPTER 1. SET THEORY

So the concentric relation is an equivalence relation. Exercise: 6 Section 1.3 Question: Let S = Z × Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m1 m2 = n1 n2 . Solution: Consider any (m1 , m2 ) ∈ S. We observe (m1 , m2 ) R (m1 , m2 ) implies m1 m2 = m1 m2 which is always true. Hence, R is reflexive. Suppose (m1 , m2 ) R (n1 , n2 ). We know m1 m2 = n1 n2 so n1 n2 = m1 m2 and therefore (n1 , n2 ) R (m1 , m2 ). Hence, R is symmetric. Suppose (m1 , m2 ) R (n1 , n2 ) and (n1 , n2 ) R (p1 , p2 ). Then, m1 m2 = n1 n2 and n1 n2 = p1 p2 . By substitution, we find m1 m2 = p1 p2 implying (m1 , m2 ) R (p1 , p2 ). Hence, R is transitive and therefore is an equivalence relation. Exercise: 7 Section 1.3 Question: Let S = Z × Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m1 n1 = m2 n2 . Solution: Consider the relation (1, 2) R (1, 2). Since 1 × 1 6= 2 × 2, then R is not reflexive. Therefore, R is not an equivalence relation. Exercise: 8 Section 1.3 Question: Let S = Z × Z and define the relation R on S by (m1 , m2 ) R (n1 , n2 ) if m1 n2 = m2 n1 . Solution: Suppose m1 , m2 ∈ Z and (m1 , m2 ) R (m1 , m2 ). Then m1 m2 = m2 m1 for any (m1 , m2 ) ∈ S, thus R is reflexive. Suppose (m1 , m2 ) R (n1 , n2 ) where n1 , n2 ∈ Z. Then, m1 n2 = m2 n1 which can be easily rearranged to show n1 m2 = n2 m1 . This implies (n1 , n2 ) R (m1 , m2 ), and therefore R is symmetric. Suppose r1 , r2 , s1 , s2 , t1 , t2 ∈ Z such that (r1 , r2 ) R (s1 , s2 ) and (s1 , s2 ) R (t1 , t2 ). Then, r1 s2 = r2 s1 and s1 t2 = s2 t1 . Using some substitution we can arrive at the desired outcome as seen below, r1 s2 = s1 r2 r1 s2 t1 = s1 r2 t1 r1 s1 t2 = s1 r2 t1 r1 t2 = t1 r2 Therefore, since R is reflexive, symmetric, and transitive, R is an equivalence relation. Exercise: 9 Section 1.3 Question: Let P3 be the set of polynomials with real coefficients and of degree 3 or less. Define the relation R on P3 by p(x) R q(x) to mean that q(x) − p(x) has 5 as a root. Solution: Notice, 5 will always be a root of f (x) − f (x) so R is reflexive. Consider the case when p(x) = x − 12 and q(x) = 2x − 7. We observe that p(x) R q(x) since q(x) − p(x) = x − 5. However, because p(x) − q(x) = −x − 5, 5 is not a root of p(x) − q(x) therefore q(x) is not related to p(x). Since R is not symmetric it cannot be an equivalence relation. Exercise: 10 Section 1.3 Question: Consider the set C 0 (R) of continuous functions over R. Define the relation R on C 0 (R) by f R g if there exist some a, b ∈ R such that g(x) = f (x + a) + b

for all x ∈ R.

Prove or disprove whether the described relation is an equivalence relation. If the relation is not an equivalence relation, determine which properties it lacks. Solution: We check the defining properties for an equivalence relation. Reflexivity: Let f be any function in C 0 (R). Setting a = b = 0 we have f (x) = f (x + a) + b for all x. Hence R is reflexive. Symmetry: Suppose that f R g. Then there exists some a, b ∈ R such that g(x) = f (x + a) + b. Then g(x) − b = f (x + a) for all x ∈ R. Setting y = x + a, we have g(y − a) − b = f (y). This holds for all y ∈ R. Hence g R f . Thus, R is symmetric.

1.3. EQUIVALENCE RELATIONS

25

Transitivity: Let f, g, h ∈ C 0 (R) and suppose that f R g and g R h. There there exist a1 , a2 , b1 , b2 ∈ R such that g(x) = f (x + a1 ) + b1 and h(x) = g(x + a2 ) + b2 for all x ∈ R. Then h(x) = f (x + a + 2 + a1 ) + b1 + b2 for all x ∈ R. Hence f R h. Thus R is transitive. Exercise: 11 Section 1.3 Question: Let Pfin (R) be the set of finite subsets of R and define the relation ∼ on Pfin (R) by A ∼ B if the sum of elements in A is equal to the sum of elements in B. Prove that ∼ is an equivalence relation. Solution: It is not hard to see A ∼ A for all A ∈ R, so ∼ is reflexive. Similarly, if A ∼ B, then the sum of elements in A is equal to the sum of elements in B. Thus, we observe that this implies that B ∼ A. Therefore, ∼ is symmetric. Moreover, given any A, B, C ∈ R where A ∼ B and B ∼ C, it quickly follows that the sum of elements in A is equal to the sum of elements in C by substitution. Hence, ∼ is transitive and therefore is an equivalence relation. Exercise: 12 Section 1.3 Question: Let `∞ (R) be the set of sequences of real numbers. Define the relation R on `∞ (R) by (an ) R (bn ) if lim (bn − an ) = 0.

n→∞

Solution: It is not hard to see that R is reflexive since lim (an − an ) = lim (0) = 0

n→∞

n→∞

Also, notice lim (bn − an ) = − lim (an − bn ) = 0

n→∞

n→∞

therefore R is symmetric. For transitivity, suppose that (an )R(bn ) and (bn )R(cn ). Then lim (bn − cn ) = 0

n→∞

and

lim (an − bn ) = 0.

n→∞

Thus lim (an − cn ) = lim ((an − bn ) + (bn − cn )) = lim (an − bn ) + lim (bn − cn ) = 0 + 0 = 0,

n→∞

n→∞

n→∞

n→∞

where the second equality holds by virtue of the addition law of limits. Note that we can apply the addition rule here because we know that each of the sequences involved in the sums converge (to 0). Exercise: 13 Section 1.3 Question: Let `∞ (R) be the set of sequences of real numbers. Define the relation R on `∞ (R) by (an ) R (bn ) if the sequence (an + bn )∞ n=1 converges. Solution: Using the properties of limits we can show limn→∞ (an + an ) = 2 limn→∞ (an ). Find any counterexample for reflexivity. Let an = 2n . Then, 2 limn→∞ (an ) = ∞ implying R cannot be reflexive. Therefore, R cannot be an equivalence relation. It should be noted, however, that R is both symmetric and transitive. Exercise: 14 Section 1.3 Question: Let S be the set of lines in R2 and let R be the relation of perpendicular. Solution: We observe that any given line cannot be perpendicular to itself, therefore R cannot be reflexive. Thus R is not an equivalence relation. Exercise: 15 Section 1.3 Question: Let W be the words in the English language (i.e., have an entry in the Oxford English Dictionary). Define the relation R on W by w1 R w2 is w1 comes before w2 in alphabetical order. Solution: This relation is not symmetric. Find any counter-example. Consider the two words, w1 =”ball” and w2 =”stick”. We observe that w1 R w2 , but the reverse is not true. Hence, R cannot be an equivalence relation.

26

CHAPTER 1. SET THEORY

Exercise: 16 Section 1.3 Question: Let C 0 ([0, 1]) be the set of continuous real-valued functions on [0, 1]. Define the relation ∼ on C 0 ([0, 1]) by Z 1 Z 1 g(x) dx. f (x) dx = f ∼ g ⇐⇒ 0

0

Show that ∼ is an equivalence relation and describe (with a precise rule) a complete set of distinct representatives of ∼. Solution: First show that ∼ is an equivalence relation. R1 R1 Reflexivity: For all f ∈ C 0 ([0, 1]), we have 0 f (x) dx = 0 f (x) dx so f ∼ f . Symmetry: Suppose that f, g ∈ C 0 ([0, 1]) with f ∼ g. Then Z 1

Z 1 f (x) dx =

0

g(x) dx. 0

Equality is reversible so g ∼ f . Hence ∼ is symmetric. Transitivity: Suppose that f, g, h ∈ C 0 ([0, 1]) with f ∼ g and g ∼ h. Then Z 1

Z 1 f (x) dx =

0

Hence

R1 0

f (x) dx =

R1 0

Z 1 g(x) dx

0

and

Z 1 g(x) dx =

0

h(x) dx. 0

h(x) dx. Thus f ∼ h. Hence ∼ is an equivalence relation.

R1 Every function f ∈ C 0 ([0, 1]) is in relation with the constant function g(x) = c, where c = 0 f (x) dx. However, constant functions that are not equal are not in relation with each other. Hence the constant functions are a complete set of distinct representatives.

Exercise: 17 Section 1.3 Question: Let C ∞ (R) be the set of all real-value functions on R such that all its derivatives exist and are continuous. Define the relation R on C ∞ (R) by f R g if f (n) (0) = g (n) (0) for all positive, even integers n. a) Prove that R is an equivalence relation. b) Describe concisely all the elements in the equivalence class [sin x]. Solution: a) It is not hard to see that for any f ∈ C ∞ (R), f R f implies f (n) (0) = f (n) (0). We conclude that R is reflexive. Similarly, for any f, g ∈ C ∞ (R) if f R g, then we observe g R f quickly follows since f (n) (0) = g (n) (0). Thus R is symmetric as well. Finally, for any given f, g, h ∈ C ∞ (R) such that f R g and g R h, we find f R h because f (n) (0) = g (n) (0) = h(n) (0) for all positive even integers n. Consequently, R is reflexive, symmetric, and transitive implying that R is an equivalence relation. b) If f (x) = sin x then for all positive and even integers n, we have f (n) (0) = 0, so this is the defining characteristic of functions in [sin x]. (If a function in [sin x] is equal to its power series in a neighborhood of 0, then [sin x] consists of functions that are a constant plus an odd function.) Exercise: 18 Section 1.3 Question: Let S = {1, 2, 3, 4} and the relation ∼ on P(S), defined by A ∼ B if and only if the sum of elements in A is equal to the sum of elements in B, is an equivalence relation. List the equivalence classes of ∼. Solution: [{1}], [{2}], [{3}] = {{1, 2}}, [{4}] = {{1, 3}}, [{1, 4}] = {{2, 3}}, [{2, 4}] = {{1, 2, 3}}, [{3, 4}] = {{1, 2, 4}}, [{1, 3, 4}], [{2, 3, 4}], [{1, 2, 3, 4}], [∅] Exercise: 19 Section 1.3 Question: Let T be the set of (non-degenerate) triangles in the plane. a) Prove that the relation ∼ of similarity on triangles in T is an equivalence relation. b) Concisely describe a complete set of distinct representatives of ∼. Solution:

1.3. EQUIVALENCE RELATIONS

27

a) Suppose you have any t ∈ T . Then, t ∼ t implies that triangle t has equal corresponding angles with itself. This is always true, thus ∼ is reflexive. Now suppose you have to triangles s, t ∈ mathcalT such that s ∼ t. Then the corresponding angles of triangle s are equal to the corresponding angles of t. It is not hard to see that the corresponding angles of triangle t are equal to the corresponding angles of triangle s implying that ∼ is symmetric. For transivity, suppose you have any r, s, t ∈ mathcalT such that r ∼ s and s ∼ t. Then the corresponding angles of triangle r are equal to the corresponding angles of triangle s which are also equal to the corresponding angles of t. Therefore the corresponding angles of triangle r are equal to those in triangle t. Hence, ∼ is transitive. Therefore, ∼ is an equivalence relation. b) The set of distinct representatives of ∼ will be equal to the unique combinations of the angles x, y, z of any triangle such that x, y, z ∈ R>0 and x + y + z = 180. For example, [30, 60, 90] is the distinct representative for all triangles with 30, 60, and 90 as their respective angles. Exercise: 20 Section 1.3 Question: Prove that the relation defined in Example 1.3.10 is an equivalence relation. Solution: Choose any (a, b) in S. We quickly observe that ab = ba which holds true for all a, b under multiplication. Thus, R is reflexive. Now suppose that R is symmetric and choose any (a, b), (c, d) ∈ S such that (a, b) ∼ (c, d). Then, ad = bc so we can show that cb = da implying that (c, d) ∼ (a, b). Hence, R is symmetric. Suppose R is also transitive. Let (a, b), (c, d), (e, f ) be in S such that (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then ad = bc and df = ec. By multiplying both sides by ab we observe, abdf = abec af bd = beac af bd = bebd af = be Consequently, (a, b) ∼ (e, f ). Thus ∼ is transitive and therefore an equivalence relation. Exercise: 21 Section 1.3 Question: Let S = {1, 2, 3, 4, 5, 6}. For the partitions of S given below, write out the equivalence relation as a subset of S × S. a) {{1, 2}, {3, 4}, {5, 6}} b) {{1}, {2}, {3, 4, 5, 6}} c) {{1, 2}, {3}, {4, 5}, {6}} Solution: a) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 2), (2, 1), (3, 4), (4, 3), (5, 6), (6, 5)} b) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 4), (3, 5), (3, 6), (4, 3), (4, 5), (4, 6), (5, 3), (5, 4), (5, 6), (6, 3), (6, 4), (6, 5)} c) {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 2), (2, 1), (4, 5), (5, 4)} Exercise: 22 Section 1.3 Question: Let S = {a, b, c, d, e}. For the partitions of S given below, write out the equivalence relation as a subset of S × S. a) {{a, d, e}, {b, c}} b) {{a}, {b}, {c}, {d}, {e}} c) {{a, b, d, e}, {c}} Solution: a) {(a, a), (b, b), (c, c), (d, d), (e, e), (a, d), (a, e), (d, a), (d, e), (e, a), (e, d), (b, c), (c, b)} b) {(a, a), (b, b), (c, c), (d, d), (e, e)}

28

CHAPTER 1. SET THEORY c) {(a, a), (b, b), (c, c), (d, d), (e, e), (a, b), (a, d), (a, e), (b, a), (b, d), (b, e), (d, a), (d, b), (d, e), (e, a), (e, b), (e, d)}

Exercise: 23 Section 1.3 Question: Let C 1 ([a, b]) be the set of continuously differentiable functions on the interval [a, b]. Define the relation ∼ on C 1 ([a, b]) as f ∼ g if and only if f 0 (x) = g 0 (x) for all x ∈ (a, b). Prove that ∼ is an equivalence relation on C 1 ([a, b]). Describe the elements in the equivalence class for a given f ∈ C 1 ([a, b]). Solution: First, we observe for any f ∈ C 1 ([a, b]) that f 0 (x) = f 0 (x) for all x ∈ (a, b). Hence, ∼ is reflexive. Similarly, given any f, g ∈ C 1 ([a, b]) we observe that if f ∼ g then f 0 (x) = g 0 (x) for all x ∈ (a, b). Thus we know g 0 (x) = f 0 (x) which establishes that ∼ is also symmetric. For transivity, consider any f, g, h ∈ C 1 ([a, b]) where f ∼ g and g ∼ h. Then f 0 (x) = g 0 (x) = h0 (x) for all x ∈ (a, b). Thus f 0 (x) = h0 (x) implying that ∼ is transitive. Therefore, ∼ is an equivalence relation. The elements of the equivalence class for a given f are the vertical translations of f . In other words, [f (x)] = {f (x) + k | k is some constant in R}. Exercise: 24 Section 1.3 Question: Let Mn×n (R) be the set of n × n matrices with real coefficients. For two matrices A, B ∈ Mn×n (R), we say that B is similar to A if there exists and invertible n × n matrix S such that B = SAS −1 . 1. Prove that similarity ∼ is an equivalence relation on Mn×n (R). 2. Prove that the function f : Mn×n (R)/ ∼ → R defined by f ([A]) = det A is a well-defined function on the quotient set Mn×n (R)/ ∼. 3. Determine with proof or counter-example whether the function g : Mn×n (R)/ ∼ → R defined by g([A]) = Tr A, the trace of A, is a well-defined function. Solution: Let Mn×n (R) be the set of n×n matrices with real coefficients and define the relation ∼ as similarity between matrices. a) We prove that ∼ is an equivalence relation. Reflexivity: For all matrices A ∈ Mn×n (R), since IAI −1 = A then A ∼ A. Symmetry: Let A, B ∈ Mn×n (R) such that A ∼ B. Then there exists an invertible matrix S such that B = SAS −1 . Hence A = S −1 BS. Thus B ∼ A so ∼ is symmetric. Transitivity: Let A, B, C ∈ Mn×n (R) such that A ∼ B and B ∼ C. Thus there exist invertible matrices A and T such that B = SAS −1 and C = T BT −1 . Then C = T (SAS −1 )T −1 = (T S)A(T S)−1 . Hence A ∼ C so ∼ is transitive. b) Suppose that B ∼ A. Then B = SAS −1 for some invertible matrix S. Then det B = (det S)(det A)(det S)−1 = det A. Therefore, the function det is well-defined on the set of ∼-equivalence classes on Mn×n (R). c) There is a property in the algebra of matrices that Tr(AB) = Tr(BA), even if the matrices do not commute. Suppose that B ∼ A with B = SAS −1 for some invertible matrix S. Then Tr(B) = Tr(SAS −1 ) = Tr(ASS −1 ) = Tr(A). Hence the function g is a well-defined function. Exercise: 25 Section 1.3 Question: Define the relation ∼ on R by a ∼ b if and only if b − a ∈ Q. a) Prove that all real x ∈ R, there exists y ∈ [x]∼ that is arbitrarily close to x. (In other words, for all ε > 0, there exists y with y ∼ x and |x − y| < ε.

1.3. EQUIVALENCE RELATIONS

29

b) (*) Prove that ∼ has an uncountable number of equivalence classes. Solution: a) Choose any q ∈ Q such that q < . Then, if |x − y| = q, y ∼ x and |x − y| < . Rearranging our equation we find, y =x+q y =x−q Notice that the domain of these functions spans R. Therefore for any x ∈ R there exists y ∈ R such that y ∼ x and |x − y| < . A graphical representation is shown below. y

y =x+ y =x+q y =x−q y =x− x

b) Since ∼ is reflexive, every value in R has its own equivalence class. Since R is uncountable, it follows that there are an uncountable amount of equivalence classes. Exercise: 26 Section 1.3 Question: Let R1 and R2 be equivalence relations on a set S. Determine (with a proof or counterexample) which of the following relations are also equivalence classes on S. (a) R1 ∩ R2 ; (b) R1 ∪ R2 ; (c) R1 4R2 . [Note that R1 ∪ R2 , and similarly for the others, is a relation as a subset of S × S.] Solution: Let R1 and R2 be equivalence relations on a set S. a) R1 ∩R2 is an equivalence relation. For all a ∈ S, the pair (a, a) is in both R1 and R2 . Hence (a, a) ∈ R1 ∩R2 . Thus R1 ∩R2 is reflexive. Suppose that (a, b) ∈ R1 ∩R2 . Then since R1 is an equivalence relation, (b, a) ∈ R1 and similarly for R2 . Hence (b, a) ∈ R1 ∩ R2 . Thus R1 ∩ R2 is symmetric. Finally, suppose that (a, b) and (b, c) are pairs in R1 ∩ R2 . Then since R1 is an equivalence relation (a, c) ∈ R1 and the same holds for R2 , so (a, c) ∈ R2 . Thus (a, c) ∈ R1 ∩ R2 and hence R1 ∩ R2 is transitive. b) R1 ∪ R2 is not an equivalence relation. Since R1 is reflexive, for all a ∈ S, the pair (a, a) ∈ R1 and thus (a, a ∈ R1 ∪ R2 . Hence, R1 ∪ R2 is reflexive. Suppose that (a, b) ∈ R1 ∪ R2 . Thus (a, b) ∈ R1 or (a, b) ∈ R2 . Both R1 and R2 are equivalence relations. If (a, b) ∈ Ri , then (b, a) ∈ Ri so (b, a) ∈ R1 ∪ R2 . Thus, R1 ∪ R2 is symmetric. Finally, suppose that (a, b) and (b, c) are in R1 ∪ R2 . In the case where (a, b) ∈ R1 and (b, c) ∈ R2 , it does not appear that transitivity would need to hold if (b, c) ∈ / R1 and (a, b) ∈ / R2 . For example, let S = P({1, 2, 3, 4, 5}) and let R1 be the equivalence relation on S of same cardinality and let R2 be the equivalence relation of elements summing to the same value. Then {1, 4} R1 ∪ R2 {5} because the elements in the set both add to the same value and {5}R1 ∪ R2 {2} because the sets are the same cardinality. However, {1, 4} is not in relation to {2} under R1 ∪ R2 . Hence, the union of two transitive relations is not necessarily transitive. c) R1 4R2 cannot be an equivalence relation since it is not reflexive. For all a ∈ S, the pair (a, a) is in both R1 and in R2 . Hence, the pair (a, a) does not occur in R1 4R2 . Exercise: 27 Section 1.3 Question: Which of the following collections of subsets of the integers for partitions? If it is not a partition, explain which properties fail. 1. {pZ | p is prime}, where kZ means all the multiples of k. 2. {{3n, 3n + 1, 3n + 2} | n ∈ Z}. 3. {{k | n2 ≤ k ≤ (n + 1)2 } |n ∈ N}.

30

CHAPTER 1. SET THEORY 4. {{n, −n} | n ∈ N}.

Solution: Testing to satisfy the properties of a partition. a) The set of subsets {pZ | p is prime} is not a partition of Z since for example 6 ∈ 2Z ∩ 3Z, so the subsets are not disjoint. Also, the union of all these sets is not all of Z but Z − {−1, 1}. b) The set of subsets {{3n, 3n + 1, 3n + 2} | n ∈ Z} is a partition of Z. Consider the function f : Z → Z defined by f (m) = bm/3c. It is easy to see that {3n, 3n + 1, 3n + 2} = f −1 (n). Since f is a function, the union of all f −1 (n) gives Z. Furthermore, since an element in Z does not map via f to distinct n values, these pre-image sets are disjoint. c) This is not a partition because for example {1, 2, 3, 4} is one set in the partition as is {4, 5, 6, 7, 8, 9} and these distinct sets are not mutually disjoint. The collection of subsets is not a partition of Z for the additional reason that it does not cover all of Z. In fact, [ {n2 , n2 + 1, . . . , (n + 1)2 } = N. n∈N

d) This collection of subsets is a partition. Consider the equivalence relation on Z defined by a ∼ b if and only if |a| = |b|. This is indeed an equivalence relation and the equivalence classes are precisely subsets of Z of the form {−n, n}. Exercise: 28 Section 1.3 Question: Let S be a set. Prove that there is a bijection between the set of partitions of S and the set of equivalence classes on S. Solution: Let f be a function from the set of partitions of S to the set of equivalence classes on S such that f (A) = R. Let R be an equivalence relation on S and assume there does not exist A over S such that f (A) = R. However, by proposition 1.3.12 we know all the distinct equivalence classes of R are disjoint and their union is equal to S. This creates a contradiction because it satisfies the definition of a partition. Hence, for every equivalence relation R on S, there must exist A where f (A) = R. Therefore, f is surjective. Now assume there exists partitions A = {Ai }i∈I and B = {Bj }j∈J such that A = 6 B and f (A) = f (B) = R. This would imply every Ai and Bj represents a distinct equivalence class of R by proposition 1.3.14. However, since A 6= B there must exist some Ai and Bj where Ai ∩ Bj 6= Ai and Ai ∩ Bj 6= ∅. This creates a contradiction by proposition 1.3.12 since the distinct equivalence classes of any equivalence relation are disjoint. Therefore, f must also be injective which establishes a bijection between the set of partitions of S and the set of equivalence classes on S.

Exercise: 29 Section 1.3 Question: Call p(n) the number of equivalence relations (equivalently, by Exercise 1.3.28, partitions) on a set of cardinality n. (The numbers p(n) are called the Bell numbers after the Scottish-born mathematician E. T. Bell.) a) (*) Prove that p(0) = 1 and that for all n ≥ 1, p(n) satisfies the condition n−1 X

n−1 p(n) = p(n − j − 1) . j j=0 b) Use the previous part to calculate p(n) for n = 1, 2, 3, 4, 5, 6, 7. Solution: Call p(n) the number of partitions that exist on the set {1, 2, . . . , n}. This will be the same number of partitions on any set of size n. a) The value p(0) = 1 comes from the comment that the empty set satisfies all the conditions for an equivalence relation on the empty set itself. Suppose that we know the value of p(k) for 0 ≤ k ≤ n − 1. To determine p(n), we count up the number of possible partitions based on how many elements are in the equivalence class of n besides n. In other words, let j = |[n]−{n}|. The index j = 0 corresponds to the equivalence class [n] being the singleton set {n} and the index j = n − 1, corresponds to the situation where the equivalence class [n] = {1, 2, . . . , n}. Now for any given j, there are n−1 ways to choose the remaining elements in the j

1.3. EQUIVALENCE RELATIONS

31

equivalence class of [n]. Furthermore, for each of those choices, there are n − 1 − j elements remaining in {1, 2, . . . , n} from which to create the remaining equivalence classes that make up the partition. Thus, for each j, there are p(n − 1 − j) n−1 partitions. Summing over j = 0 to n − 1 gives the number of possible j partitions (equivalence classes) on {1, 2, . . . , n}. b) Using this recursive formula, we get n p(n)

1 1

2 2

3 5

4 15

5 52

6 203

7 877

Exercise: 30 Section 1.3 Question: Consider the relation ∼ on R defined by x ∼ y if y − x ∈ Z. a) Prove that ∼ is an equivalence relation. b) Prove that if a ∼ b and c ∼ d, then (a + c) ∼ (b + d). c) Decide with a proof or counter-example whether ac ∼ bd, whenever a ∼ b and c ∼ d. Solution: a) First, we observe that x − x = 0 for any x ∈ R. Since 0 ∈ Z, ∼ is reflexive. Suppose ∼ is symmetric. Then for any x, y ∈ R if x ∼ y, then y − x = z where z ∈ Z. Rearranging we find that x − y = −z which establishes that ∼ is symmetric. For transivity, suppose that a, b, c ∈ R such that a ∼ b and b ∼ c. Then b − a = x and c − b = y where x, y ∈ Z. Setting c = b + y and solving we find, b−a=x b+y−a=x+y c−a=x+y Since x + y is always an integer, we conclude ∼ is transitive. Therefore, ∼ is an equivalence relation. b) Notice, b − a = x and d − c = y where x, y ∈ Z. b−a=x b−a+y =x+y b−a+d−c=x+y (b + d) − (a + c) = x + y Therefore, since x + y ∈ Z, (a + c) ∼ (b + d). c) Find any counter-example. Let a = π, b = π, c = 1, and d = 2. Then, a ∼ b and c ∼ d, yet ac ∼ bd = π. Since π 6∈ Z, ac 6∼ bd. Exercise: 31 Section1.3 Question: Let S be a set and let A = {Ai }i∈I be a partition of S. Another partition B = {Bj }j∈J is called a refinement of A if ∀j ∈ J , ∃i ∈ I, Bj ⊆ Ai . Let A and B be two partitions of a set S and let ∼A (resp. ∼B ) as the equivalence relation corresponding to A (resp. B). Prove that B is a refinement of A if and only if s1 ∼B s2 =⇒ s1 ∼A s2 . Solution: Suppose that B is a refinement of the partition A. Suppose that s1 ∼B s2 . This is equivalent to the statement that s1 and s2 are both in Bj for some index j ∈ J . Since B is a refinement of A, then Bj ⊆ Ai for some i ∈ I, and therefore, s1 and s2 are both in this Ai . Hence s1 ∼A s2 . Conversely, suppose that ∀s1 , s2 ∈ S, s1 ∼B s2 −→ s1 ∼A s2 . Consider a given subset Bj in the partition B and let s ∈ Bj . Since A is a partition of S, then s ∈ Ai for some index i ∈ I. For all s0 ∈ Bj , we have s ∼B s0 and so by our hypothesis, s ∼A s0 . Consequently, s0 ∈ Ai and this for all s0 ∈ Bj . Thus Bj ⊆ Ai . This establishes that B is a refinement of A. Exercise: 32 Section 1.3 Question: Let S be a set and let A = {Ai }i∈I and B = {Bj }j∈J be two partitions of S. Prove that the collection of sets {Ai ∩ Bj | i ∈ I and j ∈ J } − {∅}

32

CHAPTER 1. SET THEORY

is a partition of S. Solution: Call C the collection of subsets of S {Ai ∩ Bj | i ∈ I and j ∈ J } − {∅} Let s be any element in S. Since A and B are partitions of S, there exists a unique i0 ∈ I and a unique j0 ∈ J such that s ∈ Ai0 and s ∈ Bj0 . Then s ∈ Ai0 ∩ Bj0 and hence the union of subsets in C is all of S. Now consider Ai1 ∩ Bj1 and Ai2 ∩ Bj2 sets in C and suppose that (Ai1 ∩ Bj1 ) ∩ (Ai2 ∩ Bj2 ) 6= ∅. Then by associativity (Ai1 ∩ Ai2 ) ∩ (Bj1 ∩ Bj2 ) 6= ∅. However, Ai1 ∩ Ai2 6= ∅ if and only if i1 = i2 and similarly Bj1 ∩ Bj2 6= ∅ if and only if j1 = j2 . Hence, we have i1 = i2 and j1 = j2 . Thus sets in C are either equal or disjoint. Consequently, since C also covers S, C is a partition of S. Exercise: 33 Section 1.3 Question: Let S be a set and let R be any relation on S. Design an algorithm that determines the smallest equivalence relation on S that contains the relation R. Solution: There are a variety of ways to accomplish this depending on the context and, if working computationally with a finite set S, how we store the equivalence relation. From a purely theoretical standpoint, we can deduce that a smallest equivalence relation on S that contains S exists by referring to Exercise 1.3.26 and generalizing. It is not hard to prove that the intersection of any collection of equivalence relations on S is again an equivalence relation on S. Hence, we can define the smallest equivalence relation on S containing R as \ Re Re : R⊆Re

where the intersection is over all equivalence relations Re that contain R. We can call this relation the equivalence closure of R. From an algorithmic perspective, one of the challenges of this exercise is to realize that if we begin with the relation R as a subset of S × S, and adjoin pairs to force symmetry and then adjoin more pairs to force transitivity, we may need to go back and adjoin new pairs that are required for symmetry again and vice versa. It is not clear when this process will terminate. Suppose that |S| = n and label the elements of S by S = {s1 , s2 , . . . , sn }. For an n × n matrix M of nonnegative integers, we define M̃ as m̃ij = min(1, mij ). Hence, if mij ≥ 1, then m̃ij = 1 but if mij = 0, then m̃ij = 0. Here is an algorithm for determining the equivalence closure of R. • Let A = (aij ) be the n × n matrix defined by ( 1 aij = 0

if (si , sj ) ∈ R otherwise.

• Replace A with A + I. • Replace A with Ã. [These first two steps ensure reflexivity.] • While A 6= A2 ^ + (A2 )> , replace A with A2 ^ + (A2 )> . • Return A. The operation A2 will have a nonzero entry position (i, j), if there is some k such that aik and bkj are both nonzero. This process will add a nonzero entry corresponding to a pair that is required by transitivity. The operation A → A + A> adds a matrix to its transpose, which adds a nonzero entry to complete for symmetry reasons. Now if A is a matrix of 0s and 1s, then A2 ^ + (A2 )> will again be a matrix of 0s and 1s with new entries turned to 1 for transitivity or symmetry reasons.

1.4. PARTIAL ORDERS

33

The algorithm stops because if A = A2 ^ + (A2 )> then no transitivity requirement or symmetry requirement will add a new pair to the relation (nonzero entry to A). Hence, A will then correspond to an equivalence relation. Furthermore, the algorithm will terminate because before the while loop, A contains at most n2 − n entries that are 0 and at each stage of the while loop that changes A will change at least one 0 entry to a 1. Hence, the while loop can repeat at most n2 − n times.

1.4 – Partial Orders Exercise: 1 Section 1.4 Question: Let S = {a, b, c, d, e} (where we consider all the labels unique elements). In the following relations on S determine with explanation whether or not the relation is a partial order. If it fails antisymmetry then remove a least number of pairs and if it fails transitivity then add some pairs to make the relation a partial order. 1. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c)}. 2. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (a, d)}. 3. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (d, a)}. 4. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (b, c), (c, d), (d, e), (a, e)}. Solution: Consider each partial order. 1. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c)}. This relation is reflexive. The only pair in the relation not of the form (x, x) is (a, c). The pair (c, a) is not in the relation so the relation is antisymmetric. For the same reason, the relation is also transitive. 2. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (a, d)}. For the same reasons as the previous part, this relation is a partial order. 3. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (a, c), (d, a)}. In this example, the relation is again reflexive and antisymmetric but it is not transitive because of the pairs (d, a) and (a, c). In the modified relation R0 = R ∪ {(d, c)}, we now have transitivity and R0 still has antisymmetry. 4. R = {(a, a), (b, b), (c, c), (d, d), (e, e), (b, c), (c, d), (d, e), (a, e)}. This relation satisfies reflexivity and antisymmetry but fails transitivity. In order to get transitivity, we need add the pairs (b, d), (c, e), and (b, e). Then R ∪ {(b, d), (b, e), (c, e)} is a partial order. Exercise: 2 Section 1.4 Question: In microeconomics (the study of consumer behavior), one considers consumer’s utility (preference) in regards to pairs of commodities. Let (q1 , q2 ) ∈ N2 be a pair of nonnegative integers representing quantities of two commodities. Explain why, given two specific commodities and a given consumer, the relation of preferable (or equal) is a partial order. Solution: When a consumer prefers the pair of commodities in the quantities (q10 , q20 ) over (q1 , q2 ) (or these are equal), we write (q1 , q2 ) 4 (q10 , q20 ). Note that this is a partial order on N2 , where the ith entry represents the quantity of the ith commodity. By including the equality of pairs into the relation, the relation is reflexive. It is in people’s psychology that if (q1 , q2 ) 4 (q10 , q20 ) and (q10 , q20 ) 4 (q100 , q200 ), so the concept of preference is transitive. We do not have (q1 , q2 ) 4 (q10 , q20 ) and (q10 , q20 ) 4 (q1 , q2 ) at the same time unless forced by the requirement that the pairs are equal. So preference, with the additional assumption of including equality, is a partial order. Note that in the concept of preference, given two pairs (q1 , q2 ) and (q10 , q20 ) it is perfectly possible to not prefer one over the other. Such pairs would be incomparable. Exercise: 3 Section 1.4 Question: Let S = R>0 × R>0 be the positive first quadrant in the Cartesian plane. Consider the relation R on S defined by (x1 , y1 ) R (x2 , y2 ) =⇒ x1 y1 ≥ x2 y2 Prove or disprove that R is a partial order. Solution: We check the three axioms for a partial order.

34

CHAPTER 1. SET THEORY

Reflexivity: For all (x, y) ∈ S, we do have xy ≤ xy so the relation is reflexive. Antisymmetry: Consider the points (2, 1) and (1, 2). Then 1 × 2 ≤ 2 × 1 and 2 × 1 ≤ 1 × 2 so (1, 2) R (2, 1) and (2, 1) R (1, 2) even through (1, 2) 6= (2, 1) So the relation is not antisymmetric. Transitivity: We have already shown that the relation is not a partial order since it fails antisymmetry. The relation is transitive. Not a partial order. Exercise: 4 Section 1.4 √ √ Question: Prove that for any real x > 2, the inequality 2 < 12 x + x2 < x holds. √ √ Solution: Since 2 < x, then x1 < √12 , which implies x2 < 2 < x. So 1 2 Also, from x 6=

√

2, we have (x −

√

x+

2 x

<

1 (x + x) = x. 2

2)2 > 0, which is equivalent to

√ √ 1 x − 2 2x + 2 > 0 ⇐⇒ x2 + 2 > 2 2x ⇐⇒ 2 2

√ 2 x+ > 2. x

The result follows. Exercise: 5 Section 1.4 Question: Let (S, 4) be a partial order in which every element has an immediate successor. Prove that it is not necessarily true that for any two elements a 4 b that any chain between a and b has finite length. n o n o Solution: Consider the subset of (Q, ≤) with the set S = 1 − n1 n ∈ N∗ ∪ 1 + n1 n ∈ N∗ . In the poset 1 (S, ≤), every element has an immediate successor: the immediate successor of 1− n1 is 1− n+1 , and the immediate 1 1 successor of an element of the form 1 + n is 1 + n−1 . By construction S is a chain since it is a total order. Furthermore, 0, 2 ∈ S, so it is an infinite chain between 0 and 2, such that every element has an immediate successor.

Exercise: 6 Section 1.4 Question: Prove the three claims about properties of 4 in Example 1.4.8. Prove that Q≥0 is countable. Conclude that Q is countable. Solution: Consider the sets described in Example 1.4.8. Since |An | ≤ n, each set is finite. Also, the collection of sets An is countable. Furthermore, we claim that the sets An partition Q>0 . Indeed for all fractions xy , expressed in reduced form, we have xy ∈ Ax+y−1 . We can set up a bijection f : N∗ → Q>0 as follows. Define `1 = 1 and then for all positive integers k, set `k = |A1 | + · · · + |Ak |. Set f (1) = 1 and then for all integers m with `k < m ≤ `k+1 , define f (m) as the (m − `k )th element (ordered by ≤) in Ak . This function is injective since the Ak are mutually disjoint and is surjective since {Ak } cover Q>0 . We can now define a bijection between F : N → Q by F (0) = 0 and m+1 m F (m) = (−1) f . 2

Exercise: 7 Section 1.4 Question: Let S be a set. Show that the relation of refinement is a partial order on the set of partitions of S. Solution: Let S be a set. Recall that a refinement of a partition A = {Ai }i∈I of S is another partition B = {Bj }j∈J such that for all j ∈ J, there exists an i ∈ I such that Bj ⊆ Ai . We will write B 4 A if B is a refinement of A. 1. For a partition A = {Ai }i∈I of S, for each i ∈ I we have Ai ⊆ Ai so A 4 A.

1.4. PARTIAL ORDERS

35

2. Suppose that B 4 A and A 4 B. Then for all j ∈ J, there exists an i ∈ I such that Bj ⊆ Ai . However, for this index i, there exists j 0 ∈ J such that Ai ⊆ Bj 0 . Hence Bj ⊆ Ai ⊆ Bj 0 . However, since the sets in B are mutually disjoint, we conclude that j = j 0 . Hence Bj = Ai . Since j was arbitrary, we conclude that B = A. Thus the relation of refinement is antisymmetric. 3. Finally, let B be a refinement of A and let C = {Ck }k∈K be a partition of S such that C 4 B. Then for all k ∈ K, there exists j ∈ J such that Ck ⊆ Bj . Since B is a refinement of A, then there exists i ∈ I such that Bj ⊆ Ai . Hence, since ⊆ is transitive, Ck ⊆ Ai . Thus C 4 A. Hence, the relation of refinement is transitive. These three results show that refinement is a partial order on the set of partition of S. Exercise: 8 Section 1.4 Question: Draw the Hasse diagram of the partial order ⊆ on P({1, 2, 3, 4}). Solution: The Hasse diagram of (P({1, 2, 3, 4}), ⊆) is {1, 2, 3, 4}

{1, 2}

{1, 2, 3}

{1, 2, 4}

{1, 3, 4}

{2, 3, 4}

{1, 3}

{1, 4}

{2, 3}

{2, 4}

{1}

{2}

{3}

{4}

{3, 4}

∅

Exercise: 9 Section 1.4 Question: Draw the Hasse diagram for the poset ({1, 2, 3, 4, 5, 6}, ≤). Solution: The Hasse diagram of ({1, 2, 3, 4, 5, 6}, ≤) is 6 5 4 3 2 1

Exercise: 10 Section 1.4 Question: Let A = {a, b, c, d, e, f, g}. Draw the Hasse diagram for the partial order 4 given as a subset of A × A as 4 = {(a, a), (b, b), (c, c), (d, d), (e, e), (f, f ), (g, g), (a, c), (b, c), (d, g), (a, e), (b, e), (c, e), (d, h), (g, h)}

36

CHAPTER 1. SET THEORY

Solution: The Hasse diagram of (A, 4) is e

h g

c a

b

f

d

Exercise: 11 Section 1.4 Question: A person’s blood type is usually listed as one of the eight elements in the set B 0 = {o+, o−, a+, a−, b+, b−, ab+, ab−}. We define the donor relation → on B 0 as follows. The relation t1 → t2 holds if the letter portion of the blood type donates according to the way described in the examples for this section and if someone with a + designation can only give to someone else with +, while someone with − can give to anybody. 1. Draw the Hasse diagram for (B 0 , →). 2. Show that the (B 0 , →) poset does not have the lexicographic order on B × {+, −}. Solution: We study the partial order of blood type B 0 including the rhesus. a) The Hasse diagram for the poset (B 0 , →) is ab+

a+

ab−

b+

a−

o+

b−

o− b) Under the partial order on {+, −} is described completely by + → +, − → −, and − → +. So for example, in the lexicographic order on B 0 , we have (a, +) →lex (ab, −) because the pairs already differ on the first entry and a → ab in B. However, we see that in the actual donor relation on B 0 , a+ cannot donate to ab−. Hence the actual donor relation on B 0 is not the lexicographic partial order on B × {+, −}. Exercise: 12 Section 1.4 Question: Consider the set of triples of integers Z3 . Define the relation 4 on Z3 by ( a1 + a2 + a3 < b1 + b2 + b3 if a1 + a2 + a3 6= b1 + b2 + b3 ; (a1 , a2 , a3 ) 4 (b1 , b2 , b3 ) ⇐⇒ a1 + a2 + a3 4lex b1 + b2 + b3 if a1 + a2 + a3 = b1 + b2 + b3 , where 4lex is the lexicographic order on Z3 (with each copy of Z equipped with the partial order ≤). Prove that 4 is a partial order on Z3 . Prove also that 4 is a total order. Solution: We first prove that 4 is a partial order on Z3 . Reflexivity : Let (a1 , a2 , a3 ) ∈ Z3 . Then since a1 + a2 + a3 = a1 + a2 + a3 , we use the lexicographic order to compare the entries but all entries are equal and this satisfies the lexicographic order. Hence (a1 , a2 , a3 ) 4 (a1 , a2 , a3 ).

1.4. PARTIAL ORDERS

37

Antisymmetry : Suppose that (a1 , a2 , a3 ) 4 (b1 , b2 , b3 ) and (b1 , b2 , b3 ) 4 (a1 , a2 , a3 ). Assume that a1 + a2 + a3 6= b1 + b2 + b3 , then a1 + a2 + a3 < b1 + b2 + b3 and b1 + b2 + b3 < a1 + a2 + a3 , which is a contradiction. Hence we must have a1 + a2 + a3 = b1 + b2 + b3 . Thus, we compare the triples by lexicographic order. However, the lexicographic order is a partial order, which is antisymmetric so we can conclude that (a1 , a2 , a3 ) = (b1 , b2 , b3 ). Transitivity : Suppose that (a1 , a2 , a3 ) 4 (b1 , b2 , b3 ) and (b1 , b2 , b3 ) 4 (c1 , c2 , c3 ). We can break the situation into three cases. If a1 + a2 + a3 = b1 + b2 + b3 = c1 + c2 + c3 , then we compare the triples by lexicographic order, which is transitive so (a1 , a2 , a3 ) 4 (c1 , c2 , c3 ). If a1 + a2 + a3 = b1 + b2 + b3 6= c1 + c2 + c3 (resp. a1 + a2 + a3 6= b1 + b2 + b3 = c1 + c2 + c3 ) then a1 + a2 + a3 < c1 + c2 + c3 so (a1 , a2 , a3 4 (c1 , c2 , c3 ). Finally, if all thee of the sums are different, then we deduce that a1 + a2 + a3 < b1 + b2 + b3 < c1 + c2 + c3 so we can conclude that (a1 , a2 , a3 4 (c1 , c2 , c3 ). We conclude that 4 is transitive. Finally, to show that 4 is a total order, let (a1 , a2 , a3 ), (b1 , b2 , b3 ) ∈ Z3 be arbitrary, distinct elements. There are two cases. If a1 + a2 + a3 6= b1 + b2 + b3 , then either a1 + a2 + a3 < b1 + b2 + b3 , which implies that (a1 , a2 , a3 ) 4 (b1 , b2 , b3 ), or the reverse is true. If a1 + a2 + a3 = b1 + b2 + b3 , then we compare the triples by the lexicographic order. In this case, let i be the least index for which ai 6= bi . Then (a1 , a2 , a3 ) 4 (b1 , b2 , b3 ) if ai < bi and (b1 , b2 , b3 ) 4 (a1 , a2 , a3 ) if bi < ai . Hence, every pair of elements is comparable and the relation is a total order. Exercise: 13 Section 1.4 Question: Let (Ai , 4i ) be posets for i = 1, 2, . . . , n and define 4lex as the lexicographic order on A1 ×A2 ×· · · An . Prove that 4lex is a total order if and only if 4i is a total order on Ai for all i. Solution: First suppose that 4i is a total order on Ai for all i. Let (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) be elements in A1 × A2 × · · · An . If ai = bi for i = 1, . . . , n, then (a1 , a2 , . . . , an ) 4lex (b1 , b2 , . . . , bn ). Otherwise, let j be the least index for which ai = bi . Since all 4i are total orders, then 4j is a total order. Thus, either aj 4j bj or bj 4j aj . Hence, (a1 , a2 , . . . , an ) 4lex (b1 , b2 , . . . , bn ) or the reverse is true. Thus 4lex is a total order. Now suppose that 4lex is a total order. Let j be any index with 1 ≤ j ≤ n and let aj , bj ∈ Aj . Consider a pair of n-tuples (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) such that ai = bi for all i < j. Since (a1 , a2 , . . . , an ) and (b1 , b2 , . . . , bn ) are comparable in 4lex , then aj 4j bj or bj 4j aj . Since aj , bj were arbitrary, then 4j is a total order. Thus, we conclude that all the partial orders 4i are partial orders. Exercise: 14 Section 1.4 Question: Let 4 be the lexicographic order on R3 , where each R is equipped with the usual ≤. Prove or ~ if ~a 4 ~b and ~c 4 d, ~ then ~a + ~c 4 ~b + d. ~ disprove the following statement: For all vectors ~a, ~b, ~c, d, ~ ~ Solution: We first prove that if ~a 4 b, then ~a + ~c 4 b + ~c. With the lexicographic order, ~a 4 ~b if and only if the first index j ∈ {1, 2, 3} for which aj 6= bj has aj < bj . For any ~c ∈ R3 , for all indices ai = bi if and only ai + ci = bi + ci . Hence the first index for with ai + ci 6= bi + ci is j and then we have aj + cj < bj + cj . Thus ~a + ~c 4 ~b + ~c. Since the above result was arbitrary, ~a 4 ~b implies ~a +~c 4 ~b +~c and ~c 4 d~ implies ~c +~b 4 d~+~b. By transitivity of 4, we deduce that ~ ~a + ~c 4 ~b + ~c 4 ~b + d.

Exercise: 15 Section 1.4 Question: Answer the following questions pertaining to the poset described by the Hasse diagram below. 1. List all the minimal elements. 2. List all the maximal elements. 3. List all the maximal elements in the subposet with {a, b, c, d, e, f, g}. 4. Determine the length of the longest chain and find all chains of that length. 5. Find the least upper bound of {a, b}, if it exists. 6. Find the greatest lower bound of {b, c}, if it exists.

38

CHAPTER 1. SET THEORY j i

h g

f

e c

d a

b

7. List all the upper bounds of {f, d}. Solution: In the given Hasse diagram, we have the following named elements. a) The minimal elements are a and b. b) The maximal element is j. c) The maximal elements in {a, b, c, d, e, f, g} are e, f , and g. d) There are 3 chains of length 5. They are {a, d, e, i, j}, {a, d, g, h, j}, and {a, c, f, h, j}. There is no chain of greater length. e) The subposet {a, b} has only one upper bound, namely e, and this is the least upper bound. f) The subposet {b, c} does not have a lower bound. g) The set of upper bounds of {f, d} is {h, j}. Exercise: 16 Section 1.4 Question: Consider the partial order on R2 given in Example 1.4.6. Let A be the unit disk A = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}. a) Show that A has both a maximal and minimal element. Find all of them. b) Find all the upper bounds and all the lower bounds of A. Solution: The partial order on R2 is defined by (x1, y1) 4 (x2, y2) if and only if 2x1 − y1 < 2x2 − y2 or (x1, y1) = (x2, y2). a) The solution to this problem is directly related to the calculus problem of optimizing 2x − y for (x, y) ∈ A. The gradient of 2x − y is never 0 so there are no optimal values to 2x − y on the interior of A. We can parametrize the boundary of A by (cos t, sin t) with t ∈ [0, 2π]. We can optimize 2x − yon the boundary by optimizing 2 cos t − sin t. This is optimized with tan t = − 21 . With the result that sin t = − 12 cos t, we deduce that the optimum values occur at 2 1 2 1 √ , −√ √ √ and − , 5 5 5 5 with the former corresponding to a maximum and the latter being a minimum element with respect to 4. √ b) The upper bounds to A are any (x, y) with 2x − y > √55 = 5 as well as the point √25 , − √15 . The lower √ bounds of A are all the points (x, y) with 2x − y < − 5 as well as the point − √25 , √15 . Exercise: 17 Section 1.4 Question: Consider the lexicographic order on R2 coming from the standard (R, ≤). Let A be the closed disk of center (1, 2) and radius 5. 1. Show that A has both a maximal and minimal element. Find all of them. 2. Find all the upper bounds and all the lower bounds of A. 3. Show that A has both a least upper bound and a greatest lower bound.

1.4. PARTIAL ORDERS

39

Solution: The lexicographic order on R2 coming from the standard (R, ≤) is a total order. a) A maximal element of A will have a greatest x value and if there are any ties on the x value, we break them with the y value. The maximal element is the point (6, 2). No point in A has a greater x-value. Similarly, the point with the least x value is (−4, 2) and there is no other point with the same x component. Hence (−4, 2) is minimal. b) The set of upper bounds to A is {(x, y) ∈ R2 | x > 6} ∪ {(6, y) | y ≥ 2}. The set of lower bounds of A is {(x, y) ∈ R2 | x < −4} ∪ {(−4, y) | y ≤ 2}. c) The single maximal element is the unique least upper bound to A and the single minimal element is the unique greatest lower bound. Exercise: 18 Section 1.4 Question: Prove that in a finite lattice, there exists exactly one maximal element and one minimal element. Solution: Let (S, 4) be a finite lattice. We first prove that a maximal element exists. Since S is finite every chain in S is finite. Consequently, there exists a chain C of maximal length. We can consider the set of positive integers defined by {|{y ∈ C | x 4 y}| | x ∈ C}. By the well-ordering of the integers, this set must have a least element and this least element must be 1. The element M that gives this least element satisfies |{y ∈ C | M 4 y}|. Hence, the only element y such that M 4 y is M itself, which means that M is maximal. Let M1 and M2 be two maximal elements. Let M be a least upper bound to M1 and M2 . In particular, M1 4 M and M2 4 M . Since M1 and M2 are maximal, then M1 = M = M2 . Hence, the maximal element is unique. A similar reasoning holds with minimal elements and greatest lower bounds. Hence, S contains a unique minimal element. Exercise: 19 Section 1.4 Question: Let (B, →) be the poset of blood types equipped with the donor relation. (See Example 1.4.4.) a) Consider the poset ({1, 2, 3}, ≤). Show that the function f : B −→ {1, 2, 3} defined by f (o) = 1, f (a) = 2, f (b) = 2 and f (ab) = 3 is a monotonic function. b) Show that there exists no isomorphism between (B, →) and ({1, 2, 3, 4}, ≤). Solution: Let (B, →) be the poset of blood types. a) It is obvious that for all x ∈ B, we have f (x) ≤ f (x). Consequently, we only need to check the monotonic property on unequal elements x and y that satisfy x → y. The five unequal pairs of donating relation give the following output by f . in {1, 2, 3} true of false in B o→a 1≤2 T o→b 1≤2 T o → ab 1≤3 T a → ab 2≤3 T b → ab 2≤3 T Thus, we have exhaustively checked that f is monotonic. b) Let h be a bijection from ({1, 2, 3, 4}, ≤) to (B, →). Let x and y be the elements in {1, 2, 3, 4} such that h(x) = a and h(y) = (b). The elements x and y are distinct since h is a bijection. Now since ≤ is a total order, x ≤ y or y ≤ x. However, neither a → b nor b → a. Thus, no bijection can be monotonic. Exercise: 20 Section 1.4 Question: Let (S, 4), (T, 40 ), and (U, 400 ) be three posets. Let f : S → T and g : T → U be monotonic functions. Prove that the composition g ◦ f : S → U is monotonic. Solution: Let x, y ∈ S with x 4 y. Since f is monotonic, then f (x) 40 f (y). Now since g is monotonic, then g(f (x)) 400 g(f (y)). Thus we have proved that x 4 y =⇒ (g ◦ f )(x) 400 (g ◦ f )(y). Hence, the composition g ◦ f is monotonic.

40

CHAPTER 1. SET THEORY

Exercise: 21 Section 1.4 Question: Prove that the poset (R, ≤) is not isomorphic to (R − {0}, ≤). Solution: Suppose that f : R → R − {0} is an isomorphism from the poset (R, ≤) to (R − {0}, ≤). Let a, b ∈ R. For all x such that a ≤ x ≤ b, we have f (a) ≤ f (x) ≤ f (b). Furthermore, for all y with f (a) ≤ y ≤ f (b), we also have a ≤ f −1 (y) ≤ b. Hence, the interval [a, b] is mapped to the interval [f (a), f (b)]. Let a and b be such that f (a) = −1 and f (b) = 1. Then f ([a, b]) is the interval [−1, 1]. However, [−1, 1] is not a subset of R − {0}. Exercise: Question: Prove that the poset of integers greater that a fixed number k (with partial order ≤) is isomorphic to the (N, ≤). Solution: It is obvious that f : N → {k + 1, k + 2, . . .} defined by f (n) = n + k + 1 is a bijection. Furthermore, for all m, n ∈ N, we have m ≤ n implies m + k + 1 ≤ n + k + 1. The result follows. Exercise: 23 Section 1.4 Question: Let (S, 41 ) and (T, 42 ) be two partially ordered sets and let f : S → T be a monotonic function. a) Prove that if A is a subset of S with an upper bound u, then f (u) is an upper bound of f (A). b) Prove with a counterexample that f (lub(A)) is not necessarily equal to lub(f (A)). c) Prove that if f is an isomorphism, then f (lub(A)) = lub(f (A)). Solution: Let (S, 41 ) and (T, 42 ) be two partially ordered sets and let f : S → T be a monotonic function. a) Since u is an upper bound of A, then x 41 u for all x ∈ A. Hence, f (x) 42 f (u) for all x ∈ A, or in other words, y 42 f (u) for all y ∈ f (A). Thus f (u) is an upper bound of f (A). b) Consider the function tan−1 : R → R and A = R. Then lub(f (A)) = π2 whereas lub(A) does not exist so f (lub(A)) does not exist either. c) Suppose now that f is surjective and let u = lub(A). We know that f (u) is an upper bound of f (A). Let y 0 be another upper bound of f (A). Since f is surjective, there exists u0 ∈ S such that f (u0 ) = y 0 . Since b 42 y 0 for all b ∈ f (A), then since f is an isomorphism, a 41 u0 for all a ∈ A. Thus u0 is also an upper bound of A. Hence u 41 u0 . Thus, f (u) 42 y 0 . We conclude that f (u) is a least upper bound of f (A). Exercise: 24 Section 1.4 Question: Prove or disprove that (Z, ≤) and (Q, ≤) are isomorphic as posets. Solution: Suppose that h is an isomorphism from a poset (S, 41 ) to (T, 42 ). Suppose that y is an immediate successor to an element x in S. Then x 41 y and x 6= y and if z is any element such that x 41 x 41 y, then z = x or z = y. Then h(x) 6= h(y) and h(x) 42 h(y). Furthermore, for all z 0 such that h(x) 42 z 0 42 h(y), then since h−1 is also monotonic we deduce that x 41 h−1 (z 0 ) 41 y. Thus h−1 (z 0 ) = x, which implies that z 0 = h(x) or h−1 (z 0 ) = y, which implies that z 0 = h(y). Hence, if y is an immediate successor to x, then h(y) is an immediate successors to h(x). In this particular example, every element in (Z, ≤) has an immediate successor but no element in the poset (Q, ≤) has an immediate successor. Hence, they cannot be isomorphic as posets. Exercise: 25 Section 1.4 Question: Determine whether the posets corresponding to the following Hasse diagrams are lattices. If they are not, explain why. (a)

(b) g

(c)

h

(d)

h

h

f

g

f

g

f

g

f d

e

d

e

d

e

d

e

b

c

b

c

b

c

b

c

a

a

a

1.4. PARTIAL ORDERS

41

Solution: a) This is a lattice. b) This is not a lattice. Consider the pair {b, c}. The set of upper bounds is {e, f, g, h}. We note that e and f are incomparable and hence there is no upper bound u such that u 4 u0 for all other upper bounds u0 . c) This is a lattice. d) This is not a lattice. The set {d, e} does not have a greatest lower bound. Exercise: 26 Section 1.4 Question: Explain under what conditions a flow chart may be viewed as a partial order. Solution: A flow chart is a list of instructions to accomplish a certain task (or answer a certain question) where individual instructions may depend on the answers to intermediate questions. This exercise is simply asking to compare the possible structure of a flow chart to that of a Hasse diagram. If a flow chart can be viewed as a partial order, then any list of followed instructions is a chain. But in a poset, a chain cannot be a loop. Hence, a flow chart may be viewed as a partial order if the flow chart does not have any possible sequence of decisions that leads to repeating the same instruction twice. Exercise: 27 Section 1.4 Question: Let (S, 4) and (T, 42 ) be two partial orders. Suppose that ∼ is an equivalence relation on S that satisfies Condition (1.6). Prove that for any monotonic function f : S → T such that f (a) = f (b) whenever a ∼ b, there exists a unique monotonic function f 0 : S/ ∼ → T such that f = f 0 ◦ p, where p : S → S/ ∼ is the projection p(a) = [a] and the partial order on S/ ∼ is defined in (1.7). In the terminology of diagrams, prove that the diagram below is commutative. (S, 4)

p

f

(S/ ∼, 40 ) f0 (T, 42 )

Solution: Suppose that f : S → T is a monotonic function such that f (a) = f (b) whenever a ∼ b. Define the function f 0 : S/ ∼ T by f 0 ([a]) = f (a). This function is well-defined since if b ∼ a, then f (b) = f (a), so the definition of f 0 is independent of the representative of the equivalence class. This function is monotonic since [a] 40 [c] if and only if there exist a0 and c0 such that a0 ∼ a and c0 ∼ c and a0 4 c0 . But then f (a0 ) 42 f (c0 ) so f 0 ([a]) 42 f 0 ([c]). Note that f 0 is defined so that f 0 ◦ p = f . Let f˜ be any function f˜ : S/ ∼→ T satisfying the same relations as f 0 . Then for all equivalence classes [a] ∈ S/ ∼, we have f˜([a]) = f (a) = f 0 ([a]). Hence f 0 = f˜. So the function is unique. Exercise: 28 Section 1.4 Question: Let R1 and R2 be partial orders on a set S. a) Prove that R1 ∩ R2 is a partial order. b) Show by a counterexample that R1 ∪ R2 is not necessarily a partial order. Solution: Let R1 and R2 be partial orders on a set S. We consider the R1 and R2 as subsets of S × S. a) Reflexivity: Let x ∈ S; then (x, x) ∈ R1 and (x, x) ∈ R2 so (x, x) ∈ R1 ∩ R2 . Antisymmetry: suppose that (x, y) ∈ R1 ∩ R2 and (y, x) ∈ R1 ∩ R2 ; then x = y since (x, y) ∈ R1 and (y, x) ∈ R1 and R1 is a partial order. Transitivity: Let (x, y), (y, z) ∈ R1 ∩ R2 ; then (x, z) ∈ R1 since R1 is a transitive and (x, z) ∈ R2 since R2 is transitive so (x, z) ∈ R1 ∩ R2 . b) Let R1 be the partial order of ≤ on N and let R2 be the partial order of ≥ on the same set. Then R1 ∪ R2 is the relation that relates everything in N to everything else. This relation is not antisymmetric. Exercise: 29 Section 1.4 Question: Consider the poset (N, ≤) and consider the equivalence relation ∼ on N defined by j n k jmk = . n ∼ m ⇐⇒ 10 10

42

CHAPTER 1. SET THEORY 1. Describe the equivalence classes of ∼ and find a complete set of distinct representatives of ∼. 2. Show that ∼ satisfies the poset quotient condition. 3. Describe the poset (N/ ∼, ≤inh ).

Solution: a) Notice that the ∼ equivalence relation considers nonnegative integers the same if they only differ in their units digit. A complete set of distinct representatives is all integers with 0 as the units digit. In fact, we jnk note that every n ∈ N is ∼-equivalent to 10 . 10 b) Suppose n 6∼ m, which means that n and m do not have all digits equal in the 10 or above places. Suppose also that n ≤ m. Suppose also that n0 ∼ n and m0 ∼ m. Now n0 and m0 satisfy jnk jmk jmk jnk ≤ n0 ≤ 10 + 9 and 10 ≤ m0 ≤ 10 + 9. 10 10 10 10 10 n n n Now n ≤ m but n 6∼ m means that m > 10 10 + 9 so m ≥ 10 10 + 10. In particular, m0 ≥ 10 10 + 10 so m0 > n0 and hence m0 6≤ n0 . c) The quotient set N/ ∼ is in bijection with N where we have effectively ignored the units digit of n. The quotient partial order is again the usual ≤ on N. The overall quotient process ignores the units digit and manages to preserve the ≤. Exercise: 30 Section 1.4 Question: Let S = P({a, b, c, d, e}) and consider the poset (S, ⊆). Consider the equivalence relation ∼ on S that has as its partition o n o n {a} ∪ C, {b} ∪ C, {a, b} ∪ C | C ⊆ {c, d, e} ∪ C | C ⊆ {c, d, e} . (The equivalence relation has the effect of considering a and b as the same element.) Equivalence classes have either one element or three elements. a) Show that ∼ satisfies the poset quotient condition. b) Show that (S/ ∼, ⊆inh ) is isomorphic to a lattice of subgroups for a set of four elements. Solution: a) Suppose that A, B ∈ P({a, b, c, d, e}) such that A and B are inequivalent with A ⊆ B. Suppose also that A ∼ A0 and B ∼ B 0 under the equivalence relation defined in the problem. Case (1), neither A nor B contain a or b. The poset quotient condition is trivially satisfied since we must have A0 = A and B 0 = B so A 6= B and A ⊆ B so B 0 * A0 . Case (2), A contains a or b and B contains neither a nor b. But because of these conditions, we cannot have A ⊆ B since A contains a or b and B does not. Hence, this case never occurs. Case (3), A contains neither a nor b and B contains a or b. Then A0 = A and B 0 also must contain a or b so we cannot have B 0 ⊆ A0 . Case (4) both A and B contain a or b. Since A and B are inequivalent then A ∩ {c, d, e} = 6 B ∩ {c, d, e}. Also since A ⊆ B, then A ∩ {c, d, e} ( B ∩ {c, d, e}. Assume that B 0 ⊆ A0 ; then B ∩ {c, d, e} = B 0 ∩ {c, d, e} ⊆ A0 ∩ {c, d, e} = A ∩ {c, d, e}, which we just showed is not true. Hence B 0 * A0 . In all four cases, we have shown that the equivalence relation satisfies the quotient poset condition. b) Consider the function f : P({a, c, d, e}) → S/ ∼ defined by f (A) = [A], that maps a subset A to its corresponding equivalence class in S/ ∼. It is not hard to see that this is a bijection. Also, it is obvious from the definition of the inherited partial order that f is monotonic. However, we need to prove that f −1 is monotonic. We note that [A] ⊆inh [B] if and only if A ⊆ B if A and B do not both contain a or b and A ∩ {c, d, e} ⊆ B ∩ {c, d, e} otherwise. But ( A if A does not contain a or b f −1 ([A]) = (A ∩ {c, d, e}) ∪ {a} if A contains a or b. In all cases, if [A] ⊆inh [B], then f −1 ([A]) ⊆ f −1 ([B]).

2 | Number Theory 2.1 – Basic Properties of Integers Exercise: 1 Section 2.1 Question: Find the prime factorization of the following integers: (a) 56; (b) 97; (c) 126; (d) 399; (e) 255; (f) 1728 Solution: (a) 56 = 23 × 7; (b) 97=97; (c) 126 = 2 × 32 × 7; (d) 399 = 3 × 7 × 19; (e) 255 = 3 × 5 × 17; (f) 1728 = 26 × 33 . Exercise: 2 Section 2.1 Question: Find the prime factorization of the following integers: a) 111; b) 470; c) 289; d) 743; e) 2345; f) 101010 Solution: (a) 111 = 3 × 37; (b) 470 = 2 × 5 × 47; (c) 289 = 172 ; (d) 743 = 743. Note that as we check various prime factors to see if√they divide 743, the last one we need to check is 23 because it is the largest prime factor less than or equal to 743; (e) 2345 = 5 × 7 × 67; (f) 101010 = 2 × 3 × 5 × 7 × 13 × 37. Exercise: 3 Section 2.1 Question: Draw the Hasse diagram of ({1, 2, 3, . . . , 12}, |). Solution: 8

12

4

6

9

10

2

3

5

7

11

1

Exercise: 4 Section 2.1 Question: Let n be a positive integer. Show that the number of edges in the Hasse diagram of ({1, 2, 3, . . . , n}, |) is X n . p p: primes≤n

Solution: In the Hasse diagram of ({1, 2, 3, . . . , n}, |), there is an edge up from a to b if and only if b is a prime multiple of a. Hence, each edge corresponds to a unique prime number. There is an edge corresponding to a prime p out of each integer a ∈ {1, 2, . . . , n} if and only if ap ≤ n. Hence, there j k is an edge corresponding to a

prime p out of each integer a ∈ {1, 2, . . . , n} if and only if a ≤ np . There are n p . Thus the number of edges in the Hasse diagram is X p: primes≤n

n p

integers less than or equal to

n . p

Exercise: 5 Section 2.1 Question: Use the Euclidean Algorithm to find the greatest common divisor of the following pairs of integers. 1. a = 234, and b = 84. 2. a = 5241, and b = 872. 3. a = 1010101, and b = 1221. 43

44

CHAPTER 2. NUMBER THEORY

Solution: Euclidean Algorithm: a) gcd(234, 84) = 6 because 234 = 84 × 2 + 66 84 = 66 × 1 + 18 66 = 18 × 3 + 12 18 = 12 × 1 + 6 12 = 6 × 2 + 0 and 6 is the last nonzero remainder. b) gcd(5241, 872) = 1 because 5241 = 872 × 6 + 9 872 = 9 × 96 + 8 9=8×1+1 8=1×8+0 and 1 is the last nonzero remainder. c) gcd(1010101, 1221) = 1 because 1010101 = 1221 × 827 + 334 1221 = 334 × 3 + 219 334 = 219 × 1 + 115 219 = 115 × 1 + 104 115 = 104 × 1 + 11 104 = 11 × 9 + 5 11 = 5 × 2 + 1 5=1×5+0

Exercise: 6 Section 2.1 Question: Use the Euclidean Algorithm to find the greatest common divisor of the following pairs of integers. a) a = 55, and b = 34 b) a = 4321, and b = 1234 c) a = 54321, and b = 1728 Solution: a) The Euclidean Algorithm on 55 and 34 gives: 55 = 34 × 1 + 21 34 = 21 × 1 + 13 21 = 13 × 1 + 8 13 = 8 × 1 + 5 8=5×1+3 5=3×1+2 3=2×1+1 2 = 1 × 2 + 0. So the greatest common divisor is gcd(55, 34) = 1.

2.1. BASIC PROPERTIES OF INTEGERS

45

b) The Euclidean Algorithm on 4321 and 1234 gives: 4321 = 1234 × 3 + 619 1234 = 619 × 1 + 615 619 = 615 × 1 + 4 615 = 4 × 153 + 3 4 = 3 × 1 + 1. This shows that gcd(4321, 1234) = 1. c) The Euclidean Algorithm on 54321 and 1728 gives: 54321 = 1728 × 31 + 753 1728 = 753 × 2 + 222 753 = 222 × 3 + 87 222 = 87 × 2 + 48 87 = 48 × 1 + 39 48 = 39 × 1 + 9 39 = 9 × 4 + 3 9=3×3+0 This shows that gcd(54321, 1728) = 3. Exercise: 7 Section 2.1 Question: Define the Fibonacci sequence {fn }n≥0 by f0 = 0, f1 = 1 and fn = fn−1 + fn−2 for all n ≥ 2. Let fn and fn+1 be two consecutive terms in the Fibonacci sequence. Prove that gcd(fn+1 , fn ) = 1 and show that for all n ≥ 2, the Euclidean algorithm requires exactly n − 1 integer divisions (including the last one that has a remainder of 0). Solution: The successive indices f1 = f2 are the only time the Fibonacci sequence elements are equal. Hence, for n ≥ 4, the recurrence relation fn = fn−1 + fn−2 is such that 0 ≤ fn−2 < fn−1 . Thus fn = fn−1 × 1 + fn−2 is the integer division of fn by fn−1 . So when performing the Euclidean Algorithm on fn+1 and fn , the recurrence relation gives the Euclidean Algorithm step for from fn+1 down to f4 = f3 + f2 , which is 3 = 2 + 1. This accounts for n − 2 integer divisions. The last integer division is f3 = f2 × 2 + 0. Hence, there are n − 1 integer divisions in the Euclidean Algorithm and the penultimate line in the Euclidean Algorithm is 3 = 2 + 1. Consequently, gcd(fn+1 , fn ) = 1. Exercise: 8 Section 2.1 Question: Let a, b, c ∈ Z. Prove that a|b implies that a|bc. Solution: If a | b then there exists k ∈ Z such that b = ak. Hence, bc = a(kc). Thus a | bc. Exercise: 9 Section 2.1 Question: Perform the Extended Euclidean Algorithm on the three pairs of integers in Exercise 2.1.5. Solution: Extended Euclidean Algorithm: a) We have gcd(234, 84) = 6 and extending the Euclidean Algorithm gives 6 = 18 − 1 × 12 = 18 − (66 − 18 × 3) = (−1) × 66 + 4 × 18 = (−1) × 66 + 4 × (84 − 66) = 4 × 84 − 5 × 66 = 4 × 84 − 5 × (234 − 2 × 84) = −5 × 234 + 14 × 84. b) We have gcd(5241, 872) = 1 and extending the Euclidean Algorithm gives 1=9−1×8 = 9 − (872 − 96 × 9) = −872 + 97 × 9 = −872 + 97(5241 − 6 × 872) = 97 × 5241 − 583 × 872.

46

CHAPTER 2. NUMBER THEORY c) We have gcd(1010101, 1221) = 1 and extending the Euclidean Algorithm gives 1 = 11 − 2 × 5 = 11 − 2(104 − 9 × 11) = (−2) × 104 + 19 × 11 = (−2) × 104 + 19 × (115 − 104) = 19 × 115 − 21 × 104 = 19 × 115 − 21 × (219 − 115) = (−21) × 219 + 40 × 115 = (−21) × 219 + 40 × (334 − 219) = 40 × 334 − 61 × 219 = 40 × 334 − 61 × (1221 − 3 × 334) = (−61) × 1221 + 223 × 334 = (−61) × 1221 + 223 × (1010101 − 827 × 1221) = 223 × 1010101 − 184482 × 1221.

Exercise: 10 Section 2.1 Question: Perform the Extended Euclidean Algorithm on the three pairs of integers in Exercise 2.1.6 Solution: Extended Euclidean Algorithm: a) We have gcd(55, 34) = 1 and extending the Euclidean Algorithm gives 1=3−1×2 = 3 − 1 × (5 − 3) = (−1) × 5 + 2 × 3 = (−1) × 5 + 2 × (8 − 5) = 2 × 8 − 3 × 5 = 2 × 8 − 3 × (13 − 8) = (−3) × 13 + 5 × 8 = (−3) × 13 + 5 × (21 − 13) = 5 × 21 − 8 × 13 = 5 × 21 − 8 × (34 − 21) = (−8) × 34 + 13 × 21 = (−8) × 34 + 13 × (55 − 34) = 31 × 55 − 21 × 34. b) We have gcd(4321, 1234) = 1 and extending the Euclidean Algorithm gives 1=4−3 = 4 − (615 − 153 × 4) = (−1) × 615 + 154 × 4 = (−1) × 615 + 154 × (619 − 615) = 154 × 619 − 155 × 615 = 154 × 619 − 155 × (1234 − 619) = (−155) × 1234 + 309 × 619 = (−155) × 1234 + 309 × (4321 − 3 × 1234) = 309 × 4321 − 1082 × 1234. c) We have gcd(54321, 1728) = 3 and extending the Euclidean Algorithm gives 3 = 39 − 4 × 9 = 39 − 4 × (48 − 39) = (−4) × 48 + 5 × 39 = (−4) × 48 + 5 × (87 − 48) = 5 × 87 − 9 × 48 = 5 × 87 − 9 × (222 − 2 × 87) = (−9) × 222 + 23 × 87 = (−9) × 222 + 23 × (753 − 3 × 222) = 23 × 753 − 78 × 222 = 23 × 753 − 78 × (1728 − 2 × 753) = (−78) × 1728 + 179 × 753 = (−78) × 1728 + 179 × (54321 − 31 × 1728) = 179 × 54321 − 5627 × 1728.

Exercise: 11 Section 2.1 Question: Suppose that a, b ∈ Z∗ and that s, t ∈ Z∗ such that sa + tb = gcd(a, b). Show that s and t are relatively prime. Solution: Suppose that a, b ∈ Z∗ and that s, t ∈ Z∗ such that sa + tb = gcd(a, b). Since gcd(a, b) divides both a and b, we have b a s+ t = 1. gcd(a, b) gcd(a, b) Since a/ gcd(a, b) and b/ gcd(a, b) are integers, then this gives an integer linear combination of s and t that is 1. Since the greatest common divisor of two integers is the least positive integer linear combination, we deduce that gcd(s, t) = 1.

2.1. BASIC PROPERTIES OF INTEGERS

47

Exercise: 12 Section 2.1 Question: Consider the relation of “relatively prime” on Z∗ . Determine whether it is reflexive, symmetric, antisymmetric, or transitive. Solution: Reflexive. If a ≥ 2, then gcd(a, a) = a so “relatively prime” is not reflexive. Symmetric. “Relatively prime” is symmetric since gcd(a, b) = gcd(b, a). Antisymmetric. Suppose that gcd(a, b) = 1 and gcd(a, b) = 1. This does not imply that a = b. Hence “relatively prime” is not antisymmetric. Transitive. “Relatively prime” is not transitive: gcd(2, 3) = 1 and gcd(3, 4) = 1 but gcd(2, 4) = 2.

Exercise: 13 Section 2.1 Question: Let a, b, c be nonzero integers. Prove that gcd(ab, ac) = a gcd(b, c). Solution: Let s0 , t0 ∈ Z be integers such that s0 b + t0 c = gcd(b, c), which exist by virtue of Proposition 2.1.12. Then s0 ab + t0 ac = a gcd(b, c) so also by the same proposition, since a gcd(b, c) is a linear combination of ab and ac, then gcd(ab, ac) divides a gcd(b, c). Conversely, let s1 , t1 ∈ Z such that gcd(ab, ac) = s1 (ab) + t1 (ac). Then gcd(ab, ac) = a(s1 b + t1 c) and since s1 b + t1 c is a linear combination of b and c, then by Proposition 2.1.12 it must be a multiple of gcd(b, c). Hence, we deduce that gcd(ab, ac) divides a gcd(ab, ac). Two positive integers that divide each other must be equal so gcd(ab, ac) = a gcd(b, c). Exercise: 14 Section 2.1 Question: Let a and b be positive integers. Show that the set of common multiples of a and b is lcm(a, b)Z, i.e., the set of multiples of lcm(a, b). Solution: Let a and b be positive integers and let S be the set of common multiples. Let m be the least positive element of S, which exists by the well-ordering of the integers. This is lcm(a, b). (S is not empty since ab ∈ S.) Let M be any other common multiple of a and b. Consider the integer division of M by m, as M = qm + r, where 0 ≤ r < m. We then have r = M − qm. We have M = `a and m = ka so r = (` − qk)a. Hence r is a multiple of a. Similarly, r is a multiple of b. Thus r ∈ S. Since m is the least positive element in S, then r must be 0. Hence M = qm. Thus, every element of S is a multiple of m = lcm(a, b). Exercise: 15 Section 2.1 Question: Prove that any integer greater than 3 that is 1 less than a square cannot be prime. Solution: An integer that is one less than a square has the form n2 − 1 for some positive integer n. However, n2 − 1 = (n − 1)(n + 1). The condition that n2 − 1 > 3 is equivalent to n > 2. Hence, n + 1 > n − 1 > 1. Thus n2 − 1 = (n − 1)(n + 1) is a factorization of n2 − 1 in which both factors are greater than 1. Thus n2 − 1 is not a prime number. Exercise: 16 Section 2.1 Question: Prove that if 2n − 1 is prime then n is prime. Solution: We prove this with a contrapositive proof. Suppose that n is composite with n = rs with r, s > 1. Then s 2n − 1 = 2rs − 1 = (2r ) − 1. In the formula as − bs = (a − b)(as−1 + as−2 b + as−3 b2 + · · · + bs−1 ), we set a = 2r and b = 1 and get s−1 s−2 2n − 1 = (2r − 1) (2r ) + (2r ) + · · · + (2r ) + 1 .

48

CHAPTER 2. NUMBER THEORY

This expresses 2n − 1 as a product of two integers. To establish a contradiction, we simply need to show that both of these factors are greater than 1. With the assumption that r ≥ 2, we have 2r ≥ 4 so 2r − 1 ≥ 3. The s−1 s−2 first factor is greater than 1. Since s ≥ 2, the expression (2r ) + (2r ) + · · · + (2r ) + 1 involves at least two terms so it is greater than or equal to 2r + 1 ≥ 22 + 1 = 5. In particular, the second factor is greater than 1. We have proven that if n is composite then 2n − 1 is composite. The (logically equivalent) contrapositive of this statement is that if 2n − 1 is prime, then n is prime. Exercise: 17 Section 2.1 Question: Prove or disprove that p1 p2 · · · pn + 1 is a prime number where p1 , p2 , . . . , pn are the n smallest consecutive prime numbers. Solution: This hypothesis is not true. It suffices to calculate the p1 p2 · · · pn + 1 for a few n and checked their prime factorizations. n p1 p2 · · · pn + 1 1 3 =3 =7 2 7 3 61 = 61 = 421 4 421 5 2311 = 2311 6 30031 = 59 × 509

Exercise: 18 Section 2.1 Question: Prove that the product of two consecutive positive integers is even. Solution: Suppose that n is even; then n(n + 1) is even. Suppose that n is odd; then n + 1 is even so n(n + 1) is even. Hence, regardless of what n is, the product n(n + 1) is even. Exercise: 19 Section 2.1 Question: Prove that the product of four consecutive positive integers is divisible by 24. Solution: We prove this by cases that 24 | n(n + 1)(n + 2)(n + 3) for all n ∈ Z. At least one of the integers n, n + 1, n + 2, or n + 3 is divisible by 3. n has a remainder or 0 when divided by 4 . Then n is divisible by 4 and n + 2 is even. Hence, 8 | n(n + 1)(n + 2)(n + 3) so also 24 | n(n + 1)(n + 2)(n + 3). n has a remainder or 1 when divided by 4 . Then n + 3 is divisible by 4 and n + 1 is even. Hence, 8 | n(n + 1)(n + 2)(n + 3) so also 24 | n(n + 1)(n + 2)(n + 3). n has a remainder or 2 when divided by 4 . Then n + 2 is divisible by 4 and n is even. Hence, 8 | n(n + 1)(n + 2)(n + 3) so also 24 | n(n + 1)(n + 2)(n + 3). n has a remainder or 3 when divided by 4 . Then n + 1 is divisible by 4 and n + 3 is even. Hence, 8 | n(n + 1)(n + 2)(n + 3) so also 24 | n(n + 1)(n + 2)(n + 3). In all cases, 24 divides the product n(n + 1)(n + 2)(n + 3) of 4 consecutive integers. Exercise: 20 Section 2.1 Question: Suppose that the prime factorizations of a and b are αn 1 α2 a = pα 1 p2 · · · pn

and b = pβ1 1 pβ2 2 · · · pβnn ,

with pi distinct primes and αi , βi ≥ 0. min(α1 ,β1 ) min(α2 ,β2 ) n ,βn ) p2 · · · pmin(α . n max(α1 ,β1 ) max(α2 ,β2 ) max(αn ,βn ) Prove that lcm(a, b) = p1 p2 · · · pn .

a) Prove that gcd(a, b) = p1 b)

Solution: Suppose that the prime factorizations of a and b are αn 1 α2 a = pα 1 p2 · · · pn

with pi distinct primes and αi , βi ≥ 0.

and b = pβ1 1 pβ2 2 · · · pβnn ,

2.1. BASIC PROPERTIES OF INTEGERS

49

a) Let d0 be a common divisor of a and b. Since d0 k = a and d0 ` = b for some integers k and `, then the only primes occurring in the prime factorization of d0 are p1 , p2 , . . . , pn . Suppose that d0 = pγ11 pγ22 · · · pγnn . Then n −γn a = d0 p1α1 −γ1 p2α2 −γ2 · · · pα n

and b = d0 p1β1 −γ1 p2β2 −γ2 · · · pnβn −γn Since all the powers on the primes must be nonnegative, we deduce that γi ≤ αi and γi ≤ βi for i = 1, 2, . . . , n. min(α ,β ) min(α ,β )

1 1 2 2 n ,βn ) Let d = p1 p2 · · · pmin(α . We see that from the above condition that d is a common n divisor. Furthermore, every other common divisor d0 as above has γi ≤ max(αi , βi ) for i = 1, 2, . . . , n and thus min(α1 ,β1 )−γ1 min(α2 ,β2 )−γ2 n ,βn )−γn d = d0 p1 p2 · · · pmin(α . n

Consequently, d0 | d. Thus d is a greatest common divisor to a and b. b) Let m0 be a common multiple of a and b. Since m0 = ka and m0 = `b for some integers k and `, then the prime p1 , p2 , . . . , pn occur in the prime factorization of m0 . Suppose that m0 = pγ11 pγ22 · · · pγnn Q, where Q is not divisible by pi for any i = 1, 2, . . . , n. Then m0 = apγ11 −α1 pγ22 −α2 · · · pγnn −αn Q and m0 = bp1γ1 −β1 p2γ2 −β2 · · · pγnn −βn Q Since all the powers on the primes must be nonnegative, we deduce that γi ≥ αi and γi ≥ βi for i = 1, 2, . . . , n. max(α ,β ) max(α ,β )

1 1 2 2 n ,βn ) Let m = p1 p2 · · · pmax(α . We see that from the above condition that m is a common n multiple. Furthermore, every other common multiple m0 as above has γi ≥ max(αi , βi ) for i = 1, 2, . . . , n and thus γ −max(α1 ,β1 ) γ2 −max(α2 ,β2 ) m0 = mp11 p2 · · · pγnn −max(αn ,βn ) Q.

Consequently, m | m0 . Thus m is a least common multiple to a and b. Exercise: 21 Section 2.1 Question: Let a and b be positive integers. Prove that gcd(a, b) lcm(a, b) = ab using the result of Problem 2.1.20. Solution: Let a and b be integers with the prime factorization of αn 1 α2 a = pα 1 p2 · · · pn

and b = pβ1 1 pβ2 2 · · · pβnn ,

with pi distinct primes and αi , βi ≥ 0. Then according to Problem 2.1.20, min(α1 ,β1 )+max(α1 ,β1 ) min(α2 ,β2 )+max(α2 ,β2 ) n ,βn )+max(αn ,βn ) p2 · · · pmin(α . n

gcd(a, b) lcm(a, b) = p1

Now for any two real numbers x and y (and so this would also hold for integers), ( x + y if x ≤ y min(x, y) + max(x, y) = . y + x if x > y So min(x, y) + max(x, y) = x + y for all real numbers x, y ∈ R. Hence, returning to our situation, n +βn gcd(a, b) lcm(a, b) = p1α1 +β1 p2α2 +β2 · · · pα n β1 β2 αn βn 1 α2 = (pα 1 p2 · · · pn ) p1 p2 · · · pn

= ab.

50

CHAPTER 2. NUMBER THEORY

Exercise: 22 Section 2.1√ Question: Prove that 2 is not a rational number. √ Solution: Assume that 2 = ab with a and b relatively prime. Then 2b2 = a2 . Consider the prime factorization of this expression 2b2 = a2 . The prime factorization of 2b2 involves an odd number of prime factors, counting 2 separately √ repeated prime factors, whereas a has an even number of prime factors. This leads to a contradiction. Hence, 2 is irrational. Exercise: 23 Section 2.1 √ Question: Prove that for all primes p and all integers k ≥ 2, the number k p is irrational. √ √ Solution: Assume that k p is rational and that k p = ab in reduced form, i.e., gcd(a, b) = 1. Then pbk = ak . Since the prime p divides ak , we know that p | a with a = pd. Thus pbk = pk dk and hence bk = pk−1 dk . Then p is in the prime factorization of bk , so p | bk and hence p | b. This contradicts the assumption that a and b are √ relatively prime. Hence, k p is irrational. Exercise: 24 Section 2.1 Question: Determine all the nonzero ordp (n), defined in (2.4), for all primes p, where n is one of the following (a) 450;

(b) 392;

(c) 2310;

(d) 121212.

Solution: a) Since 450 = 2 × 32 × 52 , then the only nonzero ordp values are ord2 (450) = 1, ord3 (450) = 2, and ord5 (450) = 2. b) Since 392 = 23 × 72 , then the only nonzero ordp values are ord2 (4392) = 3 and ord7 (392) = 2. c) Since 2310 = 2 × 3 × 5 × 7 × 11, then the only nonzero ordp values are ord2 (2310) = ord3 (2310) = ord5 (2310) = ord7 (2310) = ord11 (2310) = 1. d) Since 121212 = 22 × 32 × 7 × 13 × 37, then the only nonzero ordp values are ord2 (121212) = ord3 (121212) = 2 and ord7 (121212) = ord13 (121212) = ord37 (121212) = 1.

Exercise: 25 Section 2.1 Question: Find ord5 (200!). Use this to determine the number of 0s to the right in the decimal expansion of 200!. Solution: There are 40 multiples of 5 occurring in the product of integers in 20!. Furthermore, 8 of them are multiples of 52 = 25 and 1 of them is a multiple of 53 = 125. Consequently, ord5 (200!) = 40 + 8 + 1 = 49. It is obvious that ord2 (200!) ≥ 100. Hence, 1049 divides 200! but 1050 - 200!. Hence, there are 49 0s in the decimal expansion of 200!. Exercise: 26 Section 2.1 Question: Let p be a prime number. Prove that the function ordp : Q → Z defined in (2.4) satisfies the following logarithmic-type properties. a) ordp (mn) = ordp (m) + ordp (n) for all m, n ∈ Z; b) ordp (mk ) = k ordp (m) for all m ∈ Z and k ∈ N∗ . Solution: Let p be a prime number. a) Suppose that ordp (m) = α and ordp (n) = β. Then pα | m and pα - m and pβ | n and pβ - n. Furthermore, pα+β | mn but pα+β+1 - mn. Hence, ordp (mn) = α + β. b) If ordp (m) = α, then m = pα Q with p - Q. Thus mk = pαk Qk and since p - Q, we still have p - Qk . Thus, pαk | mk but pαk+1 - mk . Thus, ordp (mk ) = k ordp (m).

2.1. BASIC PROPERTIES OF INTEGERS

51

Exercise: 27 Section 2.1 Question: For the following integers, calculate ϕ(n) by directly listing the set in (2.5). (a)φ(30);

(b)φ(33);

(c)φ(12).

Solution: a) {a ∈ N∗ | 1 ≤ a ≤ 30 and gcd(a, 30) = 1} = {1, 7, 11, 13, 17, 19, 23, 29}. Thus, φ(30) = 8. b) {a ∈ N∗ | 1 ≤ a ≤ 33 and gcd(a, 33) = 1} is the set {1, 2, 4, 5, 7, 8, 10, 13, 14, 16, 17, 19, 20, 23, 25, 26, 28, 28, 31, 32}. Thus, φ(33) = 20. c) {a ∈ N∗ | 1 ≤ a ≤ 12 and gcd(a, 12) = 1} = {1, 5, 7, 11}. Thus, φ(12) = 4. Exercise: 28 Section 2.1 Question: Prove that for all integers n, n=

X

φ(d),

d|n

where this summation notation means we sum over all positive divisors d of n. Solution: There are obviously n distinct fractions in the set n1 , n2 , n3 , . . . , nn . A fraction of the form na in 0 0 reduced form is na = ad where d is a divisor of n. If ad is in reduced form then a0 and d are relatively prime. For 0 each divisor d of n, there are φ(d) fractions of the form ad with gcd(a0 , d) = 1. Thus n=

X

φ(d).

d|n

Exercise: 29 Section 2.1 Question: Prove that for any prime p, the following identities hold. a) φ(p) = p − 1 b) φ(pk ) = pk − pk−1 Solution: a) The set of integers between 1 and p that are relatively prime to p is {1, 2, . . . , p − 1}. Hence, φ(p) = p − 1. b) The set of integers between 1 and pk that are relatively prime to p are the integers between 1 and pk that are not divisible by p. The number of integers between 1 and pk that are divisible by p are {pa | 1 ≤ a ≤ pk−1 }. Thus, φ(pk ) = pk − pk−1 . Exercise: 30 Section 2.1 Question: Prove that if a and b are relatively prime, then Euler’s totient function satisfies φ(ab) = φ(a)φ(b). Solution: Suppose that a and b are positive integers. Let Dn = {k | 1 ≤ k ≤ n such that gcd(k, n) > 1}. Then the sets bDa = {bk | 1 ≤ k ≤ a such that gcd(k, a) > 1} and aDb are subsets of Dab . The union aDb ∪ bDa covers Dab but aDb ∩ bDa is nontrivial. In fact, aDb ∩ bDa consists of products of elements that are not relatively prime to a and not relatively prime to b. Since gcd(a, b) = 1, this set has cardinality (a − φ(a))(b − φ(b)). By the Inclusion-Exclusion principle, |Dab | = |bDa | + |aDb | − |bDa ∩ aDb | = a(b − φ(b)) + b(a − φ(a)) − (a − φ(a))(b − φ(b)) = ab − φ(a)φ(b). But |Dab | = ab − φ(ab) so we deduce that φ(ab) = φ(a)φ(b). Exercise: 31 Section 2.1 Question: Using Exercises 2.1.29 and 2.1.30, prove Proposition 2.1.27.

52

CHAPTER 2. NUMBER THEORY α

αk αi j 1 α2 Solution: Let n = pα 1 p2 · · · pk be the prime factorization of n. By Exercise 2.1.30, since gcd(pi , pj ) = 1 for i 6= j, we have αk α2 1 φ(n) = φ((pα 1 ) φ (p2 ) · · · φ (pk ) αk −1 α1 −1 α2 −1 1 2 k = pα pα · · · pα , 1 − p1 2 − p2 k − pk

where the second equality holds by part (b) of Exercise 2.1.29.

2.2 – Modular Arithmetic Exercise: 1 Section 2.2 Question: List ten elements in the conjugacy class 3 in modulo 7. Solution: Modulo 7, the conjugacy class 3 contains among other elements . . . , −18, −11, −4, 3, 10, 17, 24, 31, 38, 45, . . . Exercise: 2 Section 2.2 Question: List all the elements in Z/13Z. Solution: Z/13Z = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Exercise: 3 Section 2.2 Question: List all the elements in Z/24Z and in U (24). Solution: The elements in Z/24Z are Z/24Z = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}. The units in this modular arithmetic are U (24) = {1, 5, 7, 11, 13, 17, 19, 23}.

Exercise: 4 Section 2.2 Question: Perform the following calculations in the modular arithmetic of the given modulus n. a) 3̄ + 5̄ · 7̄ with n = 9 b) (5̄ · 4̄ − 72 · 3̄)2 with n = 11 c) 13 · 42 · 103 with n = 15 Solution: a) With n = 9, we have 3̄ + 5̄ · 7̄ = 3̄ + 8̄ = 2̄. b) With n = 11, we have (5̄ · 4̄ − 72 · 3̄)2 = (9̄ − 7̄)2 = 2̄2 = 4̄. c) With n = 15, we have 13 · 42 · 103 = 13 · 12 · 13 = −2 · −3 · −2 = −12 = 3.

Exercise: 5 Section 2.2 Question: Write out the elements in the set U (30). Solution: U (30) = {1̄, 7̄, 11, 13, 17, 19, 23, 29}. Exercise: 6 Section 2.2 Question: In Z/17Z, solve for x in terms of y in y = 2̄x + 3̄.

2.2. MODULAR ARITHMETIC

53

Solution: In Z/17Z, the equation y = 2̄x + 3̄ is equivalent to y + 14 = 2̄x. The multiplicative inverse of 2̄ in Z/17Z is 9̄. Thus x = 9̄(y + 14) = 9̄y + 7̄. Exercise: 7 Section 2.2 Question: In Z/29Z, solve for x in terms of y in y = 17x + 20. Solution: We need to find the multiplicative inverse of 17 in Z/29Z. Whether we use the Extended Euclidean −1 Algorithm or look for the inverse by exhaustively trying elements, we find that 17 = 12. Hence we have y = 17x + 20 =⇒ y + 9 = 17x =⇒ x = 12(y + 9) = 12y + 21

Exercise: 8 Section 2.2 Question: Show that for all integers a, we have a2 ≡ 0 or 1 (mod 4). Show how this implies that for all integers a, b ∈ Z, the sum of squares a2 + b2 never has a remainder of 3 when divides by 4. Solution: Considering the four cases for a congruent modulo 4, we have (a) if a ≡ 0 (mod 4), then a2 ≡ 0 (mod 4); (b) if a ≡ 1 (mod 4), then a2 ≡ 1 (mod 4); (c) if a ≡ 2 (mod 4), then a2 ≡ 4 ≡ 0 (mod 4); and (d) if a ≡ 3 (mod 4), then a2 ≡ 9 ≡ 1 (mod 4). Hence, in all cases a2 ≡ 0 or 1 (mod 4) for all integers a. Consequently, for all integers a, b ∈ Z,  2 2  0 if a ≡ 0 and b ≡ 0 2 2 2 a + b ≡ 1 if a ≡ 1 and b2 ≡ 0ORa2 ≡ 0 and b2 ≡ 1   2 if a2 ≡ 1 and b2 ≡ 1. Hence, a2 + b2 is never congruent to 3 modulo 4. Exercise: 9 Section 2.2 Question: Prove that if d|m and a ≡ b (mod m), then a ≡ b (mod d). Solution: Suppose that d | m. Then m = dk for some integer k. The statement a ≡ b (mod m) is equivalent to m | (b − a), or in other words b − a = m` for some integer `. Thus b − a = d(k`) so d | (b − a) and therefore a ≡ b (mod d). Exercise: 10 Section 2.2 Question: Prove that if a, b, c, and m are integers with m ≥ 2 and c > 0, then a ≡ b (mod m) implies that ac ≡ bc (mod mc). Solution: Let a, b, c, and m be integers with m ≥ 2 and c > 0. Then a ≡ b (mod m) is equivalent to m | (b−a), which is equivalent to mk = b − a for some integer k. Then multiplying by c, we have (mc)k = bc − ac. Hence, mc | (bc − ac) and so ac ≡ bc (mod mc). Exercise: 11 Section 2.2 −1 Question: Perform the Extended Euclidean Algorithm to calculate 52 in Z/101Z. Solution: The Euclidean Algorithm to find that gcd(101, 52) proceeds as follows. 101 = 52 × 1 + 49 52 = 49 × 1 + 3 49 = 3 × 16 + 1 3=1×3+0 (This shows that gcd(101, 52) = 1 so 52 is invertible modulo 101. Now reversing the process to complete the Extended Euclidean Algorithm, we have 1 = 49 − 3 × 16 = 49 − (52 − 49) × 16 = 49 × 17 − 52 × 16 = (101 − 52) × 17 − 52 × 16 = 101 × 17 − 52 × 33

54

CHAPTER 2. NUMBER THEORY

Considering the linear combination 1 = 101 × 17 − 52 × 33 modulo 101, we get −52 × 33 ≡ 1 Hence 52

−1

(mod 101).

= −33 = 68.

Exercise: 12 Section 2.2 −1 Question: Perform the Extended Euclidean Algorithm to calculate 72 in Z/125Z. Solution: We first perform the Euclidean Algorithm on 125 and 72. 125 = 72 × 1 + 53 72 = 53 × 1 + 19 53 = 19 × 2 + 15 19 = 15 × 1 + 4 15 = 4 × 3 + 3 4=3×1+1 3=1×3+0 This confirms that gcd(125, 72) = 1, which means that 72 has a multiplicative inverse modulo 125. The Extended Euclidean Algorithm gives 1=4−3 = 4 − (15 − 3 × 4) = −15 + 4 × 4 = −15 + 4 × (19 − 15) = 4 × 19 − 5 × 15 = 4 × 19 − 5 × (53 − 2 × 19) = −5 × 53 + 14 × 19 = −5 × 53 + 14 × (72 − 53) = 14 × 72 − 19 × 53 = 14 × 72 − 19 × (125 − 72) = −19 × 125 + 33 × 72. The Extended Euclidean Algorithm gives the linear combination 1 = −19 × 125 + 33 × 72. Modulo 125, this −1 expression becomes 1̄ = 33 × 72. Thus 72 = 33. Exercise: 13 Section 2.2 Question: Find the smallest positive integer n such that 2n ≡ 1 (mod 17). Solution: We consider the powers of 2 modulo 17. We perform the operations in Z/17Z. We have 2̄1 = 2̄; 2̄2 = 4̄; 2̄3 = 8̄; 2̄4 = 16; 2̄5 = 32 = 15; 2̄6 = 30 = 13; 2̄7 = 26 = 9; 2̄8 = 18 = 1. Thus, the first power n for which 2n ≡ 1 (mod 17) is n = 8. Exercise: 14 Section 2.2 Question: Find the smallest positive integer n such that 3n ≡ 1 (mod 19). Solution: We consider the powers of 3 modulo 19. We perform the operations in Z/19Z. We have 3̄1 = 3̄; 3̄2 = 9̄; 3̄3 = 27 = 8; 3̄4 = 24 = 5; 3̄5 = 15; 3̄6 = 45 = 7; 3̄7 = 21 = 2; 3̄8 = 6; 3̄9 = 18. At this stage, a clever thing to do is to notice that 3̄9 = −1. Hence, for 10 ≤ k ≤ 18, we have 3̄k = −3̄k−9 . In particular, k = 18 is the first integer where 3̄k = −18 = 1. Exercise: 15 Section 2.2 Question: Show that the powers of 7 in Z/31Z account for exactly half of the elements in U (31). Solution: We first observe that |U (31)| = φ(31) = 30, since 31 is a prime number. We now must calculate 1 2 3 4 5 6 7 the powers of 7. We have 7 = 7; 7 = 49 = 18; 7 = 126 = 2; 7 = 14; 7 = 98 = 5; 7 = 35 = 4; 7 = 28; 8 9 10 11 12 13 14 7 = 196 = 10; 7 = 70 = 8; 7 = 56 = 25; 7 = 175 = 20; 7 = 140 = 16; 7 = 112 = 19; 7 = 133 = 9; 15 a 7 = 63 = 1. After this, the powers of 7 will repeat (cycle). Hence, {7 | a ∈ N} has 15 elements, which is exactly half of the elements of U (31). Exercise: 16 Section 2.2 Question: Show that a number is divisible by 11 if and only if the alternating sum of its digits is divisible by 11. (An alternating sum means that we alternate the signs in the sum + − + − . . ..)

2.2. MODULAR ARITHMETIC

55

Solution: Suppose that a number n is written in decimal as n = (dk · · · d1 d0 )10 , where di are digits with 0 ≤ di ≤ 9 and dk 6= 0. This means that n = dk × 10k + dk−1 × 10k−1 + · · · + d1 × 10 + d0 . Taking this expression modulo 11, since 10 ≡ −1 (mod 11), we have n ≡ dk × (−1)k + dk−1 × (−1)k−1 + · · · + d1 × (−1) + d0 ≡ d0 − d1 + d2 − · · · + (−1)k−1 dk−1 + (−1)k dk

(mod 11)

(mod 11).

Thus 11 | n if and only if n ≡ 0 (mod 11) if and only if d0 − d1 + d2 − · · · + (−1)k−1 dk−1 + (−1)k dk ≡ 0 if and only if the alternating sum of the digits is divisible by 11. Exercise: 17 Section 2.2 Question: Prove that if n is odd then n2 ≡ 1 (mod 8). Solution: If n is odd, then n = 2k + 1 for some integer k. Then n2 = (2k + 1)2 = 4k 2 + 4k + 1 = 4k(k + 1) + 1. Note that k(k + 1) is the product of two consecutive integers, one of which must be even. Hence, k(k + 1) is even so n2 = 8

k(k+1) 2

+ 1. Therefore, n2 ≡ 1 (mod 8) for all odd integers n.

Exercise: 18 Section 2.2 Question: Show that the difference of two consecutive cubes (an integer of the form n3 ) is never divisible by 3. Solution: Suppose that one cube is n3 . Then the next cube is (n + 1)3 . The difference between these is (n + 1)3 − n3 = n3 + 3n2 + 3n + 1 − n3 = 3(n2 + n) + 1. Hence, (n + 1)3 − n3 ≡ 1 (mod 3) for all integers n. In particular, the difference between two consecutive cubes is neve divisible by 3. Exercise: 19 Section 2.2 Question: Use Fermat’s Little Theorem to determine the remainder of 734171 modulo 13. Solution: Since 73 ≡ 8 (mod 13), we first simplify 734171 ≡ 84171 (mod 13). By Fermat’s Little Theorem, 812 ≡ 1 (mod 13). We perform the integer division of 4171 by 12 and get 4171 = 12 × 347 + 7. Thus 347 7 84171 ≡ 812×347+7 ≡ 812 8 ≡ 87 (mod 13). In this particular case of 8, we also can do the following simplification 87 ≡ 221 ≡ 212 ≡ 29 ≡ 24 · 24 · 2 ≡ 3 · 3 · 2 ≡ 5

(mod 13).

Thus, we conclude that the remainder of 734171 modulo 13 is 5. (Note that we did not use a calculator or computer or even pen and paper for any of these calculations.) Exercise: 20 Section 2.2 Question: Find the units digit of 78357 . Solution: To find the units digit of 78357 means to determine the remainder of 78357 when divided by 10. It is obvious that 78357 is even so it will suffice to determine the remainder 78357 when divided by 5. First, we make the simplification that 78357 ≡ 3357 (mod 5). Now the integer division of 357 by 4 gives 357 = 4 × 89 + 3. So by Fermat’s Little Theorem 3357 ≡ 34×89+3 ≡ (34 )8 933 ≡ 1 · 27 ≡ 2 (mod 5). So 78357 is even and has a remainder of 2 when divided by 5. The only integer between 0 and 9 to have these two properties is 2. Thus the units digit of 78357 is 2. Exercise: 21 Section 2.2 Question: Let {bn }n≥1 be the sequence of integers defined by b1 = 1, b2 = 11, b3 = 111, and in general n digits

z }| { bn = 111 · · · 1.

56

CHAPTER 2. NUMBER THEORY

Prove that for all prime numbers p different from 2 or 5, there exists a positive n such that p | bn . Solution: Solution 1: Notice that for all n ≥ 1, we have n digits

z }| { 9bn = 999 · · · 9 = 10n − 1. By Fermat’s Little Theorem, for all primes p such that p is not 2 or 5, 10p−1 ≡ 1

(mod p) =⇒ p | 10p−1 − 1.

Hence for all primes p 6= 2, 5, we have p | 9bn . If p 6= 3, then since p - 9, we have p | bn . If p = 3, it is easy to notice directly that b3 is divisible by 3, since b3 = 111 = 3 × 37. The result follows. Solution 2: Let p be a prime number distinct from 2 and 5. Let for all n ∈ N∗ , consider the set Bn defined as {1, 11, 111, . . . , bn }. By the pigeonhole (or box) principle, at least two elements in Bp+1 must be in the same congruence class in Z/pZ, say bk and b` with ` > k. Then b` − bk is divisible by p. However, b` − bk = b(`−k) × 10k+1 . Since p is a prime number and does not divide 10k+1 , then p | b(`−k) . The result follows. Exercise: 22 Section 2.2 Question: Show that 3 | n(n + 1)(n + 2) for all integers n. Solution: In three consecutive integers, there is always at least one that is divisible by 3. Hence, 3 must divide the product of three consecutive integers. Exercise: 23 Section 2.2 Question: Let p be a prime. Prove that p divides the binomial coefficient the binomial theorem to conclude that (a + b)p ≡ ap + bp

p k

for all k with 1 ≤ k ≤ p − 1. Use

(mod p)

for all integers a, b ∈ Z. Solution: The binomial formula gives (a + b)p =

p X k=0

p! ap−k bk . k!(p − k)!

Now if 1 ≤ k ≤ p−1, then k < p so no integer between 1 and k is divisible by the prime p. Hence, p - k!. Similarly p! p − k < p so p - (p − k)!. In the binomial coefficient kp = k!(p−k)! , there is a factor of p on the numerator but it p is not canceled out by the denominator, so p | k for all k with 1 ≤ k ≤ p − 1. Consequently, in modulo p, the binomial formula simplifies to (a + b)p ≡ ap + bp (mod p).

Exercise: 24 Section 2.2 Question: Let n1 and n2 be relatively prime. Prove that x ≡ a1

(mod n1 ) and x ≡ a2

(mod n2 ) ⇐⇒ x ≡ a1 t1 n2 + a2 t2 n1

(mod n1 n2 )

where t1 ≡ n−1 (mod n1 ) and t2 ≡ n−1 (mod n2 ). 2 1 Solution: Suppose that x ≡ a1 (mod n1 ) and x ≡ a2 (mod n2 ). Then x − a1 = k1 n1 for some integer k1 and x − a2 = k2 n2 for some integer k2 . Furthermore, with t1 ≡ n−1 (mod n1 ) and t2 ≡ n−1 (mod n2 ), we have 2 1 t1 n2 = 1 + `1 n1 and t2 n1 = 1 + `2 n2 for some integers `1 and `2 . Then x − (a1 t1 n2 + a2 t2 n1 ) = x − a1 (1 + `1 n1 ) − a2 t2 n1 = (k1 − a1 `1 − a2 t2 )n1 . Similarly, x − (a1 t1 n2 + a2 t2 n1 ) = (k2 − a2 `2 − a1 t1 )n2 .

2.2. MODULAR ARITHMETIC

57

Thus, x−(a1 t1 n2 +a2 t2 n1 ) is divisible by by n1 and n2 . Since n1 and n2 are relatively prime, x−(a1 t1 n2 +a2 t2 n1 ) is therefore divisible by n1 n2 . Hence, x ≡ a1 t1 n2 + a2 t2 n1

(mod n1 n2 ).

Conversely, suppose that x ≡ a1 t1 n2 +a2 t2 n1 (mod n1 n2 ) where t1 ≡ n−1 (mod n1 ) and t2 ≡ n−1 (mod n2 ). 2 1 Then x − (a1 t1 n2 + a2 t2 n1 ) = kn1 n2 for some integer k. Modulo n1 , this expression becomes x − a1 t1 n2 ≡ 0

(mod n1 ).

But since t1 ≡ n−1 (mod n1 ), we get x ≡ a1 (mod n1 ). The same result holds when considering the expression 2 modulo n2 . Exercise: 25 Section 2.2 Question: Apply the result of Exercise 2.2.24 to solve the system ( x ≡ 2 (mod 9) x ≡ 4 (mod 11). Solution: To solve the system means to find all x that solve the two congruences simultaneously. We first need to calculate t1 and t2 . Modulo n1 = 9, we have n2 ≡ 11 ≡ 2 and so t1 ≡ n−1 2 ≡ 5 (mod 9). Modulo n2 = 11, we have t2 ≡ 9−1 ≡ 5 (mod 11). Then according to Exercise 2.2.24, the solution to the system is x ≡ 2 · 5 · 11 + 4 · 5 · 9 ≡ 92

(mod 99).

Exercise: 26 Section 2.2 Question: Apply the result of Exercise 2.2.24 to solve the system ( x ≡ 15 (mod 17) x ≡ 10 (mod 169). Solution: To solve the system means to find all x that solve the two congruences simultaneously. We first need to calculate t1 and t2 . Modulo n1 = 17, we have n2 ≡ 169 ≡ 16 ≡ −1 (mod 17) and so t1 ≡ n−1 2 ≡ −1 (mod 17). Modulo n2 = 169, we have t2 ≡ 17−1 ≡ 10 (mod 169) since 17 × 10 = 169 + 1. Then according to Exercise 2.2.24, the solution to the system is x ≡ 15 · (−1) · 169 + 410 · 10 · 17 ≡ 67165 ≡ 1086

(mod 2873).

Exercise: 27 Section 2.2 Question: Prove that if ac ≡ bc (mod m) then a ≡ b (mod m d ) where d = gcd(m, c). Solution: Suppose that ac ≡ bc (mod m) and set d = gcd(m, c). Then there exists an integer k such that bc − ac = mk. We can now divide both sides by d to get b dc − a dc = m d k. Hence, c c m a ≡b (mod ). d d d Since d = gcd(m, c), then gcd(m/d, c/d) = 1. Consequently, c/d has a multiplicative inverse k modulo m/d. Multiplying the above equation by k, we get a ≡ b (mod m d ). Exercise: 28 Section 2.2 Question: Consider the sequence of integers {cn }n≥0 defined by c0 = 1,

c1 = 101,

c2 = 10101,

c3 = 1010101,

...

58

CHAPTER 2. NUMBER THEORY

Prove that for all integers n ≥ 2, the number cn is composite. Solution: We note first that c1 = 101 is a prime number. It is not hard to see that cn =

n X

100n = 100n + 100n−1 + · · · + 1002 + 100 + 1 =

i=0

100n+1 − 1 , 99

where the second equality holds by the formula for the finite geometric series. If n is odd with n ≥ 3, then the digit expansion of cn consists of an even number of 1s. Then ! k X c2k+1 = (100 + 1)1002k + · · · (100 + 1)1002 + (100 + 1)1000 = 101 × 10000i . i=0

Thus, when n is odd, 101 | cn so cn is composite. If n is even with n = 2k, then 1 1 1 (1002k+1 − 1) = (104k+2 − 1) = c2k = 99 99 11

4k+1 X

! i

10

.

i=0

There are an even number of elements in the sum on the far right so ! 2k 1 X i 10 (102k+1 + 1). c2k = 11 i=0 Note that 10 ≡ −1 (mod 11) so 102k+1 + 1 ≡ (−1)2k+1 + 1 ≡ 0 (mod 11). Hence 11 | 102k+1 + 1 and for k ≥ 1, P2k we also have (102k+1 + 1)/11 > 1. Furthermore, i=0 10i for k ≥ 1 so we have obtained a nontrivial factorization of c2k into ! 2k X 102k+1 + 1 i 10 . 11 i=0 Hence, cn is also composite for n even with n ≥ 2. The result follows.

2.3 – Mathematical Induction Exercise: 1 Section 2.3 Question: Prove that

n X i=1

i3 =

n2 (n + 1)2 for all integers n ≥ 1. 4 2

2

2

Solution: Basis Step: When n = 1, the summation is 13 = 1. If n = 1, then n (n+1) = 24 = 1. 4 n X n2 (n + 1)2 Induction Step: Suppose that i3 = for some integer n ≥ 1. Then 4 i=1 n+1 X

i3 =

i=1

n X

i3 + (n + 1)3

i=1 2

n (n + 1)2 + (n + 1)3 for the induction hypothesis 4 n2 (n + 1)2 + 4(n + 1)3 (n + 1)2 (n2 + 4(n + 1)) (n + 1)2 (n2 + 4n + 4)) = = = 4 4 4 (n + 1)2 (n + 2)2 ) . = 4 =

Thus, if the formula holds for n, it also holds for n + 1. By induction,

n X i=1

i3 =

n2 (n + 1)2 for all integers n ≥ 1. 4

2.3. MATHEMATICAL INDUCTION

59

Exercise: 2 Section 2.3 Question: Use mathematical induction to prove the geometric summation formula: n X

Ari =

i=0

A(rn+1 − 1) r−1

where r 6= 1

for all nonnegative integers n. n X

A(r − 1) . The formula holds for n = 0. r−1 i=0 Induction Step: Suppose that the formula holds for some nonnegative integer n. Then ! n+1 n X X i i Ar = Ar + Arn+1

Solution: Basis Step: If n = 0, then

i=0

Ari = A =

i=0

A(rn+1 − 1) + Arn+1 by the induction hypothesis = r−1 A(rn+2 − 1) Arn+1 − A + Arn+2 − Arn+1 = = r−1 r−1 A(r(n+1)+1 − 1) = . r−1 Hence, if the formula holds for n, then it holds for n + 1. By induction, if r 6= 1, then

n X

Ari =

i=0

A(rn+1 − 1) r−1

for all n ≥ 0. Exercise: 3 Section 2.3 Question: Prove that for every positive integer n, 1 · 2 + 2 · 3 + · · · + n(n + 1) =

n(n + 1)(n + 2) . 3

Solution: For the basis step, we note that if n = 1, the left-hand side of the formula is 1 · 2 = 2, whereas the right-hand side of the formula is 1 · 2 · 3/3 = 2. Hence, the formula holds for n = 1. Now suppose that the formula holds for some positive integer n. Then with the integer n + 1, using the induction hypothesis in the first line, we have 1 · 2 + 2 · 3 + · · · + n(n + 1) + (n + 1)(n + 2) n(n + 1)(n + 2) + (n + 1)(n + 2) 3 n(n + 1)(n + 2) + 3(n + 1)(n + 2) = 3 (n + 1)(n + 2)(n + 3) = 3 (n + 1)((n + 1) + 1)((n + 1) + 2) = . 3 =

Hence, if the formula holds for n, then it also holds for n+1. By induction, 1·2+2·3+· · ·+n(n+1) = n(n+1)(n+2) 3 for all positive integers n. Exercise: 4 Section 2.3 Question: Prove that 1 + nh ≤ (1 + h)n that for all h ≥ −1 and for nonnegative integers n ≥ 0. Solution: Note that the condition of h ≥ −1 is equivalent to h + 1 ≥ 0. When n = 0, then 1 + nh ≤ (1 + h)n reads 1 ≤ 1 so the inequality holds. Now, suppose that the inequality 1 + nh ≤ (1 + h)n holds for some n ≥ 0. Then (1 + h)n+1 = (1 + h)n (1 + h). Since 1 + h ≥ 0, using the induction hypothesis, we deduce that (1 + h)n+1 ≥ (1 + nh)(1 + h) = 1 + nh + h + nh2 ≥ 1 + (n + 1)h.

60

CHAPTER 2. NUMBER THEORY

Thus, if the inequality holds for some nonnegative n, then it also holds for n + 1. By induction, supposing h ≥ −1, then 1 + nh ≤ (1 + h)n for nonnegative integers n. Exercise: 5 Section 2.3 Question: Prove that 5|(n5 − n) for all nonnegative integers n in the following two ways: a) Using Fermat’s Little Theorem. b) By induction on n. Solution: We prove that 5|(n5 − n) for all nonnegative integers n. a) By Fermat’s Little Theorem, if 5 - n, then n4 ≡ 1 (mod 5), which means that 5 divides n4 − 1, which in turns implies that 5 divides n5 − n. If 5 | n, then n5 and n are both divisible by 5 and thus so is n5 − n. b) We note right away that if n = 0, then n5 − n = 0 so 5 divides n5 − n. Now suppose that 5 divides n5 − n for some nonnegative integer n with n5 − n = 5K. Then (n + 1)5 − (n + 1) = n5 + 5n4 + 10n3 + 10n2 + 5n + 1 − n − 1 = (n5 − n) + 5(n4 + 2n3 + 2n2 + n) = 5(K + n4 + 2n3 + 2n2 + n). Thus, 5 divides (n + 1)5 − (n + 1). By induction, 5 divides n5 − n for all nonnegative integers n. Exercise: 6 Section 2.3 Question: Prove by induction that

n X (2i + 1) = (n + 1)2 . i=0

Solution: Note that if n = 0, then

n X

(2i + 1) = 1, whereas (n + 1)2 = 12 = 1. So the formula holds for n = 0.

i=0

Now suppose that the formula holds for some nonnegative integer n. Then ! n+1 n X X (2i + 1) = (2i + 1) + (2(n + 1) + 1) = (n + 1)2 + 2(n + 1) + 1 = ((n + 1) + 1)2 . i=0

i=0

Thus, the formula holds also for n + 1. By induction,

n X

(2i + 1) = (n + 1)2 for all integers n ≥ 0.

i=0

Exercise: 7 Section 2.3 Question: Prove that fn−1 fn+1 − fn2 = (−1)n for all n ≥ 1. Solution: We prove by induction that fn−1 fn+1 − fn2 = (−1)n for all n ≥ 1. If n = 1, then f0 f2 − f12 = 0 − 12 = (−1)1 . So the formula holds for n = 1. Now suppose that the formula holds for some n ≥ 1. Then 2 2 fn fn+2 − fn+1 = fn (fn+1 + fn ) − fn+1

= fn2 + fn+1 (fn − fn+1 ) = fn2 + fn+1 (−fn−1 ) = −(−1)n = (−1)n+1

by the induction hypothesis.

Thus, if the formula holds for some n then it also holds for n + 1. By induction, fn−1 fn+1 − fn2 = (−1)n for all n ≥ 1. Exercise: 8 Section 2.3 2 Question: Prove that fn2 + fn−1 = f2n−1 for all n ≥ 1. Solution: For this exercise, we first make the following observation. fn = fn−1 + fn−2 = f3 fn−2 + f2 fn−3 = f4 fn−3 + f3 fn−4 .. . = fn = fk+1 fn−k + fk fn−k−1

2.3. MATHEMATICAL INDUCTION

61

For this exercise, it is useful to use f2n = fn+1 fn + fn fn−1 for all n. 2 2 We prove that fn2 + fn−1 = f2n−1 for all n ≥ 1 by induction. If n = 1, then fn2 + fn−1 = f12 + f02 = 1 = f1 = f2n−1 . Now suppose that the formula is true for some n ≥ 1. Then f2n+1 = f2n + f2n−1 2 = fn+1 fn + fn fn−1 + fn2 + fn−1

by the induction hypothesis

2 = fn+1 fn − fn fn−1 + fn2 + 2fn fn−1 + fn−1 = fn (fn+1 − fn−1 ) + (fn + fn−1 )2 2 = fn2 + fn+1 .

Hence, if the formula is true for some n, then it is also true for n + 1. By induction, the formula is true for all n ≥ 1. Exercise: 9 Section 2.3 Question: Prove that

n X

fi = fn+2 − 1.

i=0

Solution: We prove by induction that

n X

fi = fn+2 − 1 for all n ≥ 0. If n = 0, then

i=0

n X

fi = 0, which is equal

i=0

to 1 − 1 = f2 − 1. Thus, the formula holds for n = 0. Now suppose that the formula holds for some nonnegative n. Then ! n+1 n X X fi = fi + fn+1 i=0

i=0

= fn+2 − 1 + fn+1

by the induction hypothesis

= fn+3 − 1 = f(n+1)+2 − 1. Thus, the formula then holds for n + 1 as well. By induction, the formula holds for all nonnegative n. Exercise: 10 Section 2.3 Question: Prove 2 | fn if and only if 3 | n. Solution: We will first prove by induction that 2 | f3m for all nonnegative integers m. Let m = 0. Then f0 = 0 and 2 | f0 . Now suppose that 2 | f3m for some nonnegative integer m. Then f3m+3 = f3m+2 + f3m+1 = 2f3m+1 + f3m . Since 2 | f3m , then we see that 2 | f3m+3 . By induction, 2 | f3m for all n ≥ 0. Now we prove that 2 - f3m±1 for all m ≥ 0 (ignoring the case of f−1 implied in this notation). If m = 0 or m = 1, this takes into account f1 = 1, f2 = 1, and f4 = 3. None of these are even. Suppose that f3m±1 is odd for some m ≥ 1. Then, f3m+4 = f3m+3 + f3m+2 = 2f3m+2 + f3m+1 . Since we know that f3m+1 is odd, then f3(m+1)+1 is odd. Also, f3(m+1)−1 = f3m+2 = f3m+1 + f3m . Since we know that f3m is even from the previous part and f3m+1 is odd by the induction hypothesis, then we conclude that f3(m+1)−1 is also odd. The result is proved by induction. Comment: Though this problem is in a section on induction, a much fast proof of this same result consists in looking at the Fibonacci number modulo 2. We see that they make the pattern of 0, 1, 1, 0, 1, 1, . . .. The same result follows immediately. Exercise: 11 Section 2.3 n X Question: Prove that fi2 = fn fn+1 . i=0

62

Solution:

CHAPTER 2. NUMBER THEORY

We prove by induction that

n X

fi2 = fn fn+1 for all n ≥ 0. With n = 0, the formula claims that

i=0

f02 = 0 = f0 f1 . This is obviously true so the basis step holds. Now suppose that the formula is true for some nonnegative integer n. Then ! n+1 n X X 2 2 2 2 fi = fi + fn+1 = fn fn+1 + fn+1 = fn+1 (fn + fn+1 ) = fn+1 fn+2 . i=0

i=0

Thus, the formula holds for n + 1. By induction,

n X

fi2 = fn fn+1 for all n ≥ 0.

i=0

Exercise: 12 Section 2.3 Question: Prove that for all real numbers r 6= 1, n X

krk =

k=0

((r − 1)n − 1) rn+1 + r . (r − 1)2

Solution: We prove by induction on n with n ≥ 0, that n X k=0

krk =

((r − 1)n − 1) rn+1 + r (r − 1)2

n+1 Pn +r −r+r for all real numbers r 6= 1. If n = 0, then k=0 krk = 0 whereas ((r−1)n−1)r = (r−1) 2 = 0. Hence, the (r−1)2 formula is valid for n = 0. Now suppose that the formula holds for some n ≥ 0. Then ! n+1 n X X krk = krk + (n + 1)rn+1

k=0

k=0

((r − 1)n − 1) rn+1 + r + (n + 1)rn+1 by the induction hypothesis = (r − 1)2 ((r − 1)n − 1) rn+1 + r + (n + 1)rn+1 (r − 1)2 = (r − 1)2 nr − n − 1 + (n + 1)(r2 − 2r + 1) rn+1 + r = (r − 1)2 2 2 (nr − nr + r − 2r)rn+1 + r (nr − n + r − 2)rn+2 + r = = (r − 1)2 (r − 1)2 =

((r − 1)(n + 1) − 1)r(n+1)+1 + r . (r − 1)2

Hence, he formula then holds for n + 1. By induction, the formula holds for all n ≥ 0. Exercise: 13 Section 2.3 Question: Prove that 13 divides 3n+1 + 42n−1 for all n ≥ 1. Solution: We prove by induction that n ≥ 1, the number 3n+1 + 42n−1 is divisible by 13. For n = 1, we have 3n+1 + 42n−1 = 9 + 4 = 13, which is obviously divisible by 13. Now suppose that 13 divides 3n+1 + 42n−1 for some n ≥ 1. Then 3n+2 + 42(n+1)−1 = 3 · 3n+1 + 16 · 42n−1 = 3 · 3n+1 + (13 + 3) · 42n−1 = 3(3n+1 + 42n−1 ) + 13 · 42n−1 . By the induction hypothesis, 13 divides 3n+1 + 42n−1 so 3(n+1)+1 + 42(n+1)−1 is the sum of two multiples of 13 and hence is a multiple of 13. By induction, 3n+1 + 42n−1 is divisible by 13 for all n ≥ 1. Exercise: 14 Section 2.3 Question: A set of lines in the plane is said to be in general position if no two lines are parallel and no three

2.3. MATHEMATICAL INDUCTION

63

lines intersect at a single point. Prove that for any set {L1 , L2 , . . . , Ln } of lines in R2 in general position, the complement R2 − (L1 ∪ L2 ∪ · · · ∪ Ln ) consists of (n2 + n + 2)/2 disjoint regions in the plane. Solution: We prove the result by induction on n and we can start with n = 0. If n = 0, there are 0 lines and the complement R2 − (L1 ∪ L2 ∪ · · · ∪ Ln ) consists of a single disjoint region. Furthermore, if n = 0, then (n2 + n + 2)/2 = 2/2 = 1. So the formula holds for n = 0. Suppose that the for some n, formula is true for any set of n lines in general position. Now consider a set {L1 , L2 , . . . , Ln+1 } of n + 1 lines in general position. Then Ln+1 intersects each of the lines L1 , L2 , through Ln . The segments (possibly) rays on the line Li where Ln+1 intersects them form some of the boundary edges of n + 1 regions in R2 − (L1 ∪ L2 ∪ · · · ∪ Ln ). So Ln+1 cuts n + 1 regions in R2 − (L1 ∪ L2 ∪ · · · ∪ Ln ), adding n + 1 to the count. Hence, by induction, the complement R2 − (L1 ∪ L2 ∪ · · · ∪ Ln+1 ) consists of (n2 + n + 2)/2 + n + 1 disjoint regions in the plane. We have n2 + 2n + 1 + n + 1 + 2 (n + 1)2 + (n + 1) + 2 n2 + n + 2 + 2n + 2 = = . 2 2 2 Hence, the formula continues to hold for n + 1 lines. By induction, for all n ≥ 0, n lines in general position cut the plane into (n2 + n + 2)/2 regions. (n2 + n + 2)/2 + n + 1 =

Exercise: 15 Section 2.3 1 1 1 Question: Let Hn = 1 + + + · · · + be the nth harmonic number. Prove that H2n ≤ 1 + n. 2 3 n Solution: We prove by induction that H2n ≤ 1 + n for all n ≥ 0. For the basis step with n = 0, we notice that H1 = 1 and H1 ≤ 1 + 0. Hence, the inequality holds for n = 0. Suppose that the inequality holds for some n ≥ 0. Then 1 1 1 + + · · · . H2n+1 = H2n + 2n + 1 2n + 2 2n+1 Each of the fractions in the parentheses, is less than or equal to 21n and there are 2n of those fractions. Hence, 1 H2n+1 ≤ n + 1 + 2n n = (n + 1) + 1. 2 Hence, the inequality holds for n + 1 as well. By induction, H2n ≤ 1 + n for all n ≥ 0. Exercise: 16 Section 2.3 Question: Prove that n! ≤ nn−1 for all positive integers n. Solution: Let n = 1. Then 1! = 1 and 11−1 = 10 = 1. Hence, if n = 1, we have n! ≤ nn−1 . Now suppose that n! ≤ nn−1 for some positive n. Then, using the induction hypothesis, (n + 1)! = n!(n + 1) ≤ nn−1 (n + 1) Since n ≥ 1, then n ≤ n + 1 so nn−1 ≤ (n + 1)n−1 . We now use this with the above inequality to establish that (n + 1)! ≤ nn−1 (n + 1) ≤ (n + 1)n−1 (n + 1) = (n + 1)n . Thus, the inequality also holds for n + 1. By induction, n! ≤ nn−1 for all positive integers n. Exercise: 17 Section 2.3 n X Question: Show that i(i!) = (n + 1)! − 1 i=1

Solution: Basis Step: If n = 1, then

n X

i(i!) = 1 · 1! = 1 and also (1 + 1)! − 1 =. Hence, the formula holds for

i=1

n = 1. Induction Step: Suppose that the formula holds for some positive integer n. Then ! n+1 n X X i(i!) = i(i!) + (n + 1)(n + 1)! i=1

i=1

= (n + 1)! − 1 + (n + 1)(n + 1)! = (1 + n + 1)(n + 1)! − 1 = (n + 2)! − 1 = ((n + 1) + 1)! − 1.

64

CHAPTER 2. NUMBER THEORY

So the formula then also holds for n + 1. By induction,

n X

i(i!) = (n + 1)! − 1 for all n ≥ 1.

i=1

Exercise: 18 Section 2.3 Question: Show that any amount of postage of value 48 cents or higher can be formed using just 5-cent and 12-cent stamps. Solution: The question can be rephrased to say that for all integers n ≥ 48, there exist s, t ∈ N such that n = 5s + 12t. We prove this using strong induction. We can list a first few possibilities n 48 49 50 51 52 53

= 5s + 12t = 5 × 0 + 12 × 4 = 5 × 5 + 12 × 2 = 5 × 10 + 12 × 0 = 5 × 3 + 12 × 3 = 5 × 8 + 12 × 1 = 5 × 1 + 12 × 4

We see that there are solutions up to n = 54. (Strong Induction hypothesis) Suppose that n ≥ 53 and that for all 48 ≤ k ≤ n, there exist s, t ∈ N such that k = 5s + 12t. Consider now n + 1. Then by the induction hypothesis, there exists s0 , t0 ∈ N such that n + 1 − 5 = 5s0 + 12t0 . Then n + 1 = 5(s0 + 1) + 12t0 . Consequently, by strong induction, for all n ≥ 48, there exist s, t ∈ N such that n = 5s + 12t. Exercise: 19 Section 2.3 Question: Let A1 , A2 , . . . , An and B be sets. Use mathematical induction to prove that (A1 − B) ∩ (A2 − B) ∩ · · · (An − B) = (A1 ∩ A2 ∩ · · · ∩ An ) − B.

Solution: Let A1 , A2 , . . . , An and B be sets. Using induction on n, we prove that for all sets A1 , A2 , . . . , An and set B, (A1 − B) ∩ (A2 − B) ∩ · · · (An − B) = (A1 ∩ A2 ∩ · · · ∩ An ) − B. If n = 1 then A1 − B = A1 − B. Now suppose that for a given positive integer n, the formula (A1 − B) ∩ (A2 − B) ∩ · · · (An − B) = (A1 ∩ A2 ∩ · · · ∩ An ) − B is true for all sets A1 , A2 , . . . , An and B. Consider a collection A1 , A2 , . . . , An+1 of sets. Then (A1 − B) ∩ (A2 − B) ∩ · · · (An+1 − B) = ((A1 − B) ∩ (A2 − B) ∩ · · · (An − B)) ∩ (An+1 − B) = ((A1 ∩ A2 ∩ · · · ∩ An ) − B) ∩ (An+1 − B) = ((A1 ∩ A2 ∩ · · · ∩ An ) ∩ B) ∩ (An+1 ∩ B) = (A1 ∩ A2 ∩ · · · ∩ An+1 ) ∩ (B ∩ B) = (A1 ∩ A2 ∩ · · · ∩ An+1 ) ∩ B = (A1 ∩ A2 ∩ · · · ∩ An+1 ) − B. Hence, by induction, for all n ≥ 1 and all sets A1 , A2 , . . . , An and set B, we have (A1 − B) ∩ (A2 − B) ∩ · · · (An − B) = (A1 ∩ A2 ∩ · · · ∩ An ) − B.

Exercise: 20 Section 2.3 Question: Let α be any real number such that α + α1 ∈ Z. Prove that for all nonnegative integers n, αn +

1 ∈ Z. αn

2.3. MATHEMATICAL INDUCTION

65

Solution: Suppose that α ∈ R such that α + α1 ∈ Z. Obviously, for n = 0 or n = 1, we have αn + Now let n ≥ 2 suppose that αk +

1 . αn

1 ∈ Z for all k with 0 ≤ k < n. Then αk n X n 1 n n−2j α+ = α . α j j=0

By grouping together the jth element from the first and jth element from the last in the above binomial expansion, we have n (Pbn/2c n n−2j 1 α + αn−2j if n is odd 1 j=0 j Pn/2−1 n n−2j = α+ n 1 α α + αn−2j if n is even. j=0 n/2 + j Solving for αn + α1n we get 1 α + n = α n

1 α+ α

n −

(Pbn/2c n 1 αn−2j + αn−2j j=1 j P n/2−1 n n 1 αn−2j + αn−2j j=1 n/2 + j

if n is odd if n is even

! .

By the induction hypothesis, all terms in the summations on the right-hand side are integers. Hence, αn + α1n ∈ Z.

66

CHAPTER 2. NUMBER THEORY

3 | Groups 3.1 – Symmetries of the Regular n-gon Exercise: 1 Section 3.1 Question: Use diagrams to describe all the dihedral symmetries of the equilateral triangle. Solution: The equilateral triangle has 6 dihedral symmetries.

identity

rotation 120◦

rotation 240◦

reflection through x-axis

reflection

reflection

Exercise: 2 Section 3.1 Question: Write down the composition table for D4 . Solution: Composition table for D4 where the entries give a ◦ b. a\b 1 r r2 r3 s sr sr2 sr3

1 1 r r2 r3 s sr sr2 sr3

r r r2 r3 1 sr sr2 sr3 s

r2 r2 r3 1 r sr2 sr3 s sr

r3 r3 1 r r2 sr3 s sr sr2

s s sr3 sr2 sr 1 r3 r2 r

sr sr s sr3 sr2 r 1 r3 r2

sr2 sr2 sr s sr3 r2 r 1 r3

sr3 sr3 sr2 sr s r3 r2 r 1

Exercise: 3 Section 3.1 Question: Determine what r3 sr4 sr corresponds to in dihedral symmetry of D8 . Solution: In dihedral symmetry of D8 , we have the following algebraic identities on r and s: r8 = 1,

s2 = 1,

rk s = sr−k .

So for our element, progressively change it to put all the s terms to the left: r3 sr4 sr = r3 s(r4 s)r = r3 s2 r−4 r = r3 1r−3 = 1.

Exercise: 4 Section 3.1 Question: Determine what sr6 sr5 srs corresponds to as a dihedral symmetry of D9 . 67

(3.1)

68

CHAPTER 3. GROUPS

Solution: Recall from Corollary 3.5 that srk = rn−k s where in our case n = 9. So, sr6 sr5 srs = sr6 sr5 ssr8 = ssr3 r5 (1)r8 = (1)r8 r8 = r9 r7 = 1r7 = r7 .

Exercise: 5 Section 3.1 Question: Let n be an even integer with n ≥ 4. Prove that in Dn , the element rn/2 satisfies rn/2 w = wrn/2 for all w ∈ Dn . Solution: From the paragraph above Proposition 3.1.4 we can write w ∈ Dn as w = sa rb where a is either 0 or 1. Consider rn/2 sa rb . We have two cases. Case 1: a = 0 So we have rn/2 rb = rn/2+b = rb+n/2 = rb rn/2 . Case 2: a = 1 Now, rn/2 srb = srn−n/2 rb = srn/2 rb = srn/2+b = srb+n/2 = srb rn/2 . In both cases, we see that rn/2 w = wrn/2 . Exercise: 6 Section 3.1 Question: Let n be an arbitrary integer n ≥ 3. Show that an expression of the form ra sb rc sd · · · is a rotation if and only if the sum of the powers on s is even. Solution: For any numbers l and m we have rl sm = sm rl−m . So we can move all powers of s around without changing the exponent’s value. Since we can rewrite any element as sj rk , we have ra sb rc sd · · · = sb+d+··· rm for some m. Now, if b+d+· · · is an even number then sb+d+··· rm = s2 s2 · · · s2 rm = (1)(1) · · · (1)rm = 1rm = rm and our element is a rotation. If b+d+· · · is an odd number then sb+d+··· rm = s1 s2 · · · s2 rm = s(1)(1) · · · (1)rm = srm and our elements is not a rotation. Exercise: 7 Section 3.1 Question: Use linear algebra to prove that Rα ◦ Fβ = Fα/2+β ,

Fα ◦ Rβ = Fα−β/2

, and Fα ◦ Fβ = R2(α−β) .

Solution: As linear transformations on R2 → R2 , the matrices of the rotation Rα and of the reflection Fβ with respect to the standard basis are respectively cos α − sin α cos 2β sin 2β and . sin α cos α sin 2β − cos 2β The matrix for Rα ◦ Fβ is cos α − sin α cos 2β sin α cos α sin 2β

sin 2β − cos 2β

cos α cos 2β − sin α sin 2β cos α sin 2β + sin α cos 2β sin α cos 2β + cos α sin 2β sin α sin 2β − cos α cos 2β cos(α + 2β) sin(α + 2β) = . sin(α + 2β) − cos(α + 2β)

=

This matrix corresponds to the reflection Fα/2+β .

3.1. SYMMETRIES OF THE REGULAR N -GON The matrix for Fα ◦ Rβ is cos 2α sin 2α cos β sin 2α − cos 2α sin β

− sin β cos β

69

cos 2α cos β + sin 2α sin β − cos 2α sin β + sin 2α cos β = sin 2α cos β − cos 2α sin β − sin 2α sin β − cos 2α cos β cos(2α − β) sin(2α − β) = sin(2α − β) − cos(2α − β)

This matrix corresponds to the reflection Fα−β/2 . The matrix for Fα ◦ Fβ is cos 2α sin 2α cos 2β sin 2β cos 2α cos 2β + sin 2α sin 2β cos 2α sin 2β − sin 2α cos 2β = sin 2α − cos 2α sin 2β − cos 2β sin 2α cos 2β − cos 2α sin 2β sin 2α sin 2β + cos 2α cos 2β cos(2α − 2β) − sin(2α − 2β) = sin(2α − 2β) cos(2α − 2β) This matrix corresponds to the reflection R2(α−β) . Exercise: 8 Section 3.1 Question: Describe the symmetries of an ellipse with unequal half-axes. Solution: The ellipse with unequal half-axes has 4 symmetries. Supposing that the axes of the ellipse are on the x and y axes, then the ellipse has for symmetries: the identity, reflection through the x axis, reflection through the y axis, and rotation by 180◦ , which is the composition of the two reflections. Exercise: 9 Section 3.1 Question: List all the symmetries of the circle and describe the compositions between them. Solution: Assume that a circle has the origin as its center. The (dihedral) symmetries of the circle consist of all rotation Rα , where α ∈ R, and reflections Fβ , through a line through the origin that makes and angle β ∈ R with respect to the x-axis. he rotations are uniquely defined if α ∈ [0, 2π) and the reflections are uniquely defined if β ∈ [0, π). The compositions between these symmetries are Rα ◦ Rβ = Rα+β along with the other three compositions described in Exercise 3.1.7. Exercise: 10 Section 3.1 Question: List all the symmetries and describe the compositions between them for the infinitely long sine curve shown below: ...

...

Solution: It is useful to sketch the axes and the vertical line x = −π/2 and x = π/2. b ...

a ...

We see the the sine curve is symmetric with respect to reflection through the line marked a. We denote the reflection by a as well. The sine curve is also symmetric with respect to reflection through b. The sine function is also invariant under reflection through any line x = π2 + kπ. The function on R2 is fk (x, y) = ((2k + 1)π − x, y), where k can be an integer. The sine curve also stays invariant under a rotation by an angle π through the origin. (The sine function is odd.) The sine curve is also invariant under rotation by π around any point of (kπ, 0). The function on R2 is rk (x, y) = (2kπ − x, −y). The sine function is also periodic to it is invariant under a translation by 2π. The function for this is tk (x, y) = (x + 2kπ, y).

70

CHAPTER 3. GROUPS

The sine function also is invariant under a flip-translation described by the function on R2 given by ak = (x, y) = (x + (2k + 1)π, −y). There is a briefer way to list all the possible symmetries of the sine curve, but we will use this intuitive ones. We can summarize all compositions by the following chart. ◦ fm rm tm am

fn tm−n am−n−1 fm+n rm+n+1

rn am−n tm−n rm+n fm+n

tn fm−n rm−n tm+n am+n

an rm−n fm−n−1 am+n tm+n

Exercise: 11 Section 3.1 Question: List all the symmetries and describe the compositions between them for the infinitely long pattern shown below: ...

...

Solution: In the diagram, consider the point O and the vector ~v . . . . ~v O

...

This figure has the following symmetries, with k any integer: • reflection Fx through the x-axis; • translations Tk by any integer multiple k of the vector ~v ; • glide reflections Gk defined by Gk = Tk ◦ Fx ; −→ • rotations Rk/2 by 180◦ about any point A located on the x-axis at position OA = 21 k~v from O; −→ • reflections Fk/2 through a vertical line through any point A located on the x-axis at position OA = 12 k~v from O. (It is convenient to use vector notation to locate points because of the composition calculations that come below.) The compositions between these are given by the table the shows f ◦ g with f along the rows and g along the columns. (Note that I = T~0 is the identity.) Fx Tk Gk Rk/2 Fk/2

Fx I Gk Tk Fk/2 Rk/2

Tr Gk

Gr Tk

Rr/2 Fr/2

Fr/2 Rr/2

T(k+r) G(k+r) R(k−r)/2 F(k−r)/2

G(k+r) T(k+r) F(k−r)/2 R(k−r)/2

R(k+r)/2 F(k+r)/2 Tk−r Gk−r

F(k+r)/2 R(k+r)/2 Gk−r Tk−r

Exercise: 12 Section 3.1 Question: List all the symmetries and describe the compositions between them for the infinitely long pattern shown below: ...

Solution: We draw an x-axis through the pattern and fix an origin.

...

3.1. SYMMETRIES OF THE REGULAR N -GON

71

... O

...

This pattern is periodic and it has only two types of symmetries • translation Tk corresponding to a translation by k~v , where k ∈ Z and ~v is the displacement of unit periodicity of the shape; • rotations Rk/2 which are rotations by 180◦ about any point A located on the x-axis a displacement of k2 ~v from the origin. The composition of these types of symmetries is given by the following chart that shows f ◦ g, where f is listed in the row and g in the column.

Tk Rk/2

Tr Tr+k R(k−r)/2

Rr/2 R(r+k)/2 Tk−r

Exercise: 13 Section 3.1 Question: Determine the set of symmetries for each of the following shapes (ignoring shading):

(c)

(a)

(b)

(d)

(e)

(f)

Solution: a) This shape has square rotational symmetry. b) This shape has triangular dihedral symmetry, D3 . c) This flower shape has dodecahedral dihedral symmetry, D12 . d) This shape has octagonal rotational symmetry. e) This shape has pentagonal dihedral symmetry, D5 . f) This shape has 180◦ degree rotational symmetry.

Exercise: 14 Section 3.1 Question: Sketch a pattern/shape (possibly a commonly known logo) that has D8 symmetry but does not have Dn symmetry for n > 8. Solution: Here is an example of D8 symmetry:

72

CHAPTER 3. GROUPS

Exercise: 15 Section 3.1 Question: Sketch a pattern/shape (possibly a commonly known logo) that has rotational symmetry of angle π 2 but does not have full D4 symmetry. Solution: The logo for Chase Bank has the desired symmetry.

Exercise: 16 Section 3.1 Question: Consider a regular tetrahedron. We call a rigid motion of the tetrahedron any rotation or composition of rotations in R3 that map a regular tetrahedron back into itself, though possibly changing specific vertices, edges, and faces. Rigid motions of solids do not include reflections through a plane. a) Prove that there are exactly 12 rigid motions of the tetrahedron. Call this set R. b) Using a labeling of the tetrahedron, explicitly list all rigid motions of the tetrahedron. c) Explain why function composition ◦ is a binary operation on R. d) Write down the composition table of ◦ on R. Solution: a) Every rigid motion of a tetrahedron will take a vertex and map it into another vertex. There are 4 possibilities for correspondences. Furthermore, for every function that maps a vertex A to a vertex B, there are three ways to rotate the tetrahedron around the axis OB, where O is the center of the tetrahedron. Hence, in total there are 4 × 3 = 12 possible rigid motions of a tetrahedron. b) Let us label the vertices of the tetrahedron as 1, 2, 3, 4. The rigid motions of the tetrahedron are described by functions f : {1, 2, 3, 4} → {1, 2, 3, 4}. In fact, these functions will be bijections. We can encode such functions as 3241 to mean that f (1) = 3, f (2) = 2, f (3) = 4, and f (4) = 1. In this manner of encoding the rigid motions, a complete list is R ={1234, 1342, 1423, 3241, 4213, 2431,

4132, 2314, 3124, 2143, 3412, 4321}.

c) The definition of a rigid motion allows for a composition of two rigid motions to again be a rigid motion. Any face will be mapped to any other face the through composition of to rigid motions.

3.1. SYMMETRIES OF THE REGULAR N -GON

73

d) The composition table of this group has f ◦ g with g along the columns and f along the rows. 1234 1342 1423 3241 4213 2431 4132 2314 3124 2143 3412 4321

1234 1234 1342 1423 3241 4213 2431 4132 2314 3124 2143 3412 4321

1342 1342 1423 1234 3412 4132 2314 4321 2143 3241 2431 3124 4213

1423 1423 1234 1342 3124 4321 2143 4213 2431 3412 2314 3241 4132

3241 3241 4321 2431 4213 1234 3412 3124 1342 2143 4132 1423 2314

4213 4213 2314 3412 1234 3241 1423 2143 4321 4132 3124 2431 1342

2431 2431 3241 4321 2143 2314 4132 1234 3412 1423 1342 4213 3124

4132 4132 2143 3124 1342 3412 1234 2431 4213 4321 3241 2314 1423

2314 2314 3412 4213 2431 2143 4321 1342 3124 1234 1423 4132 3241

3124 3124 4132 2143 4321 1423 3214 3412 1234 2314 4213 1342 2431

2143 2143 3124 4132 2314 2431 4213 1423 3241 1342 1234 4321 3412

3412 3412 4213 2314 4132 1342 3124 3241 1423 2431 4321 1234 2143

4321 4321 2431 3241 1423 3124 1342 2314 4132 4213 3412 2143 1234

Exercise: 17 Section 3.1 Question: Consider the hexagonal tiling pattern on the plane drawn below.

P

O

Define r as the rotation by 60◦ about O and t as the translation that moves the whole plane one hexagon to the right. We denote by ◦ the operation of composition on functions R2 → R2 and we denote by r−1 and t−1 the inverse functions to r and t. 1. Show that r ◦ t is not equal to t ◦ r. 2. Show that r ◦ t ◦ r ◦ t−1 ◦ r−1 has the effect of rotating the plane by 60◦ about the point P . 3. Prove or disprove that there exists a composition of functions involving only r, t, and their inverses that −−→ has the effect of translating the whole plane in the direction OP ? If there is, give it. Solution: a) r ◦ t has the effect of sending the point O to the point P . On the other hand t ◦ r sends the point O to the center of the hexagon immediately to the right of the hexagon containing O. (We call this point Q since it is useful later.) b) The function composition r ◦ t ◦ r ◦ t−1 ◦ r−1 maps P to itself. Furthermore, if we follow how the function acts on the hexagon around P , we see that it must map the hexagon around P back into the hexagon around P . If we call A the vertex in the hexagon directly above P and follow how it changes through r ◦ t ◦ r ◦ t−1 ◦ r−1 , we see that it transforms to the vertex of the hexagon around P rotated by 60◦ . Hence, r ◦ t ◦ r ◦ t−1 ◦ r−1 applied to the whole plane has the effect of rotating the plane by 60◦ about P . c) Consider the function r ◦ t ◦ r−1 maps O to O then to Q and then to P . By visually following the diagram, we can see that it also sends the vertical edge to the right of O to the vertical edge tot he right of P . Hence, −−→ the hexagon around O is simply translated by OP to the hexagon around P . Hence, the overall effect of −−→ r ◦ t ◦ r−1 on the whole plane is a translation by OP . Exercise: 18 Section 3.1 Question: Consider the diagram S 0 below. Sketch the diagram S that has S 0 as a fundamental region with a) D4 symmetry; b) only rotational square symmetry. [Assume reflection through the x-axis is one of the reflections.]

74

CHAPTER 3. GROUPS

S0

Solution: a) The fundamental region acted on by D4 symmetry:

b) The fundamental region acted on by rotational symmetry only:

Exercise: 19 Section 3.1 Question: Consider the diagram S 0 below. Sketch the diagram S that has S 0 as a fundamental region with (a) D6 symmetry; (b) D3 symmetry. [Assume reflection through the x-axis is one of the reflections.]

S0

Solution: a) With D6 symmetry.

3.2. INTRODUCTION TO GROUPS

75

b) With D3 symmetry.

3.2 – Introduction to Groups Exercise: 1 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (N, +). Solution: First we check whether + is a binary operation. For any a, b ∈ N, a + b is also in N. So + is a binary operation on N. We will check the group axioms. Associativity Let a, b, c ∈ (N, +). Then (a + b) + c = a + (b + c) since addition is associative in the natural numbers. So associativity holds. Identity For any a ∈ (N, +), 0 + a = a + 0 = a. So 0 is the identity in our group. Inverses For any nonzero a ∈ (N, +), there is no b ∈ (N, +) where a + b = 0 since the natural numbers do not include the negative numbers.

76

CHAPTER 3. GROUPS

So we do not have inverses in (N, +) and it is not a group. Exercise: 2 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (Q − {−1}, × +), where × + is defined by a+ ×b = a + b + ab. Solution: We first check that × + is indeed a binary operation. Certainly for any a, b ∈ (Q − {−1}, × +), a+ ×b ∈ Q, but we must check that a+ ×b 6= −1. Suppose a+ ×b = −1 ⇒ a+b+ab = −1 ⇒ a+b(1+a) = −1 ⇒ 1+a+b(1+a) = 0 ⇒ (1 + a)(1 + b) = 0. Which can only happen if either a = −1 or b = −1, however, -1 is not in our set so this can never occur. So × + is a binary operation. Next we check the group axioms. Associativity Let a, b, c ∈ (Q − {−1}, × +) and consider (a+ ×b)+ ×c = (a + b + ab)+ ×c = (a + b + ab) + c + (a + b + ab)c = a + b + ab + c + ac + bc + abc = a + b + c + bc + ab + ac + abc = a + (b + c + bc) + a(b + c + bc) = a+ ×(b + c + bc) = a+ ×(b+ ×c). So associativity holds. Identity Let a ∈ (Q − {−1}, × +), we will solve for e where a+ ×e = e+ ×a = a (Note that × + is commutative so we need only solve for one of those equations). So a + e + ae = a ⇒ e + ae = 0 ⇒ e(1 + a) = 0. So either e is 0 or a is −1, however −1 is not in our set so e = 0 holds as an identity for every element. Inverses Let a ∈ (Q − {−1}, × +), we will solve for b where a+ ×b = 0. So a + b + ab = 0 ⇒ a + b(a + 1) = 0 ⇒ −a b(a + 1) = −a ⇒ b = a+1 . Now since we have something in our denominator we need to worry about whether or not it could equal 0. This only could happen if a = −1 which is not in our set. The other property of inverses is that the inverse of b should be a. So if we use our previous formula for the inverse −b of b, c = b+1 we hope that when we plug in our formula for b in terms of a we will find that c = a. So −(−a)/(a+1) a/(a+1) a/(a+1) −b a b+1 = (−a/(a+1))+1 = (−a+a+1)/(a+1) = 1/(a+1) = 1 = a. So our inverse formula works in both directions. −a So for a ∈ (Q − {−1}, × +), a−1 = a+1 .

Since all of the groups axioms hold and × + is a binary operation on our set, the pair (Q − {−1}, × +) is a group. Exercise: 3 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (Q, ÷), with a ÷ b = ab . Solution: ÷ is not a binary operation since, for any nonzero a ∈ Q, a ÷ 0 does not exist in Q. Since we do not have a binary operation, there is no need to check the rest of the group axioms and (Q, ÷) is not a group. Exercise: 4 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (A, +), where A = {x ∈ Q | |x| < 1}. Solution: We first check that + is a binary operation on A. Consider .5 ∈ A. .5 + .5 = 1, however 1 ∈ / A since it doesnt meet the conditions. This shows that + is not a binary operation. So we do not need to check the group axioms and (A, +) is not a group. Exercise: 5 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (Z ⊕ Z, ∗), where (a, b) ∗ (c, d) = (ad + bc, bd). Solution: We check the group axioms. Associativity Let (a, b), (c, d), (e, f ) be three elements in Z ⊕ Z. Then ((a, b) ∗ (c, d)) ∗ (e, f ) = (ad + bc, bd) ∗ (e, f ) = (adf + bcf + bde, bdf ) and (a, b) ∗ ((c, d) ∗ (e, f )) = (a, b) ∗ (cf + de, df ) = (adf + bcf + bde, bdf ). Associativity holds.

3.2. INTRODUCTION TO GROUPS

77

Identity The operation has an identity (0, 1) since (a, b) ∗ (0, 1) = (0, 1) ∗ (a, b) = (a, b). Inverses Not all elements have inverses with this binary operation. For example, with the element (2, 3) and any other element (a, b), we have (2, 3) ∗ (a, b) = (2b + 3a, 3b). There is no value of b such that 3b = 1. Hence, the pair (Z ⊕ Z, ∗) is not a group. Exercise: 6 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair ([0, 1), ), where x y = x + y − bx + yc. Solution: We first check whether is a binary operation or not. Prior to this we mention a property of taking the floor of a sum. You can pull out any integers in the sum of a floor. For instance b4 + xc = 4 + bxc. Now we will show that is a binary operation by proving that 0 ≤ a b < 1. For any number c, consider the expression bc − bccc. By the previously mentioned property of floors, we can pull out the −bcc since it is an integer. So this is equal to bc + 0c + (−bcc) = bcc − bcc = 0. Now, the expression c − bcc represents a b by replacing c with (a + b). So that bc − bccc = b(a + b) − b(a + b)cc = b(a b)c = 0. This implies that 0 ≤ a b < 1 or a b ∈ [0, 1). So is a binary operation on [0, 1). Now, we check the group axioms. Associativity Let a, b, c ∈ [0, 1) and consider a (b c) = a (b+c−bb+cc) = a+b+c−bb+cc−ba+b+c−bb+ccc. By the previously mentioned properties of floors we can pull out the bb + cc from the other floor. So a + b + c − bb + cc − (−bb + cc) − ba + b + cc = a + b + c − ba + b + cc = a + b + c − ba + bc − (−ba + bc) − ba + b + cc = a + b − ba + bc + c − ba + b − ba + bc + cc = (a b) + c − b(a b) + cc = (a b) c. This shows that is associative on [0, 1). Identity Since 0 ∈ [0, 1), for any a ∈ [0, 1) consider a 0 = a + 0 − ba + 0c = a − bac = a − 0 = a. Likewise, 0 a = 0. So the Identity axiom holds. Inverses For any a ∈ [0, 1), 1 − a ∈ [0, 1). Consider a (1 − a) = a + (1 − a) − ba + (1 − a)c = 1 − b1c = 1 − 1 = 0. Furthermore, since is a commutative operation, we have a (1 − a) = (1 − a) a = 0. So the Inverse axiom holds. Since the group axioms hold and is a binary operation on [0, 1), the pair ([0, 1), ) is a group. Exercise: 7 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (A, +), where A is the set of rational numbers that when reduced have a denominator of 1 or 3. Solution: First we check if + is a binary operation on A. For any two elements a, b ∈ A, these elements can be represented by c/3 and d/3. If it’s denominator is already 3 this is trivial, if the denominator is 1, multiply the top and bottom by three. Then consider c/3 + d/3 = (c + d)/3. Now, if (c + d)/3 cannot be reduced, it must exist in A since it has a denominator of 3. If it can be reduced, then the only number the denominator can reduce to is 1 and then the element will still exist in A. This shows that + is a binary operation on A. Now we check the group axioms. Associativity Since addition is associative in Q, it is also associative in the subset A of Q. Identity Since 0/1 ∈ A, for any a ∈ A, 0/1 + a = a + 0/1 = a + 0 = a. So A has an identity. Inverses For any a ∈ A, −a also exists in A since it will have the same valued denominator. Then a + (−a) = a − a = 0 and likewise −a + a = 0. So the Inverse axiom holds.

78

CHAPTER 3. GROUPS

Since all the axioms hold and + is a binary operation on A, This proves that the pair (A, +) is a group. Exercise: 8 Section 3.2 Question: Decide whether the given set and the operation√pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (A, +), where A = {a + b 5 | a, b ∈ Q}. √ √ Solution: We first check that + is a binary operation over A. Consider any two elements (a + √ b 5), (c + d√ 5) ∈ A. We know that√the sums of√any two rational numbers is another rational number. Then, (a+b 5)+(c+d 5) = (a + c) + (b + d) 5 = e + f 5 for some e, f ∈ Q, which implies that the sum exists in A. Then + is a binary operation over A. Now, we check the group axioms. Associativity Since addition is associative in R and A is a subset of R, addition is also associative in A. √ √ √ √ √ consider the expression, √ (0 + 0 5) + (a + b 5) = (a + b 5) + (0 + 0 5) = Identity For any a +√b 5 ∈ A √ (a + 0) + (b + 0) 5 = a + b 5. So the element 0 + 0 5 ∈ A acts as the identity. √ √ √ √ 5) = (a− Inverses For any √ a+b 5 ∈√A, we know that −a+(−b) 5 ∈ A as well. Consider (a+b √5)+(−a+(−b) √ √ a)+(b−b) 5 = 0+0 5. Since addition is commutative we also have (−a+(−b) 5)+(a+b 5) = 0+0 5. So the Inverse axiom holds. So all the axioms hold and + is a binary operation on A. This proves that the pair (A, +) is a group. Exercise: 9 Section 3.2 Question: Decide whether the given set and the operation√pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (A, ×), where A = {a + b 5 | a, b ∈ Q}. √ √ Solution: First we √ √ will check whether × is a binary √ operation √ on A. For any two elements a+b 5, c+d 5 ∈ A, (a + b 5) × (c + d 5) = (ac + 5bd) + (ad + bc) 5 = e + f 5 for some e, f ∈ Q. So × is a binary operation on A. Now, we check the group axioms. Associativity Since × is associative over R and A is a subset of R, × is also associative over A. √ √ √ √ √ √ Identity For √ any element a + b 5 ∈√A, consider (1 + 0 5) × (a + b 5) = (a + b 5) × (1 + 0 5) = a1 + b1 5 = a + b 5. So the element 1 + 0 5 ∈ A functions as the identity. √ Inverses Consider the element 0 + 0 5 ∈ A. This is an example of an element that will not have an inverse. So the axiom does not hold. We have seen that the inverse axiom does not hold for the pair (A, ×). So it is not a group. Exercise: 10 Section 3.2 √ Question: The pair (A, ×), where A = {a + b 5 | a, b ∈ Q and (a, b) 6= (0, 0)}.

√ √ Solution: First we will check whether √ × is a binary √ operation on A. For any two √ elements√a+b 5, c+d 5 ∈ A, such that (a,b),(c,d) 6= (0,0), (a + b 5) × (c + d 5) = (ac + 5bd) + (ad + bc) 5 = e + f 5 for some e, f ∈ Q. Now √ we must check that (e, f ) 6= (0, 0). Suppose (e, f ) = (0, 0). Then ac = -5bd and ad = bc, implying a = b −5. Then a is not an element of Q. Contradiction. Thus (e, f ) 6= (0, 0). So × is a binary operation on A. Now, we check the group axioms. √ √ √ √ √ √ Associativity Let a + b 5, c + d 5,√ e + f 5 ∈ A. × (c + d 5))√× (e + f 5) = ace + 5adf + 5bcf + √ ((a + b 5) √ 5bde + (acf + ade + bce + 5bdf ) 5 = (a + b 5) × ((c + d 5) × (e + f 5)). So × is associative over A. √ √ √ √ √ √ Identity For √ any element a + b 5 ∈√A, consider (1 + 0 5) × (a + b 5) = (a + b 5) × (1 + 0 5) = a1 + b1 5 = a + b 5. So the element 1 + 0 5 ∈ A functions as the identity. √ √ √ √ Inverses Consider a + b 5 ∈ A. Suppose there exists a c + d 5 ∈ A such that ac + 5bd + (ad + bc) 5 = 1 + 0 5. Then ac + 5bd = 1 and ad + bc = 0. Hence c = a/(a2 − 5b2 ) and d = −b/a2 − 5b2 . Note that c and d must be in A because a, b ∈ Q and that c and d are always defined because (a, b) 6= (0, 0). Thus there exists inverses for all elements in A.

3.2. INTRODUCTION TO GROUPS

79

Because × is a binary operation, there exists an identity, and associativity and inverses hold, the pair (A, ×) is a group. Exercise: 11 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (U (20), +). Solution: ¯ 13, ¯ 17, ¯ 19}. ¯ We first check whether + is a binary operation on U (20). Not Recall that U (20) = {1̄, 3̄, 7̄, 9̄, 11, that 1̄ + 3̄ = 4̄ ∈ / U (20). So + is not a binary operation and the pair (U (20), +) cannot be a group. Exercise: 12 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (P(S), 4), where S is any set and 4 is the symmetric difference of two sets. Solution: We check the group axioms. Associativity We had already showed that that 4 is associative. No matter how we place the parentheses, A4B4C consists of the elements that occur in an odd number of the sets A, B, and C. Identity The operation has an identity, namely ∅, because for all A ∈ P(S), A4∅ = A = ∅4A. Inverses For all A ∈ P(S), we have A4A = ∅ so every element is its own inverse. Hence (P(S), 4) is a group. Exercise: 13 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (G, ×), where G = {z ∈ C |z| = 1}. Solution: We will show that × is a binary operation over G. Consider two arbitrary elements a + bi, c + di ∈ G. Since both exist in G, we know that a2 + b2 = c2 + d2 = 1. Then (a + bi) × (c + di) = (ac − bd) + (ad + bc)i and p |(ac − bd) + (ad + bc)i| = (ac)2 − 2abcd + (bd)2 + (ad)2 + 2abcd + (bc)2 p = (ac)2 + (ad)2 + (bc)2 + (bd)2 − 2abcd + 2abcd p = a2 (c2 + d2 ) + b2 (c2 + d2 ) p p √ = a2 (1) + b2 (1) = a2 + b2 = 1 = 1. Which shows that an arbitrary product of two elements from G ends up in G. Now, we check the group axioms. Associativity Multiplication is associative in C and G is a subset of C so it is also associative in G. √ Identity We will show that 1 ∈ G and it functions as the identity. Consider |1| = 12 + 02 = 1. So 1 ∈ C. Consider any arbitrary element a + bi ∈ G. Then, (a + bi) × 1 = 1 × (a + bi) = a + bi. So G has an identity. Inverses We will find the inverse for an arbitrary element in G and then see if it also belongs in G. Consider an arbitrary element a + bi ∈ G. Consider it’s conjugate a − bi. Now (a + bi) × (a − bi) = (a − bi) × (a + bi) =pa2 + b2 = 1. √Then (a − bi) is the inverse. Now we will show that a − bi belongs to G. |a − bi| = a2 + (−b)2 = a2 + b2 = |a + bi| = 1. Which shows that a − bi ∈ G. So G contains inverses. So all the axioms hold and × is a binary operation on G which proves that the pair (G, ×) is a group. Exercise: 14 Section 3.2 Question: Decide whether the given set and the operation pair forms a group. If it is, prove it. If it is not, decide which axioms fail. The pair (R3 , ×), where × is the vector cross product. Solution: First we check if the cross product is a binary operation. We do know that × is a binary operation over all of R3 since it takes in two vectors and outputs another vector in the same space. We check the group axioms.

80

CHAPTER 3. GROUPS

Associativity The cross product is not associative. We exploit the fact that for any vector ~a ∈ R3 , ~a × ~a = ~0. Let ~i, ~j, and ~k represent three standard basis vectors in R3 . Consider the expression (~i ×~i) × ~j = ~0 × ~j = ~0. However, grouping differently we find ~i × (~i × ~j) = ~i × ~k = −~j 6= ~0. This is an example of how the cross product is not associative in R3 . Identity Suppose that there was some ~e that functioned as the identity. Due to properties of cross products though, ~e × ~e = ~0. However, we know that this should give us the identity vector, so the only option then is that ~0 functions as the identity vector. However for any vector ~v ∈ R3 we have ~0 × ~v = ~0. This shows that ~0 is not the identity vector and that no identity vector can exist. Inverses Since we do not have an identity vector we do not have inverses. Since every single group axiom has failed. This shows that the pair (R3 , ×) is not a group. Exercise: 15 Section 3.2 Question: Show that Z5 ⊕ Z2 is cyclic. Solution: This can be done by finding an element that generates the entire group. We will let Z5 ⊕ Z2 = {(xi , y j ) x5 = 1 and y 2 = 1}. For a generating element, any element of the form (xi , y j ) where both i and j are nonzero will work. We will show it is cyclic using the element (x, y). The set of all different elements generated by (x, y) is {(x, y), (x2 , 1), (x3 , y), (x4 , 1), (1, y), (x, 1), (x2 , y), (x3 , 1), (x4 , y), (1, 1), (x, y)...}. We found ten different elements before repeating (x, y). Since the size of the group is ten this shows that the element (x, y) generated the entire group which proves that the group is cyclic. Exercise: 16 Section 3.2 Question: Show that Z4 ⊕ Z2 is not cyclic. Solution: To prove that Z4 ⊕Z2 is not cyclic, we must prove that there does not exist an element that generates Z4 ⊕ Z2 . It is easy to see that in order for an element to be a generator, it must be in the form (xi , y) for some 1 ≤ i ≤ 3. (x2 , y) can be ruled out as (x2 , y)2 is (1,1). Similarly, (x, y)4 and (x3 , y)4 are (1,1), showing that they do not generate all eight elements in Z4 ⊕ Z2 . Thus there is no generator for Z4 ⊕ Z2 , and Z4 ⊕ Z2 is not cyclic.

Exercise: 17 Section 3.2 Question: Prove Proposition 3.2.13. Solution: We must show that xm xn = xm+n and (xm )n = xmn for all x in a group G and all integers m, n. First, suppose m = 0. Then xm xn = x0 xn = exn = xn = x0+n and (xm )n = (x0 )n = en = e = x0n . Similarly, if n = 0, xm xn = xm x0 = xm e = xm = xm+0 and (xm )n = (xm )0 = e = x0m . Therefore the proposition holds for m = 0 or n = 0. Next, suppose m, n > 0. Then m times n times

m+n times

z }| { z }| {z }| { x x = xx · · · xxx · · · x = xx · · · x = xm+n , m n

n times

z }| { m times mn times z }| { z }| { (xm )n = xx · · · x = xx · · · x = xmn . Next, suppose without loss of generality than m > 0, n < 0. Then |n| times

m times }| { z }| {z xm xn = xx · · · xx−1 x−1 · · · x−1 .

Now if |m| > |n|, we are left with |m| − |n| = m + n multiples of x, or xm+n . If |m| < |n|, we are also left with |n| − |m| = m + n multiples of x, or xm+n . We must treat separately the cases of m > 0, n < 0 and m < 0, n > 0 for the second formula. Suppose that m > 0, n < 0. Then |n| times

z }| { m|n| times m times z }| { z }| { (xm )n = (xx · · · x)−1 = x−1 x−1 · · · x−1 = xmn .

3.2. INTRODUCTION TO GROUPS

81

Suppose alternatively that m < 0, n > 0. Then n times

z

}|

|m| times

{

|m|n times

z }| { z }| { (xm )n = x−1 x−1 · · · x−1 = x−1 x−1 · · · x−1 = xmn . Finally, suppose m, n < 0. Then |n| times

|m| times

|m|+|n| times

}| {z }| { z }| { z xm xn = x−1 x−1 · · · x−1 x−1 x−1 · · · x−1 = x−1 x−1 · · · x−1 = xm+n , |n| times

z

|n| times

}|

{

|m| times

z }| {

|m||n| times |m| times z }| { z }| { z }| { (xm )n = (x−1 x−1 · · · x−1 )−1 = xx · · · x = xx · · · x = xmn .

By cases, we have showed that the proposition holds for all integers m, n, whether positive, negative, or any combination. Exercise: 18 Section 3.2 Question: Is U (11) a cyclic group? Solution: Determine U (11)={1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Consider 2. As every element in U (11) can be generated by a power of 2, 2 is a generator of U (11). Thus U (11) is a cyclic group. Exercise: 19 Section 3.2 Question: Is U (10) a cyclic group? Solution: First, note that U (10) = {1, 3, 7, 9}. The powers of 3 are 3, 9, 27, 81, which are 3, 9, 7, 1 taken modulo 10. Therefore U (10) is a cyclic group generated by the element 3. Exercise: 20 Section 3.2 Question: Prove that (Q, +) is not a cyclic group. Solution: Consider pq ∈ Q such that pq ∈ Z and q 6= 0. Consider the possibility that pq is a generator of (Q, +). 1 . However, we know from arithmetic Then some number of + operations on pq would have to be able to get to q+1 p 1 that no number of + operations on q will result in a number with denominator q+1. Hence q+1 can never be p achieved. Contradiction. Thus q is not a generator for (Q, +), and (Q, +) is not a cyclic group. Exercise: 21 Section 3.2 Question: Construct the Cayley table for U (15). Solution: × 1 2 4 7 8 11 13 14

1 2 1 2 2 4 4 8 7 14 8 1 11 7 13 11 14 13

4 7 8 4 7 8 8 14 1 1 13 2 13 4 11 2 11 4 14 2 13 7 1 14 11 8 7

Exercise: 22 Section 3.2 Question: Construct the Cayley table for Z3 ⊕ Z3 .

11 11 7 14 2 13 1 8 4

13 13 11 7 1 14 8 4 2

14 14 13 11 8 7 4 2 1

(3.2)

82

CHAPTER 3. GROUPS

Solution: × (e, e) (e, z) (e, z 2 ) (z, e) (z, z) (z, z 2 ) (z 2 , e) (z 2 , z) (z 2 , z 2 )

(e, e)

(e, z)

(e, z 2 )

(e, e) (e, z) (e, z 2 ) 2 (e, z) (e, z ) (e, e) (e, z 2 ) (e, e) (e, z) (z, e) (z, z) (z, z 2 ) 2 (z, z) (z, z ) (z, e) (z, z 2 ) (z, e) (z, z) (z 2 , e) (z 2 , z) (z 2 , z 2 ) (z 2 , z) (z 2 , z 2 ) (z 2 , e) (z 2 , z 2 ) (z 2 , e) (z 2 , z)

(z, e)

(z, z)

(z, z 2 )

(z, e) (z, z) (z, z 2 ) 2 (z, z) (z, z ) (z, e) (z, z 2 ) (z, e) (z, z) (z 2 , e) (z 2 , z) (z 2 , z 2 ) (z 2 , z) (z 2 , z 2 ) (z 2 , e) (z 2 , z 2 ) (z 2 , e) (z 2 , z) (e, e) (e, z) (e, z 2 ) 2 (e, z) (e, z ) (e, e) (e, z 2 ) (e, e) (e, z)

(z 2 , e)

(z 2 , z)

(z 2 , z 2 )

(z 2 , e) (z 2 , z) (z 2 , z 2 ) (z 2 , z) (z 2 , z 2 ) (z 2 , e) (z 2 , z 2 ) (z 2 , e) (z 2 , z) (e, e) (e, z) (e, z 2 ) 2 (e, z) (e, z ) (e, e) (e, z 2 ) (e, e) (e, z) (z, e) (z, z) (z, z 2 ) 2 (z, z) (z, z ) (z, e) (z, z 2 ) (z, e) (z, z)

(3.3)

Exercise: 23 Section 3.2 Question: Prove that a group is abelian if and only if its Cayley table is symmetric across the main diagonal. Solution: First, suppose a group is abelian. Then ab = ba for all a, b in the group. But then the entries for ab and ba in the Cayley table are equal, and the position of these entries is such that they are reflected across the main diagonal. (Trivially the elements on the main diagonal are equal to themselves, regardless of whether the group is abelian.) Next, suppose the Cayley table for a group is symmetric across the main diagonal. Then the product in the i, j entry is equal to the product in the j, i entry for all i and j. But this means that for all elements a, b in the group, ab = ba, so the group is abelian. Exercise: 24 Section 3.2 Question: Prove that S = {2a 5b | a, b ∈ Z} as a subset of rational numbers is a group under multiplication. Solution: First we must be sure that multiplication is a binary operation on this set; that is, that the product of two numbers in S always remains in S. Let 2a 5b , 2c 5d ∈ S. Then their product, (2a 5b )(2c 5d ) = 2a+c 5b+d by properties of rational numbers, and since a + c and b + d remain in the integers, S is closed under multiplication. We now check the three group axioms. For associativity, let 2e 5f ∈ S also. Then (2a 5b )(2c 5d ) × (2e 5f ) = a+c+e b+d+f 2 5 = (2a 5b ) × (2c 5d )(2e 5f ). The identity element is 20 50 , and the inverse of any element 2a 5b is −a −b 2 5 , since −a and −b remain integers. Therefore S is a group under multiplication. Exercise: 25 Section 3.2 Question: Prove that S = {2a 5b | a, b ∈ Z} as a subset of rational numbers is a group under multiplication. Solution: We will show closure using a Cayley table. Clearly every product remains in the set: ×

1

13 29

41

1 13 29 41

1 13 29 13 1 41 29 41 1 41 29 13

41 29 13 1

(3.4)

Since every product remains in the set after reducing modulo 42, associativity follows from the associativity of natural numbers and properties of the reduction operation . Let a, b, c ∈ S. Then (ab)c = (ab)overlainc = abc = a(bc) = a(bc. As for the identity, it is clearly 1, and the inverse of every element is itself. Exercise: 26 Section 3.2 Question: Prove that if xn = e, then x−1 = xn−1 . Solution: We know that xn = e. We know that x−1 exists and is unique by the group axioms. Multiply both sides by x−1 ; then x−1 xn = x−1 e = x−1 . By Proposition 3.2.13, we can simplify to x−1 xn = x−1+n = xn−1 = x−1 , the desired result. Exercise: 27 Section 3.2 Question: Let A and B be groups. Prove that the direct sum A ⊕ B is abelian if and only if A and B are both abelian.

3.2. INTRODUCTION TO GROUPS

83

Solution: Suppose that A and B are abelian. Then a1 a2 = a2 a1 and b1 b2 = b2 b1 for all a1 , a2 ∈ A and b1 , b2 ∈ B. Let (a1 , b2 ), (a2 , b2 ) be arbitrary elements of A ⊕ B. Then (a1 , b2 )(a2 , b2 ) = (a1 a2 , b1 b2 ) = (a2 a1 , b2 b1 ) = (a2 , b2 )(a1 , b2 ), so A ⊕ B is abelian. Conversely, suppose that A ⊕ B is abelian. Then (a1 , b2 )(a2 , b2 ) = (a2 , b2 )(a1 , b2 ) for all a1 , a2 ∈ A and b1 , b2 ∈ B. But (a1 , b2 )(a2 , b2 ) = (a1 a2 , b1 b2 ) and (a2 , b2 )(a1 , b2 ) = (a2 a1 , b2 b1 ), so by the transitivity of equality (a1 a2 , b1 b2 ) = (a2 a1 , b2 b1 ). But these are equal precisely if a1 a2 = a2 a1 and b1 b2 = b2 b1 . As a1 , a2 , b1 , b2 are arbitrary, this proves that A and B are abelian. Exercise: 28 Section 3.2 Question: Prove that if a group G satisfies x2 = e for all x ∈ G, then G is abelian. Solution: Let x, y ∈ G. Then xy ∈ G, so (xy)2 = e. But then xyxy = e, and x−1 xyxyy −1 = x−1 ey −1 , and yx = x−1 y −1 . But since x2 = y 2 = 1, x−1 = x and y −1 = y. Therefore yx = xy for arbitrary x, y ∈ G, making G abelian. Exercise: 29 Section 3.2 Question: Prove that if a group G satisfies (xy)−1 = x−1 y −1 for all x, y ∈ G, then G is abelian. Solution: Suppose that G satisfies (xy)−1 = x−1 y −1 for all x, y ∈ G. We have the following chain of implications (xy)−1 = x−1 y −1 =⇒ (xy)−1 y = x−1 =⇒ (xy)−1 yx = 1 =⇒ yx = (xy).

Hence G is abelian. Exercise: 30 Section 3.2 Question: Let g1 , g2 , g3 ∈ G. What is (g1 g2 g3 )−1 ? Generalize your result. Solution: We have (g1 g2 g3 )−1 = g3−1 g2−1 g1−1 since g3−1 g2−1 g1−1 g1 g2 g3 = e. In general, (g1 g2 · · · gn )−1 = −1 −1 −1 gn gn−1 · · · g1 . Exercise: 31 Section 3.2 Question: Prove that every cyclic group is abelian. Solution: Let G be a cyclic group of order n generated by z, so that z n = e. Let x, y ∈ G. Then x = z a and y = z b for some integers a and b between 0 and n − 1. So xy = z a z b = z a+b = z b+a = z b z a = yx, making G abelian. Since G was an arbitrary cyclic group, all cyclic groups are abelian. Exercise: 32 Section 3.2 Question: Prove that GLn (Fp ) contains (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ) elements. [Hint: Use the fact the GLn (Fp ) consists of n × n matrices with coefficients in Fp that have columns that are linearly independent.] Solution: Considering each n × n matrix by the linearly independant columns, we can count the number of matrics in GLn (Fp ). Without restrictions note that each column has p possiblites in n indices, making an upper bound of pn . The only restriction on the first column is that it cannot be the zero vector, making (pn − 1) possiblities. As there are p multiples of the first column and the second column must be linearly independant of the first column, there are (pn − p) possibilities (note the zero vector is one of the multiples of the first column). Generally, any column i must be linearly independent from the i − 1 columns before it which remove the pi−1 linearly combinations of the previous columns leaving a total of (pn − pi−1 ). In total multiplying the possibilites for each column together, there are (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ) elements in GLn (Fp ).

84

CHAPTER 3. GROUPS

Exercise: 33 Section 3.2 Question: In the given general linear group, for the given matrices A and B, calculate the products A2 , AB and B −1 . 0 2 1 1 1. GL2 (F3 ) with A = and B = . 2 1 1 2 1 3 0 1 and B = . 2. GL2 (F5 ) with A = 4 1 2 3 4 6 5 4 3. GL2 (F7 ) with A = and B = . 3 2 3 2 Solution: The desired matrix operations are a) In GL2 (F3 ), 0 2 0 2 1 2 2 = , A = 2 1 2 1 2 2 b) In GL2 (F5 ), 1 3 1 A2 = 4 1 4

3 3 = 1 1

3 , 3

c) In GL2 (F7 ), 4 6 4 2 A = 3 2 3

6 5 = 2 4

1 , 1

AB =

AB =

AB =

0 2

2 1 1 1

1 , 1

B

1 4

3 1

0 2

1 3

1 2

0 , 2

B −1 = 3

6 5 2 3

4 3 = 2 0

0 , 2

−1

4 3

1 2 = 2 0

=

−1

B

=

3 4 3 0

2 4

−1

=5

2 2

−1

3 5

2 1

1 1

3 . 0

6 = 5

2 1

=

Exercise: 34 Section 3.2 Question: In the given general linear group, for the given matrices A and B, calculate the products A2 , AB, and B −1 . 0 2 1 1 a) GL2 (F3 ) with A = and B = . 2 1 1 2 1 3 0 1 and B = . b) GL2 (F5 ) with A = 4 1 2 3 4 6 5 4 c) GL2 (F7 ) with A = and B = . 3 2 3 2 Solution: a) With A =

0 2

2 1 and B = 1 1 A2 =

b) With A =

1 4

3 0 and B = 1 2

1 in GL2 (F3 ), we have 2 1 2

3 A = 3 c) With A =

4 3

6 5 and B = 2 3

AB =

2 0

1 1

B −1 =

2 2

2 . 1

1 3

AB =

1 2

0 2

B

−1

1 1

3 . 0

B

−1

6 5

2 . 1

=

4 in GL2 (F7 ), we have 2

6 A = 4 2

1 in GL2 (F5 ), we have 3

2

2 2

1 1

AB =

3 0

0 2

=

3.3. PROPERTIES OF GROUP ELEMENTS

85

Exercise: 35 Section 3.2 Question: Let F be Q, R, C, or Fp . The Heisenberg group with coefficients in F is     1 a b  H(F ) = 0 1 c  ∈ GL3 (F ) a, b, c ∈ F .   0 0 1 1. Show that H(F ) is a group under matrix multiplication. 2. Explicitly show the inverse of an element in H(F ). Solution: We work with the set H(F ) of matrices as described in the problem. a) We first need to show that multiplication is a binary operation on H(F ). We have      1 a1 b1 1 a2 b2 1 a1 + a2 b1 + b2 + a1 c2 0 1 c1  0 1 c2  = 0 . 1 c1 + c2 0 0 1 0 0 1 0 0 1 This shows that multiplication is a binary operation on H(F ). Matrix multiplication is associative in general so it is associative in H(F ). The identity matrix is in H(F ) so H(F ) has an identity. From the multiplication above, we see that for all   1 a b 0 1 c  0 0 1 in H(F ), the multiplicative inverse is 

1 0 0

−a 1 0

 ac − b −c  . 1

Hence H(F ) has inverses and thus H(F ) is a group under matrix multiplication. b) We already explicitly calculated the inverse in the above part.

3.3 – Properties of Group Elements Exercise: 1 Section 3.3 Question: Find the orders of 5̄ and 6̄ in (Z/21Z, +). Solution: There are different ways to do this. We are going to find the first element that is of the form 21k that is divisible by 5. So we find {21, 42, 63, 84, 105}. So 105/5 = 21 and so |5̄| = 21. We will use the same technique for 6̄. We get {21, 42} and then we can stop since 42 is divisible by 6. So 42/6 = 7 and that implies that |6̄| = 7. Exercise: 2 Section 3.3 Question: Find the orders of all the elements in (Z/18Z, +). ¯ = 8, Solution: |0̄| = 1, |1̄| = 18, |2̄| = 9, |3̄| = 6, |4̄| = 9, |5̄| = 18, |6̄| = 3, |7̄| = 18, |8̄| = 9, |9̄| = 2, |10| ¯ = 18, |12| ¯ = 3, |13| ¯ = 18, |14| ¯ = 8, |15| ¯ = 6, |16| ¯ = 9, |17| ¯ = 18. |11| Exercise: 3 Section 3.3 Question: Find the orders of all the elements in U (21). Solution: The elements of U (21) are 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20. By calculation, |1| = 1, |2| = 6, |4| = 3, |5| = 6, |8| = 2, |10| = 6, |11| = 6, 13| = 2, |16| = 3, |17| = 6, |19| = 6, |20| = 2. Exercise: 4 Section 3.3 Question: Find the orders of all the elements in GL2 (F3 ).

86

CHAPTER 3. GROUPS

Solution: There are 48 elements in GL2 (F3 ) but calculating the orders will not be as tedious as we might first expect if we take advantage of Proposition 3.3.7. 1 0

0 =1 1

1 0

0 2

1 0 2 0 0 1 0 1 1 1 2 1 2 1 0 2 1 2 1 2 2 2

2 =3 1 1 =2 1 1 =2 0 1 =8 2 0 =3 1 1 =8 0 0 =6 2 1 =6 1 1 =6 0 0 =2 2 0 =2 1

1 0 2 0 0 1 0 1 1 1 2 1 2 1 0 2 1 2 1 2 2 2

2 2

1 2

2 0

2 2

2 1

2 0

2 2

2 1

2 0

2 2

2 1

=2 =6 =6 =4 =3 =8 =3 =8 =8 =8 =4 =4

1 0

1 1

2 0

0 1

2 0

2 1

0 1

1 1

1 1

1 0

1 1

0 2

2 1

0 1

0 2

1 0

0 2

1 2

1 2

0 1

2 2

1 0

2 2

0 2

=3 =2 =2 =8 =8 =2 =2 =4 =3 =3 =3 =6

1 0

1 2

2 0

0 2

2 0

2 2

0 1

2 1

1 1

2 0

1 1

1 2

2 1

1 1

0 2

2 0

0 2

2 2

1 2

1 1

2 2

2 0

2 2

1 2

=2

=2 =6 =6 =6 =4 =4 =2 =8 =8 =8 =8

Exercise: 5 Section 3.3 Question: Find the orders of all the elements in Q8 . Solution: The orders of the elements of the quaternion group are g |g|

1 1

−1 2

i −i j 4 4 4

−j 4

k 4

−k 4

Exercise: 6 Section 3.3 Question: Find the orders of all the elements in Z4 ⊕ Z2 . Solution: We will represent Z4 ⊕ Z2 as {(xi , y j ) x4 = 1 and y 2 = 1}. We will start by finding the orders of individual elements in Z4 and Z2 and then repeatedly apply Theorem 3.3.9. For Z4 , {|1| = 1, |x| = 4, |x2 | = 2, |x3 | = 4} and for Z2 , {|1| = 1, |y| = 2}. Since all the elements in Z4 or Z2 are of order 1, 2, or 4, by Theorem 3.3.9 all of elements in Z4 ⊕ Z2 are of order 1, 2, or 4. Then in Z4 ⊕ Z2 all of the elements of order 1 are {(1, 1)}, all the elements of order 2 are {(1, y), (x2 , 1), (x2 , y)}, and finally all the element of order 4 are {(x, 1), (x, y), (x3 , 1), (x3 , y)}. Exercise: 7 Section 3.3 Question: Prove Proposition 3.3.4 which states, ”Let G be any group and let x ∈ G. Then x−1 = |x|”. Solution:

3.3. PROPERTIES OF GROUP ELEMENTS

87

Case 1: Suppose |x| = k is some finite number. Then, xk = 1. But xk = (x−k )−1 = ((x−1 )k )−1 = 1. Taking the inverse of both sides we get, (((x−1 )k )−1 )−1 = 1−1 ⇒ ((x−1 )k )(−1)(−1) = 1 ⇒ ((x−1 )k )1 = 1 ⇒ (x−1 )k = 1 Which implies that |(x−1 )| divides |x|. Now, let y = x−1 and assume that |y| = n, by going through the same computation as above we will arrive at the fact that |y −1 | divides |y| which, in terms of x, implies that |x| divides |x−1 |. Since both orders divide one another, this implies that |(x−1 )| = |x|. Case 2: Now suppose that the order of x is infinite. Then suppose for sake of contradiction that |(x−1 )| = k is finite. Well by the above result, |x| = k as well, which is a contradiction. So our assumption must be wrong and the order of x−1 must also be infinite. This covers all cases and shows that |x−1 | = |x|. Exercise: 8 Section 3.3 Question: Show that for all integers n ≥ 1, the element − sin 2π cos 2π n n R= sin 2π cos 2π n n in GL2 (R) has order n. Solution: Recall the double angle formulas for sin and cos. sin(α + β) = sin(α) cos(β) + sin(β) cos(α) and cos(α + β) = cos(α) cos(β) − sin(α) sin(β). Now consider the multiplication of these two matrices: cos α − sin α cos β − sin β cos α cos β − sin α sin β − cos α sin β − cos β sin α = . sin α cos α sin β cos β cos β sin α + cos α sin β cos α cos β − sin α sin β By the double angle formulas above, this is equal to: cos(α + β) − sin(α + β) . sin(α + β) cos(α + β) So if our matrices have this format we simply add the angles. Then, k cos 2π − sin 2π cos 2kπ − sin 2kπ n n n n Rk = = sin 2π cos 2π cos 2kπ sin 2kπ n n n n 2kπ 2kπ 2kπ Now, cos n = 1 when k is a multiple of n. Likewise, sin n = − sin n = 0 when k is a multiple of n/2. Then |R| = lcm{n, n/2} = n. Exercise: 9 Section 3.3 Question: Calculate the order 20 in Z/52Z. Solution: We want to find the smallest positive integer k so that 20k = 52m for some integer m. Then we have k = lcm(20, 52)/20 = 260/20 = 13. So |20| = 13. Exercise: 10 Section 3.3 Question: Calculate the order of 285 in the group Z/360Z. Solution: We want to find the smallest positive integer k such that 285k = 360m for some integer m. The smallest k is lcm(285, 360)/285 = 6840/285 = 24. So the order is |285| = 24. Exercise: 11 Section 3.3 Question: Calculate the order of r16 in D24 . Solution: In D24 , the element r has order |r| = 24. Consequently, the order of r16 is |r16 | =

24 24 = = 3. gcd(24, 16) 8

88

CHAPTER 3. GROUPS

Exercise: 12 Section 3.3 Question: What is the largest order of an element in Z75 ⊕ Z100 ? Illustrate with a specific element. Solution: By Theorem 3.3.9 we know that for any two elements xi ∈ Z75 and y j ∈ Z100 , |(xi , y j )| = i j lcm{|x |, |x |} which by proposition 3.3.7 is equal to lcm{75/(gcd{i, 75}), 100/(gcd{j, 100})}. Now, no matter our choice for i and j, we will end up with lcm{c, d} where c|75 and d|100. We will prove that lcm{c, d} ≤ lcm{75, 100} when c|75 and d|100. Let m = lcm{75, 100}. Since c|75 and 75|m, this implies that c|m. In the same way, we also know that d|m. This implies that lcm{c, d} ≤ m. Therefore, in order to get the largest order possible, we should choose i and j so that gcd{i, 75} = gcd{j, 100} = 1. We will let i = j = 1 (There are many other options). Then our element is (x, y) and |(x, y)| = lcm{75, 100} = 300. Exercise: 13 Section 3.3 Question: Find an element of order 3 in GL3 (F2 ). Solution: There are many solutions.  0 0 1

1 0 0

Exercise: 14 Section 3.3 Question: Revisit Example 3.3.3. Calculate the order of

 0 1 0

0̄ 2̄

2̄ 2̄

in GL2 (F3 ) with performing a single matrix

operation. 0̄ Solution: In Example 3.3.3, we see a matrix g which has order 8. We also see that g = 2̄ 8 = 81 = 8. Proposition 3.3.7, |g 3 | = gcd{8,3} 3

2̄ . By 2̄

Exercise: 15 Section 3.3 Question: Find all the generators of the cyclic group Z40 . Solution: We will let Z40 = {xi x40 = 1}. We know that |x1 | = 40. Now consider |xa | = 40/ gcd{40, a}. Now, xa is a generator when |xa | = 1, which only occurs when gcd{40, a} = 40. So, for any a, relatively prime to 40, xa generates Z40 . So all the generators are {x, x3 , x7 , x9 , x11 , x13 , x17 , x19 , x21 , x23 , x27 , x29 , x31 , x33 , x37 , x39 }.

Exercise: 16 Section 3.3 Question: Find all the generators of the cyclic group Z/36Z. Solution: Understanding that the notation xa from Proposition 3.3.7 represents operating x with itself a times regardless of the group operation, we can apply the results of Proposition 3.3.7 to our cyclic group as follows. We know that |1̄| = 36. And from Proposition 3.3.7, |1̄a | = 36/ gcd{a, 36}. Now under the group operation for Z/36Z, 1̄a = 1̄ + 1̄ + · · · + 1̄ (a times) = a ∗ 1̄ = ā. So that, |ā| = 36/ gcd{a, 36}. Now, a group element, ā, is a generator when it has order 36. And ā has order 36 when gcd{a, 36} = 1. So that any ā, relatively prime to 36, is a generator. So all our generators are: ¯ 13, ¯ 17, ¯ 19, ¯ 23, ¯ 25, ¯ 29, ¯ 31, ¯ 35}. ¯ {1̄, 5̄, 7̄, 11,

Exercise: 17 Section 3.3 Question: Let p be an odd prime. n a) Use the Binomial Theorem to prove that (1 + p)p ≡ 1 (mod pn+1 ) for all positive integers n. b) Prove also that (1 + p)p

n−1

6≡ 1 (mod pn+1 ) for all positive integers n.

3.3. PROPERTIES OF GROUP ELEMENTS

89

c) Conclude that 1 + p has order pn in U (pn+1 ). Solution: Let p be an odd prime. 0 a) We prove this result by induction on n. If nn = 0, then (1 + p)p = 1 + p and this is indeed congruent to 1 modulo p. Now suppose that (1 + p)p ≡ 1 (mod pn+1 ) for some nonnegative integer n. Then n (1 + p)p = 1 + kpn+1 for some integer k. Thus n+1 n p p p 2 2n+2 (1 + p)p = (1 + p)p = (1 + kpn+1 )p = 1 + kpn+1 + k p + · · · + k p ppn+p . 1 2 The first term in this binomial expansion is 1 and since p1 = p, we see that all the successive terms are n n+1 divisible by pn+2 . Hence, if (1 + p)pn ≡ 1 (mod pn+1 ) for some nonnegative integer n then (1 + p)p ≡1 n+2 p n+1 (mod p ). By induction, (1 + p) ≡ 1 (mod p ) for all nonnegative integers n. n−1

b) We prove by induction that (1 + p)p ≡ 1 + pn (mod pn+1 ) for all n ≥ 1. This formula is trivial for n−1 n = 1. Suppose that the formula is true for some positive integer n. Then (1 + p)p = 1 + pn + kpn+1 for some integer k. Then by the Binomial Theorem p X j n p (1 + p)p = (1 + pn + kpn+1 )p = (1 + pn )p − j kpn+1 j j=0 Considering congruence modulo pn+2 we have (1 + p)

pn

p ≡ (1 + p ) + (1 + pn )p−1 kpn+1 1 n p

(mod pn+2 ) p 2n n ≡1+p·p + p + · · · + ppn 2

(mod pn+2 )

≡ (1 + pn )p

≡ 1 + pn+1 By induction, (1 + p)p

n−1

(mod pn+2 )

(mod pn+2 ).

≡ 1 + pn (mod pn+1 ) for all n ≥ 1. In particular, (1 + p)p

c) We now consider the element 1 + p in U (pn+1 ). We have shown that 1 + p n

p

n−1

6≡ 1 (mod pn+1 )

n

= 1. Hence, the order of 1 + p

n

must divide p . However, the only divisors of p are powers of p and since we proved that 1 + p we deduce that 1 + p has order pn .

pn−1

6= 1,

Exercise: 18 Section 3.3 Question: Let G be a group such that for all a, b, c, d, x ∈ G, the identity axb = cxd implies ab = cd. Prove that G is abelian. Solution: For any a, b ∈ G we know that aba = aba. We then write, (1)a(ba) = (ab)a(1), which implies by the identity, where x corresponds to the first occuring a on the left-hand side and the last a on the right hand side, that 1ba = ab1 ⇒ ba = ab. This shows that G is abelian. Exercise: 19 Section 3.3 Question: Consider the infinite dihedral group as presented in Example 3.3.8. a) Show that y(xy)n = (xy)−n−1 x for all integers n. b) Conclude that every element in D∞ can be written as (xy)n or (xy)n x for some n ∈ Z. Solution: First, we recall that x2 = y 2 = ι, so x−1 = x and y −1 = y. Then we consider (xy)−n−1 x = y −1 x−1 y −1 x−1 · · · y −1 x−1 x = yxyx · · · yxx = yxyx · · · yxy, where there are n pairs of yx followed by y. This is equal to y(xy)n . Next, we recall that all elements in D∞ can be written in the form (xy)n , y(xy)n , (xy)n x, or y(xy)n x according to the example, intending to show that all four forms can be reduced to (xy)m or(xy)m x for some m not necessarily equal to n. We observe that (xy)n works, y(xy)n = (xy)−n−1 x = (xy)m x, (xy)n x works, and y(xy)n x = y(xy)n = (xy)−n−1 xx = (xy)m . Exercise: 20 Section 3.3 Question: Let a and b be elements in a group G. Prove that if a and b commute, then the order of ab divides lcm(|a|, |b|).

90

CHAPTER 3. GROUPS

Solution: Let m = lcm(|a|, |b|). Consider (ab)m = (ab)(ab) · · · (ab) (m times). Now since a and b commutes this is equal to am bm . Since m is a multiple of both |a| and |b| this is equal to 1. By Corollary 3.3.6, this implies that the order of ab divides lcm(|a|, |b|). Exercise: 21 Section 3.3 Question: Find a nonabelian group G and two elements a, b ∈ G such that |ab| does not divide lcm(|a|, |b|). Solution: Let G be Dn where n is an odd number greater than one. Now consider the elements rs and s. (rs)2 = rsrs = rr−1 ss = 1 so that |rs| = 2 and of course |s| = 2. Then lcm(|rs|, |s|) = lcm(2, 2) = 2. However (rs)(s) = r(ss) = r(1) = r, and |r| = n which is odd, so in this case, |(rs)(s)| > lcm(|rs|, |s|). Exercise: 22 Section 3.3 Question: Let G and H be two finite groups. Prove that G ⊕ H is cyclic if and only if G and H are both cyclic with gcd(|G|, |H|) = 1. Solution: Let G = {xi xn = 1} and H = {y i y m = 1}. Then |G| = n and |H| = m. (=⇒) We will assume that G ⊕ H is cyclic and show that the other two conditions follow. Since G ⊕ H is cyclic there exists some generator (xi , y j ). Consider any two elements xa ∈ G and y b ∈ H. Since we have a generator, there must be a number k where (xi , y j )k = ((xi )k , (y j )k ) = (xa , y b ). Since for any xa , there exists an k so that (xi )k = xa , this shows that xi also functions as a generator for G. By the exact same reasoning y j also functions as a generator for H, which shows that both G and H are cyclic. Now, since we have a generator we know that |(xi , y j )| = nm. By Theorem 3.3.9, |(xi , y j )| = lcm(|xi |, |y j |) and by Proposition 3.3.7 this is equal to lcm(n/ gcd(i, n), m/ gcd(j, m)). Since xi and y j function as generators of G and H respectively, we know that n/ gcd(i, n) = n and m/ gcd(j, m) = m. Then |(xi , y j )| = lcm(n, m) = nm/ gcd(n, m). Since this must equal nm, we know that gcd(n, m) = gcd(|G|, |H|) = 1. (⇐=) Now we will assume that G and H are both cyclic with gcd(|G|, |H|) = 1 and show that G⊕H is also cyclic. Consider the element (x, y) ∈ G ⊕ H. We will show that this element functions as a generator of G ⊕ H by finding it’s order. So, |(x, y)| = lcm(n/ gcd(1, n), m/ gcd(1, m)) = lcm(n, m) = nm/ gcd(n, m) = nm. Since the order of this element is equal to the size of our group, it must be a generator and G ⊕ H is cyclic. This completes the proof and shows the if and only if statement holds in both directions. Exercise: 23 Section 3.3 Question: Let G1 , G2 , . . . , Gn be n finite groups. Prove that G1 ⊕ G2 ⊕ · · · ⊕ Gn is cyclic if and only if Gi is cyclic for all i = 1, 2, . . . , n and gcd(|Gi |, |Gj |) = 1 for all i 6= j. Solution: Exercise 3.3.22 establishes this result for n = 2. Now suppose that for some integer n ≥ 2, any collection of finite groups G1 , G2 , . . . , Gn has the property that G1 ⊕ G2 ⊕ · · · ⊕ Gn is cyclic if and only if Gi is cyclic for all i = 1, 2, . . . , n and gcd(|Gi |, |Gj |) = 1 for all i 6= j. Now consider a collection of n + 1 finite groups H1 , H2 , . . . , Hn+1 . By Exercise 3.3.22, H1 ⊕ H2 ⊕ · · · ⊕ Hn+1 is cyclic if and only if H1 ⊕ H2 ⊕ · · · ⊕ Hn and Hn+1 are both cyclic and gcd(|H1 ||H2 | · · · |Hn |, |Hn+1 |) = 1. By the induction hypothesis, H1 ⊕ H2 ⊕ · · · ⊕ Hn is cyclic if and only if H1 , H2 , . . . , Hn are all cyclic and gcd(|Hi |, |Hj |) = 1 for 1 ≤ i < j ≤ n. The condition gcd(|H1 ||H2 | · · · |Hn |, |Hn+1 |) = 1 is equivalent to gcd(|Hi |, |Hn+1 |) = 1 for 1 ≤ i ≤ n. Hence, H1 ⊕H2 ⊕· · ·⊕Hn+1 is cyclic if and only if H1 , H2 , . . . , Hn+1 are all cyclic and gcd(|Hi |, |Hj |) = 1 for 1 ≤ i < j ≤ n + 1. By induction, the theorem holds for all n. Exercise: 24 Section 3.3 Question: Let x ∈ G be an element of finite order n. Prove that 1, x, x2 , . . . , xn−1 are all distinct. Deduce that |x| ≤ |G|. Solution: Assume that there is some xi and xj where xi = xj and i 6= j and 0 ≤ i, j ≤ n − 1. Without loss of generality, we will assume that i < j. Then since xi = xj we have xi x−i = xj x−i ⇒ 1 = xj−i . Now, 1 ≤ (j − i) ≤ n − 1, which contradicts the facts that |x| = n. So all the elements must be distinct. So, since there are as many distinct elements as |x| and since all those elements are also a a part of G, this implies |x| ≤ |G|. Exercise: 25 Section 3.3 Question: Prove that for elements x and y in any group G we have |x| = |yxy −1 |.

3.3. PROPERTIES OF GROUP ELEMENTS

91

Solution: We remark first that for all x, y ∈ G, (yxy −1 )n = (yxy −1 )(yxy −1 ) · n times · · · (yxy −1 ) = yx(yy −1 )x(yy −1 )x · · · xy −1 . Since there are n − 1 pairs (yy −1 ), these cancel to give (yxy −1 )n = yxn y −1 . Now suppose that te order of x is infinite. Then there is no positive value of k such that xk = 1. Assume there could exist a positive value ` of yxy −1 such that (yxy −1 )` = yx` y −1 = 1. Then x` = y −1 1y = 1. This contradicts the assumption that x has infinite order. Thus x has infinite order implies yxy −1 has infinite order. Now suppose that |x| = n. Then (yxy −1 )n = yxn y −1 = yy −1 = 1. So the order of yxy −1 is finite and we set |yxy −1 | = m. Since (yxy −1 )n = 1, then m divides n. Conversely, xm = (y −1 yxy −1 )y)m = y −1 (yxy −1 )m y = y −1 y = 1. Hence n divides m. Hence, |x| = n = m = |yxy −1 |. Exercise: 26 Section 3.3 Question: Use the preceding exercise to show that for any elements g1 and g2 in a group |g1 g2 | = |g2 g1 |, even if g1 and g2 do not commute. Solution: We will show that g2 g1 is a conjugate of g1 g2 . Consider g1−1 (g1 g2 )g1 = (g1−1 g1 )g2 g1 = 1(g2 g1 ) = g2 g1 . Since g1 g2 and g2 g1 are conjugates. By excercise 3.3.25, we have |g1 g2 | = |g2 g1 |. Exercise: 27 Section 3.3 Question: Prove that the group of rigid motions of the cube has order 24. Solution: Every rigid motion of a cube maps a given reference face into one of six of the other faces. Furthermore, for each of these options, there are 4 ways to rotate the face. Each of these 6 × 4 = 24 options gives all possible rigid motions and futhermore, these are all distinct. Hence, there are 24 rigid motions of the cube. Exercise: 28 Section 3.3 Question: Prove that the group of rigid motions of the octahedron has order 24. Solution: When considering rigid motions of an octahedron, a given reference face (an equilateral triangle) can be mapped to any other face. There are 8 options for this. However, each mapped face can be oriented (rotated) in 3 ways. Each of these options corresponds to a given rigid motion. Furthermore, once these choices are made the rigid motion is completely determined. Hence, there are 3 × 8 = 24 rigid motions of the octahedron. Exercise: 29 Section 3.3 Question: Prove that the group of rigid motions of a classic soccer ball has order 60. Solution: In a rigid motion of a soccer ball, the ball moves (rotates) in Euclidean space but in such a way that the structure of the polyhedral surface returns to the same structure (pentagons return to pentagons and hexagons back onto hexagons). By inspection, we can count off that there are 12 pentagons and 20 hexagons. Fix one of the pentagons P1 . In any rigid motion, since a pentagon must go to another pentagon, P1 can be mapped back to any of the 12 pentagons. However, each of the twelve pentagons can be oriented in 1 of 5 ways (rotated around its center). This amounts to 12 × 5 = 60 possible rigid motions so far. This is all of them because as soon as we know where one of the pentagons go, the rest of the polygons are determined in reference to this pentagon. [Note that this just gives the order of the group of rigid motions without describing any of the internal structure or properties of any group elements.] Exercise: 30 Section 3.3 Question: Find all groups of order 5. Solution: Let G have order 5. We know e ∈ G is the only element of order 1.. Also, we know that no element in G has order greater than 5 because then G would have more than 5 elements. Suppose that G contains an element g of order 4; then G = {e, g, g 2 , g 3 , h} where h is an additional element. What is gh? It cannot be a power g n of g because then h = g n−1 , a contradiction, and it cannot be h because then g = e, a contradiction. Therefore it is impossible for G to contain an element of order 4. Suppose now that G contains an element g of order 3; then G = {e, g, g 2 , a, b} where a, b are additional elements. What is ga? As before, it can be neither a power of g nor a, so ga = b; similarly, gb = a necessarily.

92

CHAPTER 3. GROUPS

Then a = g −1 b and a = gb, so g −1 b = gb. But then g = g −1 , which contradicts that g has order 3. Therefore it is impossible for G to contain an element of order 3. Suppose finally that G contains an element g or order 2; then G = {e, g, a, b, c}. We know that G can contain no elements of order 3 or 4; therefore |a|, |b|, |c| must be 2 or 5. If, without loss of generality, a has order 5, then G = {e, a, a2 , a3 , a4 }; but then there is no element g with order 2. Therefore a, b, c must all have order 2. It is now helpful to relabel the group elements as G = {e, a, b, c, d} to emphasize that a, b, c, d are four elements of order 2. Consider ab. It cannot be a or b because then b or a would be the identity. Therefore assign ab = c. This forces ac = b because ac = aab = eb = b. What, now, is ad? It cannot be a or d, so it must be b or c. If ad = b, then ad = ac, so d = c, a contradiction. If ad = c, then ad = ab, so d = b, a contradiction. Therefore it is impossible for G to contain an element of order 4. Suppose, then, that G contains an element g of order 5. Then G contains all powers of G, so G = {e, g, g 2 , g 3 , g 4 }, and G is the cyclic group of 5 elements. By exhaustion this is the only possible group of order 5. Exercise: 31 Section 3.3 Question: We consider groups of order 6. We know that Z6 is a group of order 6. We now look for all the others. Let G be any group of order 6 that is not Z6 , i.e., does not contain an element of order 6. a) Show that G cannot have an element of order 7 or greater. b) Show that G cannot have an element of order 5. c) Show that G cannot have an element of order 4. d) Show that the nonidentity elements of G have order 2 or 3. e) Conclude that exist only two subgroups of order 6. In particular, there exists one abelian group of order 6 (the cyclic group Z6 ) and one nonabelian group of order 6 (D3 is such a group). Solution: We consider groups of order 6. a) If G contained an element g of order 7 or greater, then the elements e, g, g 2 , g 3 , g 4 , g 5 , and g 6 would all be distinct and therefore G would contain more than 6 elements. Hence, if |G| = 6, it cannot contain elements of order 7 or greater. b) Assume the |G| = 6 and that G contains an element x of order 5. Then e, x, x2 , x3 , and x5 are distinct elements in G. Let y be the one remaining element in G. We consider possibilities for the value of xy. If xy = xn for some 1 ≤ n ≤ 5, then y = xn−1 , which is a contradiction. If xy = y, then x = e, which is also a contradiction. Thus, by contradiction, G cannot contain an element of order 5. c) Assume the |G| = 6 and that G contains an element x of order 4. Then e, x, x2 , and x3 are all distinct. Let y and z be the two remaining distinct elements in G. We consider the values of xy and xz. By the same reasoning in the previous part, we cannot have xy = xn for any n and we cannot have xy = y. Thus xy = z. By the identical reasoning, we must have xz = y. Thus x2 y = y, which implies that x2 = 1. This is again a contradiction. Thus G cannot contain an element of order 4. d) By the above two parts, G cannot contain elements of order 4 or 5 so it can only contain nonidentity elements of order 2 or 3. Furthermore, assume that G contains only elements of order 3. Let x be an element of order 3. Then e, x, and x2 are distinct. Let y be another element of order 3. Then y and y 2 are two more elements (of order 3) in G. Let z be the remaining element. By similar reasoning as above, xy cannot be a power of x or of y so xy = z. But for the same reason, x2 y cannot be a power of x or of y. Hence, x2 y = z = xy, which implies that x = 1, a contradiction. Assume that G contains only elements of order 2. Let x and y be distinct elements of order 2. Then e, x, y, xy are all distinct elements of order 2. Let z be another element of order 2. Then z, zx, zy, zxy must all be distinct elements of order 2. (E.g., if zy = xy, then z = x, which is a contradiction.) Hence, G cannot be a group of order 6 and contain only elements of order 2. We conclude that G must contain at least one element of order 2 and one element of order 3. e) Let G be a group of order 6. Suppose that x is an element of order 3 and y is an element of order 2. Then e, x, x2 , y, yx, yx2 are all distinct elements in G. Consider the product xy. It must be one of these six elements. However, it cannot be 1, x, x2 , or y, without leading to a contradiction. Hence, there are exactly two possibilities: (i) xy = yx, in which case xy is an element of order 6, so the group is cyclic of order 6; (ii) xy = yx2 , in which case the group is nonabelian. Exercise: 32 Section 3.3

3.3. PROPERTIES OF GROUP ELEMENTS

93

Question: Let G = {e, v, w, x, y, z} be a group of order 6. For the following partial table, decide if it can be completed to the Cayley table of some G and if so fill it in. e v w x y z

e − − − − − −

v − − − y − −

w − − − − − −

x − − z − − −

y − − e − − −

z − w − − − −

Solution: This partial Cayley table cannot be completed to a Cayley table of a group. e v w x y z

e e v w x y z

v v − − y − e

w w − − − − −

x x − z − − −

y y − e − − −

z z w − − x −

We use (i) wy = e implies yw = e; (ii) x = (yw)x = y(wx) = yz; (iii) e = wy = w(xv) = (wx)v = zv; (iv) zv = e implies vz = e. However, we were given vz = w, which leads to a contradiction. Exercise: 33 Section 3.3 Question: Let G = {e, t, u, v, w, x, y, z} be a group of order 8. For the following partial table, decide if it can be completed to the Cayley table of some G and if so fill it in. e t u v w x y z

e − − − − − − − −

t − − − − x − − −

u − − e − v − − −

v − − − u − z t −

w − − − − − − z x

x − − y t − − − −

y − − x − − − − −

z − e t − y − − u

Solution: The partial Cayley table can be completed to the Cayley table of a group. e t u v w x y z

e e t u v w x y z

t t u z y x v w e

u u z e w v y x t

v v x w u e z t y

w w y v e u t z x

x x w y t z u e v

y y v x z t e u w

z z e t x y w v u

Explanation/Steps: After filling in the row and column corresponding to operating by the identity, we used: (i) e = u2 implies w = w(uu) = (wu)u = vu; (ii) uz = t implies z = u2 t = ut; tz = e implies zt = e. At this stage the u-row has two vacancies, for uv and uw and these must be either v or w. However, uv 6= v so uv = w and also uw = v. Now we have also (iv) x = wt = (vu)t = v(ut) = vz; (v) u2 = e = zt implies u = uzt = t2 ; (vi) vt = v(uz) = (vu)z = wz = y; (vii) t = (zt)t = zt2 = zu. At this point, there are two vacancies in the v row and they can be e or z in the w or y columns. However, a z already occurs in the w column so we must have vw = e and vy = z. Then (viii) vw = e implies that wv = e. Now the v column has two blank spaces, which must be x or y in the t or z rows. But there already is an x in the z row so we conclude that zv = y and tv = x. We can now fill out the u column using the requirement that each row and column must contain only one of each element

94

CHAPTER 3. GROUPS

of the group. By the same condition, we now have tw = y. Now we continue (ix) xt = (vz)t = v(zt) = ve = v. Using the requirement that each row and column contains one of each element allows us to fill out three more entries. (x) ww = (vu)w = v(uw) = vv = u; (xi) wx = w(uy) = (wu)y = vy = z, which also forces wy = t; (xii) xx = x(tv) = (xt)v = v 2 = u, which allows us to fill out three more entries; (xiii) tx = t(tv) = (tt)v = uv = w, which allows us to complete the table. Exercise: 34 Section 3.3 Question: Let {Gi }i∈I be a collection of groups, indexed by a set I that is not necessarily finite. We define the direct sum of this collection, denoted by M Gi i∈I

as the set of choice functions f that for each i ∈ I associates f (i) ∈ Gi , such that f (i) = 1 for all but a L finite number of indices i. We define an operation · on i∈I Gi by f · g as the choice function such that (f · g)(i) = f (i)g(i) in each group Gi . Prove that this direct sum of the collection {Gi }i∈I is itself a group. [In the special case that I = N, the direct sum consists of infinite sequences (g0 , g1 , g2 , . . .) such that gi ∈ Gi and gi = 1 for all but a finite number of indices i. In the case that I is a finite set, then this definition is identical to Definition 3.2.12 and its generalization to a finite number of groups.] Solution: We first need to verify that that operation is indeed a binary operation. Let f and g be two choice functions. The images of f are not the identity on a finite set If and the images of g are not the identity on a finite set Ig . Thus, as defined f ġ returns the identity except on a subset of If ∪ Ig , which is finite. Hence, the the choice function f · g is not the identity except on a finite number of indices. Consequently, the defined operation is indeed a binary operation. M Let f, g, h be three choice functions in Gi that are the identity on all indices but a set If , Ig , and Ih i∈I

respectively. For all i ∈ If ∪ Ig ∪ Ih we have (f (i)g(i))h(i) = f (i)(g(i)h(i)). Hence, (f · g) · h = f · (g · h). The identity in this arbitrary direct sum is the choice function ι such that ι(i) = ei for all i ∈ I. For each choice function f in the arbitrary direct sum that is the identity on all but the finite set If of indices, we consider the choice function f −1 defined by f −1 (i) = f (i)−1 in Gi for all i ∈ If . This is the inverse of f . M We have shown · is a binary operation on Gi and that this operation endows the set of choice functions i∈I

with the structure of a group.

3.4 – Symmetric Groups Exercise: 1 Section 3.4 Question: Write standard cycle notation the following permutation expressed in chart notation. 1 2 3 4 5 6 7 a) σ = . 6 3 2 4 7 1 5 1 2 3 4 5 6 7 b) τ = . 3 5 7 2 1 6 4 Solution: a) σ = (1 6)(2 3)(5 7) b) τ = (1 3 7 4 2 5) Exercise: 2 Section 3.4 Question: Suppose the permutation σ ∈ S8 given in n-tuple notation is (3, 2, 6, 4, 5, 8, 7, 1). Depict σ with a directed graph and express it in standard cycle notation. Solution: The directed graph representation of this permutation is

3.4. SYMMETRIC GROUPS

95

3 4

2

5

1

6

8 7

The cycle notation of this permutation is σ = (1 3 6 8). Exercise: 3 Section 3.4 Question: Suppose the permutation σ ∈ S9 given in n-tuple notation is (4, 6, 5, 2, 3, 1, 8, 7, 9). Depict σ with a directed graph and express it in standard cycle notation. Solution: The directed graph representation of this permutation is 4

3 2

5 1 6 9 7

8

The cycle notation of this permutation is σ = (1 4 2 6)(3 5)(7 8). Exercise: 4 Section 3.4 Question: In S6 , with σ = (1 3 5)(2 6) and τ = (1 3 4 5 6), calculate: a) στ ; b) τ σ; c) σ 2 ; d) τ −1 ; e) στ σ −1 . Solution: a) στ = (1 5 2 6 3 4). b) τ σ = (1 4 5 3 6 2). c) σ 2 = (1 5 3). d) τ −1 = (1 6 5 4 3). e) First we find σ −1 = (1 5 3)(2 6). Using our solution from part a) we get, στ σ −1 = (1 5 2 6 3 4)(1 5 3)(2 6) = (1 2 3 5 4).

Exercise: 5 Section 3.4 Question: In S7 , with σ = (1 4)(2 6)(3 5 7) and τ = (1 6 7), calculate: a) στ ; b) τ σ; c) τ −1 σ 2 ; d) στ σ −1 . Solution: a) στ = (1 2 6 3 5 7 4).

96

CHAPTER 3. GROUPS b) τ σ = (1 4 6 2 7 3 5). c) First we find τ −1 = (1 7 6), and then σ 2 = (3 7 5). Now we can find the complete solution, τ −1 σ 2 = (1 7 6)(3 7 5) = (1 7 5 3 6). d) First we find σ −1 = (1 4)(2 6)(3 7 5). Using our solution from part a we get, στ σ −1 = (1 2 6 3 5 7 4)(1 4)(2 6)(3 7 5) = (2 3 4).

Exercise: 6 Section 3.4 Question: List all the cycle types in S7 . Solution: All the cycle types are {1, (a b), (a b c), (a b)(c d), (a b c d), (a b)(c d e), (a b c d e), (a b c)(d e f ), (a b c d)(e f ), (a b)(c d)(e f ), (a b c d e f ), (a b c d)(e f g), (a b)(c d)(e f g), (a b c d e)(f g), (a b c d e f g)}.

Exercise: 7 Section 3.4 Question: Let σ = (a0 a1 a2 · · · am−1 ) be an m cycle in Sn . Prove that σ k (ai ) = a(i+k) mod m . Conclude that the order of σ is m. Solution: By definition of the cycle notation σ k (ai ) = σ k−1 (σ(ai )) = σ k−1 (ai+1 =σ

k−2

= = ai+k

(σ(ai+1 .. .

mod m )) = σ

mod m ) k−2

(ai+2

mod m )

mod m .

Exercise: 8 Section 3.4 Question: Let σ be an m-cycle in Sn . Prove that σ k is an m-cycle if and only if gcd(k, m) = 1. Solution: (=⇒) Suppose that σ k is an m-cycle. By Proposition 3.3.7, we know that |σ k | = |σ|/ gcd(k, m). Since both σ and σ k are m-cycles, we also know that |σ k | = m and |σ| = m. Therefore the previous equation becomes m = m/ gcd(k, m), which implies that gcd(k, m) = 1. (⇐=) Now we will suppose that gcd(k, m) = 1. We will be using the results of Exercise 4.7 where σ k (ai ) = a(i+k) mod m . Consider (σ k )n (ai ) = σ k σ k . . . σ k (ai ) = a(i+nk) mod m . Now, for any n and ai , if n is the smallest number where (σ k )n (ai ) = ai , this implies that there exists a cycle of size n that looks like (ai a(i+k) mod m . . . a(i+(n−1)k) mod m) . Therefore σ k is not an m-cycle if we can find an i and n such that i + nk = ī and n < m. This is equivalent to finding a n < m such that, nk = 0̄. The smallest possible value of such an n would be lcm(k, m)/k = m/ gcd(k, m) = m by our assumption. So this shows that there is no n < m and i such that i + nk = ī which implies that σ k is an m-cycle. This completes the proof of both directions of the if and only if statement. Exercise: 9 Section 3.4 Question: Suppose that ai are distinct positive integers of i = 1, 2, . . . , m. Prove that (a1 a2 a3 · · · am ) = (a1 am ) · · · (a1 a4 )(a1 a3 )(a1 a2 ). Use this to show that an m-cycle is even if and only if m is odd. Solution: First we let σi = (a1 ai ) and then we let σ = σm σm−1 . . . σ2 = (a1 am ) · · · (a1 a3 )(a1 a2 ). Then for any σi and aj :   a1 if i = j σi (aj ) = ai if j = 1   aj elsewise.

3.4. SYMMETRIC GROUPS

97

Now, by the lefthand side of the original equation we can recognize that a1 is the lowest positive integer since it comes first in the cycle. So σ = (a1 . . .?). Now σ(a1 ) = σm σm−1 . . . σ3 σ2 (a1 ) = σm σm−1 . . . σ3 (a2 ) = a2 since the remaining σi ’s have no effect. So then σ = (a1 a2 . . .?). Now we will show that σ(ak ) = ak+1 where 1 < k ≤ (m − 1). Now σ(ak ) = σm σm−1 . . . σ2 (ak ). Now, since k > 1 we know that while i < k, σi (ak ) = ak . Then the equation reduces to σm σm−1 . . . σk+1 σk (ak ) = σm σm−1 . . . σk+2 σk+1 (a1 ) = σm σm−1 . . . σk+2 (ak+1 ). And the resulting σi ’s have no effect and we end with σ(ak ) = ak+1 . We can now fill in most of the permutation, σ = (a1 a2 a3 . . . am ?) and we just need to prove that σ(am ) = a1 . Since the first m − 1 σ’s have no impact, σ(am ) = σm (am ) = a1 . So that σ = (a1 am ) · · · (a1 a4 )(a1 a3 )(a1 a2 ) = (a1 a2 a3 . . . am ) and this proves the equality. By Theorem 3.4.16, if an m-cycle is an even permutation, this implies that the m-cycle can only be written as an even number of transpositions. Since we can write the m-cycle as m − 1 transpositions, m − 1 must be even and so m must be odd. If m is odd, then we can write the m-cycle as m − 1 tranpositions which is an even number and so by Theorem 3.4.16, the m-cycle must be an even permutation. Exercise: 10 Section 3.4 Question: Suppose that ai are distinct positive integers of i = 1, 2, . . . , m. Prove that (a1 a2 a3 · · · am ) = (a1 a2 ) · · · (am−2 am−1 )(am−1 am ). Use this to show that an m-cycle is even if and only if m is odd. Solution: Reading the product of transpositions σ = (a1 a2 ) · · · (am−2 am−1 )(am−1 am ) from right to left, a1 is only affected only by the last transposition and σ(a1 ) = a2 . Next, reading σ from right to left, the first transposition to affect a2 is the second to last transposition and maps a2 to a3 . Then (a1 a2 ) does not affect a3 so σ(a2 ) = a3 . More generally, if i < m, the first transposition to affect ai is (ai ai+1 ) and it maps ai to ai+1 . Then no other transpositions affect ai+1 . Hence σ(ai ) = ai+1 . Finally, reading σ from right to left, we see that (am−1 am ) maps am to am−1 ; then (am−2 am−1 ) maps am−1 to am−2 ; and so forth down to a1 . Hence σ(am ) = a1 . Thus, we have shown that (a1 a2 ) · · · (am−2 am−1 )(am−1 am ) = (a1 a2 a3 · · · am ).

Exercise: 11 Section 3.4 Question: Use Exercise 3.4.9 or Excercise 3.4.10 to show that every permutation can be written as a product of transpositions. Solution: Consider a permutation written in standard cycle notation. For every cycle that is written, decompose it in the way shown by either Exercise 3.4.9 or Exercise 3.4.10. Then all that remains is transpositions and we have shown it to be true. Exercise: 12 Section 3.4 Question: Describe the Shell Game using the concept of products of transpositions in S3 . Solution: In the Shell Game, an object is hidden under one of three shells, which are then rearranged in a deceptive manner, and the player must guess which position the shell is in. The initial position of the object can be labeled 1, 2, or 3. The rearrangement of the shells corresponds to a product of transpositions in S3 because the person rearranging has only two hands, so he can only swap two shells at once. Therefore the entire rearrangement comes down to a product of transpositions of shells. For example, if first shells 1 and 2 were swapped, then 1 and 3, then 2 and 3, then 1 and 3, this could be represented by the product (1 3)(2 3)(1 3)(1 2), which simplifies to (1 3 2). Therefore the final position of the object can be obtained by applying the permutation (1 3 2) to the initial position of the object. Exercise: 13 Section 3.4 Question: Let σ be an m-cycle and suppose that d < m divides m. Prove that the standard cycle notation of σ d is the product of d disjoint cycles of length m d. Solution: Let σ = (a1 a2 . . . am ). By Excercise 3.4.7, we know that σ d (ai ) = a(i+d) mod m . For an arbitrary i, we will examine the cycle that ai is a member of in σ d , (ai σ d (ai ) (σ d )2 (ai ) . . . (σ d )k−1 (ai )) where k is the smallest positive integer such that (σ d )k (ai ) = ai . From above we know that k is also the smallest positive

98

CHAPTER 3. GROUPS

integer where a(i+kd) mod m = ai which implies i + kd = ī or kd = 0̄. Since d divides m, the smallest such value for k is m/d. Since this also represents the cycle length, for any i, ai is a member of a cycle of length m/d. Since there are m different ai ’s we must have d cycles of length m/d. Since the cycles are constructed from iteratively applying the bijection σ d , if some ai were a member of two different cycles, this would imply that σ d was not a bijection! So we have d disjoint cycles of length m/d. Exercise: 14 Section 3.4 Question: What is the highest possible order for permutations in S11 ? Illustrate your answer with a specific element having that order. Solution: The highest possible order for permutations in S11 is 30, there are many examples but they should all come in one of the following formats: {(a b c d e)(f g h i j k ), (a b)(c d e)(f g h i j)}. A concrete example is (1 2 3)(4 5)(6 7 8 9 10). Exercise: 15 Section 3.4 Question: What is the highest possible order of an element in each of the following groups. Illustrate with a specific element. a) S5 ⊕ D11 b) S5 ⊕ S5 Solution: a) The highest possible order of an element is 66. One possibility is (1 2 3)(4 5), r . b) The highest possible order of an element is 30. One possibility is (1 4)(2 5 3), (1 5 4 2 3) . Exercise: 16 Section 3.4 Question: What is the highest possible order of an element in S7 ⊕ S7 ⊕ S7 ? Illustrate your answer with a specific element having that order. Solution: It is possible to show (another exercise) that the order of (g1 , g2 , . . . , gn ) in G1 ⊕ G2 ⊕ · · · ⊕ Gn is lcm(|g1 |, |g2 |, . . . , |gn |). The orders of non-identity elements in S7 are (when written in reduce cycle notation): • type (a b): order 2; • type (a b)(c d): order 2 • type (a b)(c d)(e f ): order 2 • type (a b c): order 3; • type (a b c)(d e f ): order 3; • type (a b c d): order 4; • type (a b c d)(e f ): order 4; • type (a b c d e): order 5; • type (a b c d e f ): order 6; • type (a b c)(d e): order 6; • type (a b c)(d e)(f g): order 6; • type (a b c d e f g): order 7; • type (a b c d e)(f g): order 10; • type (a b c d)(e f g): order 12.

3.4. SYMMETRIC GROUPS

99

We get an element (σ1 , σ2 , σ3 ) of highest order in S7 ⊕ S7 ⊕ S7 by taking one of the σi to be of type (a b c d)(e f g), another to be of type (a b c d e)(f g), and the third to be of type (a b c d e f g). The order will be lcm(12, 10, 7) = 420. We can confirm this is the highest possible order by noticing that 420 is the least common multiple of all the orders of elements in S7 . Exercise: 17 Section 3.4 Question: Prove that a permutation σ ∈ Sn satisfies σ −1 = σ if and only if σ is the identity or, in standard cycle notation, consists of a product of disjoint 2-cycles. Solution: We will start by writing out σ as a composition of disjoint cycles, σ1 σ2 . . . σm . If σ −1 = σ ⇒ σ 2 = 1. In terms of the disjoint cycles, σ 2 = (σ1 σ2 . . . σm )(σ1 σ2 . . . σm ) and since disjoint cycles commute this is equal to 2 σ12 σ22 . . . σm = 1. Since the inverse of a cycle cannot be disjoint to the original cycle, we must have σi2 = 1 for all i. This implies that either σi is a 2-cycle or 1. Which shows that σ is either the composition of disjoint 2-cycles or 1. Exercise: 18 Section 3.4 Question: Prove Proposition 3.4.7. [Hint: Use Exercise 3.4.7] Solution: Let σ ∈ Sn . We first write σ = σ1 σ2 . . . σn in standard cycle notation where all the σi ’s are disjoint m-cycles. Let si be the length of σi and k = lcm(s1 , s2 , . . . sn ). Consider σ k = (σ1 σ2 . . . σn )k and since disjoint cycles commute this equals σ1k σ2k . . . σnk . For any σi , we know from Exercise 3.4.7 that σisi = 1. Since k is a multiple of every si we have σ1k σ2k . . . σnk = 1. Suppose that another l < k satisfied σ l = 1. Since this would imply that σil = 1 for every i. Then l would have to be a multiple of all of the lengths of the disjoint cycles and therefore would be a multiple of k contradicting the fact that l < k. This show that the least common multiple of all of the lengths of the disjoint cycles of the standard cycle notation of σ is the order of σ. Exercise: 19 Section 3.4 Question: In some Sn , find two elements σ and τ such that |σ| = |τ | = 2 and |στ | = 3. Solution: There are many solutions. We will work in S3 . Consider the elements σ = (1 2) and τ = (1 3). Since both are 2-cycles, |σ| = |τ | = 2. |στ | = |(1 2)(1 3)| = |(1 3 2)| = 3 since it is a 3-cycle. Exercise: 20 Section 3.4 Question: In some Sn , find two elements σ and τ such that |σ| = |τ | = 2 and |στ | = 4. Solution: There are many possible solutions. We will work in S4 . Consider σ = (1 2)(3 4) and τ = (2 3). Since both σ and τ are made up of disjoint 2-cycles, we know that |σ| = |τ | = 2. Consider στ = (1 2)(3 4)(2 3) = (1 2 4 3). Since στ is a 4-cycle, |στ | = 4. Exercise: 21 Section 3.4 Question: How many permutations of order 4 does S7 have? Solution: First, we identify the cycle types that have order 4. They are {(a b c d), (a b c d)(e f )}. Now we will count the possibilities. For a cycle of type (a b c d), we must choose 4 numbers out of 7, then the lowest gets put first and the remaining three can be ordered in 3! = 6 ways. So we have 74 · 6 = 210 possibilities. For a cycle of type (a b c d)(e f ), we still have 74 · 6 possible ways to form the 4-cycle. Then we must choose 2 out of the remaining 3 numbers and they can only be put in the transposition in one manner. So we have 32 possibilities for the 2-cycle. In total we have 74 · 6 · 32 = (35)(6)(3) = 630 possibilities. So the total number of permutations of order 4 is the sum, 630 + 210 = 840 (Note that this is out of the 7! = 5, 040 possibilities of S7 ). Exercise: 22 Section 3.4 Question: How many even permutations of order 5 does S8 have? odd permutations? Solution: Consider some σ of order 5, where in standard cycle notation σ = σ1 σ2 · · · σn . Then, if si = |σi |, |σ| = 5 = lcm(s1 , s2 , . . . , sn ) which implies that si = 1 or 5 and there must be at least one cycle of length 5. Therefore the only cycle type of order 5 has the form (a b c d e). By Exercise 3.4.9, every 5-cycle (and every permutation of order 5) is an even permutation since 5 is odd. Now we just count all of the 5-cycles. We first choose 5 different numbers out of 8, then the lowest goes first and we can order the remaining four in 4! = 24 different ways. So there are 85 · 24 = 56 · 24 = 1344 even permutations of order 5 and 0 odd permutations of order 5.

100

CHAPTER 3. GROUPS

Exercise: 23 Section 3.4 Question: Suppose that m ≤ n. Prove that the number of m-cycles in Sn is n(n − 1)(n − 2) · · · (n − m + 1) . m n Solution: First we choose m different numbers out of n possibilities or m . Then we are forced to put the lowest first by the rules of standard cycle notation, and we can order the remaining m − 1 numbers in (m − 1)! n different ways. So the total number of m-cycles is m · (m − 1)! = n(n−1)...(n−m+1) · (m − 1)! = n(n−1)...(n−m+1) . m(m−1)! m

Exercise: 24 Section 3.4 Question: Suppose that n ≥ 4. Prove that the number of permutations of cycle type (ab)(cd) in Sn is n(n − 1)(n − 2)(n − 3) . 8 Solution: First we choose 4 different numbers out of the n possibilities, or n4 . Then we set the lowest number as a in our cycle and we choose one number out of the remaining 3 to be in the transposition with a, or 31 . After that, the remaining two go in the other not matter. So the total number of transposition and order does n(n−1)(n−2)(n−3) permutations of cycle type (ab)(cd) is n4 · 31 = n(n−1)(n−2)(n−3) · 3 = . 4·3·2·1 8 Exercise: 25 Section 3.4 Question: Show that the function f : Z/10Z → Z/10Z defined by f (a) = a3 is a permutation on Z/10Z and write f in cycle notation as an element of S10 . (Use the bijection g : {1, 2, . . . , 10} → Z/10Z with g(a) = a to set up numerical labels for elements in Z/10Z.) Solution: We will utilize the bijection g : {1, 2, . . . , 10} → Z/10Z with g(a) = a. We will work in S10 and define σ = g −1 ◦ f ◦ g : S10 → S10 , we will show that σ ∈ S10 by showing σ is a bijection through observation: {σ(1) = 1, σ(2) = 8, σ(3) = 7, σ(4) = 4, σ(5) = 5, σ(6) = 6, σ(7) = 3, σ(8) = 2, σ(9) = 9, σ(10) = 10}. So σ is a bijection. Since σ = g −1 ◦ f ◦ g, we can solve for f by g ◦ σ ◦ g −1 = f . Since f is a composition of bijections, this implies that f is also a bijection, which shows that f is a permutation on Z/10Z. Based on our list of where sigma sends elements of {1, 2, . . . 10}, we can write out the cycle in standard cycle notation, σ = (2 8)(3 7). Exercise: 26 Section 3.4 Question: Show that the function f : Z/11Z → Z/11Z defined by f (a) = a3 is a permutation on Z/11Z and write f in cycle notation as an element of S11 . (Use the bijection g : {1, 2, . . . , 11} → Z/11Z with g(a) = a to set up numerical labels for elements in Z/11Z.) Solution: We will utilize the bijection g : {1, 2, . . . , 11} → Z/11Z with g(a) = a. We will work in S11 and define σ = g −1 ◦ f ◦ g, we will show that σ ∈ S11 by showing σ is a bijection through observation: {σ(1) = 1, σ(2) = 8, σ(3) = 5, σ(4) = 9, σ(5) = 4, σ(6) = 7, σ(7) = 2, σ(8) = 6, σ(9) = 3, σ(10) = 10, σ(11) = 11}. So σ is a bijection. Since σ = g −1 ◦ f ◦ g, we can solve for f by g ◦ σ ◦ g −1 = f . Since f is a composition of bijections, this implies that f is also a bijection, which shows that f is a permutation on Z/11Z. Based on our list of where sigma sends elements of {1, 2, . . . 11}, we can write out the cycle in standard cycle notation, σ = (2 8 6 7)(3 5 4 9). Exercise: 27 Section 3.4 Question: Calculate the set {|σ| σ ∈ S7 }, i.e., the set of orders of elements in S7 . Solution: Let σ ∈ S7 and write σ = σ1 σ2 · · · σn in standard cycle notation and define si = |σi |. We know that |σ| = lcm(s1 , s2 , · · · , sn ) and that s1 + s2 + · · · + sn ≤ 7. So we seek combinations of the number {2, 3, 4, 5, 6, 7} (since 1 does nothing to change the order of σ). We will also impose the condition that si > si+1 to avoid counting orders twice. So we have {lcm(7), lcm(6), lcm(5), lcm(5, 2), lcm(4), lcm(4, 3), lcm(3), lcm(3, 2), lcm(2)} = {7, 6, 5, 10, 4, 12, 3, 2} = {2, 3, 4, 5, 6, 7, 10, 12}.

3.4. SYMMETRIC GROUPS

101

Exercise: 28 Section 3.4 Question: We work in the group Sn for n ≥ 3. Prove or disprove that if σ1 and σ2 have the same cycle-type and τ1 and τ2 have the same cycle-type, then σ1 τ1 has the same cycle-type as σ2 τ2 . Does your answer depend on n? Solution: This is not true. We will disprove it with a counter-example that is applicable for all n. Let σ1 = (1 2), σ2 = (1 2), τ1 = (1 2) and τ2 = (1 3). In our case, all the permutation are 2-cycles. Now σ1 τ1 = (1 2)(1 2) = 1 is a 1-cycle and σ2 τ2 = (1 2)(1 3) = (1 3 2) is a 3-cycle. So the statement is false. Exercise: 29 Section 3.4 Question: In a six contestant steeple race, the horses arrived in the order C, B, F, A, D, E. Suppose someone predicted they would arrive in the order, F, E, C, B, D, A. How many inversions are in the guessed ordering? Solution: We will first find the permutation in S6 that corresponds to the prediction. Consider the chart notation where the top column is the actual order and the bottom is the predicted order: C B F A D E σ= . F E C B D A Now we will rewrite the chart replacing the following letters with the corresponding numbers, (C, 1), (B, 2), (F, 3), (A, 4), (D, 5), (E, 6): 1 2 3 4 5 6 σ= . 3 6 1 2 5 4 Now we can write the permutation, σ = (1 3)(2 6 4). Now we calculate the amount of inversions: (i, j) (σ(i), σ(j)) (i, j) (σ(i), σ(j))

(1, 2) (3, 6)

(2, 5) (6, 5)

(1, 3) (3, 1)

(2, 6) (6, 4)

(1, 4) (3, 2)

(3, 4) (1, 2)

(1, 5) (3, 5)

(3, 5) (1, 5)

(1, 6) (3, 4)

(3, 6) (1, 4)

(2, 3) (6, 1)

(4, 5) (2, 5)

(2, 4) (6, 2)

(4, 6) (2, 4)

(5, 6) . (5, 4)

So inv(σ) = 7. Exercise: 30 Section 3.4 Question: In S5 , count the number of inversions of the following permutations: a) σ = (1 4 2 5); b) τ = (1 4 3)(2 5); c) ρ = (1 5)(2 3). Solution: a) For σ = (1 4 2 5) we have: (i, j) (σ(i), σ(j))

(1, 2) (4, 5)

(1, 3) (4, 3)

(1, 4) (4, 2)

(1, 5) (4, 1)

(2, 3) (5, 3)

(2, 4) (5, 2)

(2, 5) (5, 1)

(3, 4) (3, 2)

(3, 5) (3, 1)

(4, 5) (2, 1)

(1, 3) (4, 1)

(1, 4) (4, 3)

(1, 5) (4, 2)

(2, 3) (5, 1)

(2, 4) (5, 3)

(2, 5) (5, 2)

(3, 4) (1, 3)

(3, 5) (1, 2)

(4, 5) (3, 2)

(1, 3) (5, 2)

(1, 4) (5, 4)

(1, 5) (5, 1)

(2, 3) (3, 2)

(2, 4) (3, 4)

(2, 5) (3, 1)

(3, 4) (2, 4)

(3, 5) (2, 1)

(4, 5) (4, 1)

So inv(σ) = 9. b) For τ = (1 4 3)(2 5) we have: (i, j) (σ(i), σ(j))

(1, 2) (4, 5)

So inv(τ ) = 7. c) For ρ = (1 5)(2 3) we have: (i, j) (σ(i), σ(j)) So inv(ρ) = 8.

(1, 2) (5, 3)

102

CHAPTER 3. GROUPS

Exercise: 31 Section 3.4 Question: In S6 , count the number of inversions of the following permutations: a) σ = (1 3 5 6 2); b) τ = (1 6)(2 3 4); c) ρ = (1 3 5)(2 4 6). Solution: We create a table that shows how σ, τ and ρ act on the 15 pairs in T6 . From this, we will read off the count of how many inversions there are. (i, j) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6)

(σ(i), σ(j)) (3, 2) (3, 5) (3, 4) (3, 6) (3, 1) (1, 5) (1, 4) (1, 6) (1, 2) (5, 4) (5, 6) (5, 2) (4, 6) (4, 2) (6, 2)

(τ (i), τ (j)) (6, 3) (6, 4) (6, 2) (6, 5) (6, 1) (3, 4) (3, 2) (3, 5) (3, 1) (4, 2) (4, 5) (4, 1) (2, 5) (2, 1) (5, 1)

(ρ(i), ρ(j)) (3, 4) (3, 5) (3, 6) (3, 1) (3, 2) (4, 5) (4, 6) (4, 1) (4, 2) (5, 6) (5, 1) (5, 2) (6, 1) (6, 2) (1, 2)

By reading off this table, we see that inv(σ) = 6, that inv(τ ) = 11, and inv(ρ) = 8. Exercise: 32 Section 3.4 Question: Consider a 2-cycle in Sn of the form τ = (a b). Prove that inv(τ ) = 2(b − a) − 1. Solution: We consider the 2-cycle σ = (a b) (where we assume that a < b) and consider how it acts on the pairs (j, k) ∈ Tn (where by definition j < k). Note that if neither j nor k are equal to a or b, then (σ(j), σ(k)) = (j, k) so (a b) would never invert such pairs. We break the remaining situations into 5 mutually disjoint cases: • j < a and k = a. Then pairs of the form (j, a) ∈ Tn have (σ(j), σ(a)) = (j, b). Since j < b, none of these pairs in Tn lead to inversions. • j = a and a < k < b. Then pairs of the form (a, k) ∈ Tn have (σ(a), σ(k)) = (b, k). Since b > k, all of these lead to inversions. Since j is fixed, there are b − a − 1 integers k satisfying a < k < b. This situation accounts for b − a − 1 inversions. • j = a and k = b. This accounts for a single pair in Tn . We have (σ(a), σ(b)) = (b, a). Since b > a, this accounts for 1 inversion. • a < j < b and k = b. Then pairs of the form (j, b) ∈ Tn have (σ(j), σ(b)) = (j, a). Since j > a, all of these lead to inversions. Since k is fixed, there are b − a − 1 integers j satisfying a < j < b. This situation accounts for b − a − 1 inversions. • j = b and k > b. Then pairs of the form (b, k) ∈ Tn have (σ(b), σ(k)) = (a, k). Since a < b < k, none of these pairs in Tn lead to inversions. This approach gives us a stronger result that what was asked for. Namely the exact number of inversions of (a b) is inv((a b)) = 2(b − a − 1) + 1 = 2(b − a) − 1.

Exercise: 33 Section 3.4 Question: Let A be an n × n matrix and let σ ∈ Sn . Suppose that A0 (respectively A00 ) is the matrix

3.4. SYMMETRIC GROUPS

103

obtained from A by permuting the columns (respectively rows) of A according to the permutation σ. Prove that det(A0 ) = det(A00 ) = sign(σ) det(A). Solution: Let A be an n × n matrix and let σ ∈ Sn . We know from the Laplace expansion (discussed in linear algebra) formula to obtain the determinant of a matrix, that if A0 is obtained from A by interchanging two adjacent rows, then det(A0 ) = − det(A). This accounts for transpositions of the form (a a + 1) for 1 ≤ a ≤ n − 1. However, it is easy to check that if a < b, then (a b) = (b − 1 b) · · · (a + 1 a + 2)(a a + 1)(a + 1 a + 2) · · · (b − 1 b). Hence, every transposition can be expressed as an product of an odd number of transpositions of the form (a a + 1). Hence, any transposition acting on a matrix changes its determinant by a sign. Every permutation σ can be expressed as a product of transpositions σ = τ1 τ2 · · · τk . If A0 is obtained from A by permuting the columns by σ, then det(A0 ) = (−1)k det(A) = sign(σ) det(A). Similarly for A00 . Exercise: 34 Section 3.4 Question: Prove that the sum of inversions of all the permutations in Sn is n!n(n − 1)/4. In other words, prove that X n!n(n − 1) . inv(σ) = 4 σ∈Sn

Solution: We prove by induction on n that X

inv(σ) =

σ∈Sn

n!n(n − 1) . 4

If n = 2, then S2 contains only two permutations, namely id and (1 2). So inv(id) + inv((1 2)) = 0 + 1. Furthermore, when n = 2 we have n!n(n − 1)/4 = 2!2 · 1/4 = 1. Hence, the formula holds for n = 2. Now suppose that the formula works for some n ≥ 2. We consider permutations in Sn+1 . Note that |Tn | = n(n−1)/2 but that Tn+1 adds to Tn pairs of the form (a, n+1), where 1 ≤ a ≤ n. Hence, |Tn+1 | = |Tn |+n. We partition Sn+1 into subsets based on where the permutations send n+1. Call Ak = {σ ∈ Sn+1 | σ(n+1) = k}. Note that |Ak | = n! for all k. • If k = n + 1, then all the permutations in Ai fix n + 1 and the sum of all the inversions of permutations in P An+1 is σ∈Sn inv(σ) = n!n(n−1) by the induction hypothesis. 4 • If k = n, then the sum of all the inversions on the pairs (i, j) ∈ Tn+1 with i < j < n + 1 is still n!n(n−1) 4 but for each σ ∈ An , there is only one inversion on the pairs (i, n + 1), so this adds one inversion for each P n! element in An . Hence, An contributes n!n(n−1) + n! inversions to the sum σ∈Sn+1 inv(σ). 4 • More generally, for any k, then the sum of all the inversions on the pairs (i, j) ∈ Tn+1 with i < j < n + 1 but for each σ ∈ Ak , there is exactly n +P 1 − k inversion on the pairs (i, n + 1). Hence, this is still n!n(n−1) 4 adds an additional (n + 1 − k)n! inversions to the sum σ∈Sn+1 inv(σ). Consequently, we have X

inv(σ) =

σ∈Sn+1

n+1 X k=1

=

n!n(n − 1) n!(n + 1)n(n − 1) n(n + 1) + kn! = + n! 4 4 2

n!n(n + 1)(n − 1 + 2) (n + 1)!(n + 1)n = . 4 4

Consequently, we have established the induction step. Hence, by induction for all n, X σ∈Sn

inv(σ) =

n!n(n − 1) . 4

Exercise: 35 Section 3.4 Question: Show by example that a permutation can be written in more than one way as a product of

104

CHAPTER 3. GROUPS

transpositions. Prove that if σ = τ1 τ2 · · · τm and σ = ε1 ε2 · · · εn are two different expressions of σ as a product of transpositions, then m and n have the same parity. Solution: As an example, consider the element (1 2 3 4) in S4 that has the following different expressions (1 4)(1 3)(1 2)

(1 2)(2 3)(3 4)

(1 4)(2 4)(1 2)(1 4)(2 3).

Suppose that σ = τ1 τ2 · · · τm . Then sign(σ) = (−1)inv(τ1 τ2 ···τm ) = (−1)inv(τ1 )+inv(τ2 )+···+inv(τm ) by Proposition 3.4.13. Since inv(τi ) is odd by Exercise 3.4.32, we deduce that sign(σ) = (−1)m . Hence, if σ = ε1 ε2 · · · εn is another expression of σ as a product of transpositions, then (−1)n = sign(σ) = (−1)m . Therefore, m and n have the same parity. Exercise: 36 Section 3.4 Question: Show that for all σ ∈ Sn , the number of inversions inv(σ −1 ) = inv(σ). Conclude that σ and σ −1 have the same parity. Solution: For a given permutation σ, let A and B be a partition of Tn according to Aσ = {(i, j) ∈ Tn | σ(i) > σ(j)} Bσ = {(i, j) ∈ Tn | σ(i) < σ(j)}. By definition, inv(σ) = |Aσ |. Define σ(Aσ ) as {(σ(j), σ(i) | (i, j) ∈ Aσ } and σ(Bσ ) as {(σ(i), σ(j) | (i, j) ∈ Bσ }. These sets are defined asymmetrically so that the pairs in Aσ and in Bσ are in increasing order. We know that {σ(Aσ ), σ(Bσ )} is also a partition of Tn . Furthermore, |σ(Aσ )| = |Aσ | and |σ(Bσ )| = |Bσ |. Applying σ −1 to a pair (σ(j), σ(i)) in σ(Aσ ) gives (j, i), where j > i. Hence, σ −1 inverts all the pairs in σ(Aσ ). On the other hand, applying σ −1 to a pair (σ(i), σ(j)) in σ(Bσ ) gives (i, j), where i < j. Hence, σ −1 does not invert any pair in σ(Bσ ). Thus inv(σ −1 ) = |σ(Aσ )| = |Aσ | = inv(σ). Exercise: 37 Section 3.4 Question: Show that for all σ, τ ∈ Sn , the element στ σ −1 has the same parity as τ . Solution: The parity of a permutation τ depends on the sign of (−1)inv(τ ) , odd if the sign is −1 and even if the sign is 1. But sign(στ σ −1 ) = (−1)inv(στ σ

−1

)

= (−1)inv(σ)+inv(τ )+inv(σ

= (−1)inv(σ)+inv(τ )+inv(σ)

−1

)

by the previous exercise

= (−1)inv(τ ) = sign(τ ).

Hence, στ σ −1 has the same parity as τ .

3.5 – Subgroups Exercise: 1 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = Z with addition and A is the multiples of 5. Solution: Let x, y ∈ A. Then there exist n, m ∈ Z such that x = 5n, y = 5m. We will use the One-Step Subgroup Criterion. Clearly A is nonempty; for example, 5 = 5(1) ∈ A. Notice that the inverse of y in A is −y = −5m. Then x − y = 5n − 5m = 5(n − m). Since Z is closed under subtraction, n − m ∈ Z, and x − y is a multiple of 5. Therefore x − y ∈ A and A is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 2 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = (Q, +) and A is the set of rational numbers with odd denominators (when written in reduced form).

3.5. SUBGROUPS

105

Solution: Let xa , yb ∈ A. Then a, b, x, y ∈ Z, x and y are odd, and the fractions are in reduced form. We will use the One-Step Subgroup Criterion. Clearly A is nonempty; for example, 13 ∈ A. Notice that the inverse of yb in A is − yb . Then xa − yb = ay−bx xy . Since the product of two odd integers is odd, xy is odd, which means it has no factor of 2. Therefore, even after reducing the fraction, there will be no factor of 2 in the denominator, causing the denominator to remain odd. Therefore xa − yb ∈ A and A is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 3 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = (Q∗ , ×) 2 and A is the set of rational numbers of the form pq2 . Solution:

2

2

Let pq2 , rs2 ∈ A. Then p, q, r, s ∈ Z. We will use the One-Step Subgroup Criterion. Clearly A is 2

2

2

2

2

2

2

2

nonempty; for example, 94 = 232 ∈ A. Notice that the inverse of rs2 in A is rs2 . Then pq2 ( rs2 )−1 = pq2 rs2 = (ps) (qr)2 . 2

2

Since Z is closed under multiplication, ps, qr ∈ Z. Therefore pq2 ( rs2 )−1 ∈ A and A is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 4 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = (C∗ , ×) and A = {a + ai | a ∈ R}. 1 can be found by multiplying top and Solution: Let a + ai ∈ A. Then a ∈ R. The inverse of a + ai in A, a+ai 1 a−ai a−ai 1−i 1 1 bottom by the conjugate of a + ai, a − ai. So a+ai a−ai = 2a2 = 2a = 2a − 2a i. But this number is not in A, so A is not even closed under taking inverses. Therefore A is not even a group on its own, let alone a subgroup of G. Exercise: 5 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = Z/12Z and A = {0, 4, 8}. Solution: We will use the Finite Subgroup Test. Clearly A is nonempty and finite; it is given explicitly as {0, 4, 8}. We can easily see that 0 + 0 = 0, 0 + 4 = 4, 0 + 8 = 8, 4 + 4 = 8, 4 + 8 = 0, 8 + 8 = 4 are all in A, so A is closed under the operation +. Therefore A is a subgroup of G by the Finite Subgroup Test. Exercise: 6 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = U (11) and A = {1, 2, 9, 10}. Solution: We will show that A is not a subgroup by a counterexample. Consider the product 2 × 2 = 4. The number 4 is clearly not an element of the set A. Therefore A is not even closed under the operation, so it cannot be a subgroup. Exercise: 7 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = S5 and A is the set of transpositions. Solution: We will show that A is not a subgroup by a counterexample. Consider the product (1 2)(1 3) = (1 2 3). The cycle (1 2 3) is clearly not an element of the set A, since it is a 3-cycle and not a transposition. Therefore A is not even closed under the operation, so it cannot be a subgroup. Exercise: 8 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = D6 and A = {ι, s, r2 , sr2 }. Solution: We will show that A is not a subgroup by a counterexample. Consider the composition r2 r2 = r4 . The symmetry r4 is distinct in D6 and is not an element of A. Therefore A is not even closed under the operation, so it cannot be a subgroup. Exercise: 9 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = D6 and A = {ι, s, r3 , sr3 }.

106

CHAPTER 3. GROUPS

Solution: We will use the Finite Subgroup Test. Clearly A is nonempty and finite; it is given explicitly as {ι, s, r3 , sr3 }. We can easily see that composing composing s with r3 and sr3 in either order merely switches from one to the other, composing r3 with sr3 in either order gives ι, and composing s, r3 , sr3 with themselves gives ι in every case. Also, ι is of course the identity symmetry. Therefore any composition within A stays in A, so A is closed under the operation. Therefore A is a subgroup of G by the Finite Subgroup Test. Exercise: 10 Section 3.5 Question:q Prove or disprove that the given subset A of the given group G is a subgroup, where G = (R, +) and A = { pq | pq ∈ Q>0 }. Solution: We will show that A is not a subgroup by a counterexample. Consider the composition r2 r2 = r4 . The symmetry r4 is distinct in D6 and is not an element of A. Therefore A is not even closed under the operation, so it cannot be a subgroup. Exercise: 11 Section 3.5 Question:q Prove or disprove that the given subset A of the given group G is a subgroup, where G = (R, +) and A = { pq | pq ∈ Q>0 }. q q Solution: We will show that A is not a subgroup by a counterexample. We observe that 1 = 11 and 2 = 41 are in A. But 1 − 2 = −1 is not the square root of any positive rational number and so is not an element of A. Therefore A is not even closed under the operation, so it cannot be a subgroup. Exercise: 12 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = U (30) and A = {1, 7, 13, 19}. 2

Solution: We remark at the outset that the order U (30) is φ(30) = 8. We have 7 = 49 = 19. Furthermore, 3 4 7 = 7 × 19 = 133 = 13. Finally, 7 = 7 × 13 = 91 = 1. Hence, we notice that A = h7i so A is a subgroup of U (30). Exercise: 13 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = GL2 (R) and A is the subset of matrices with integer coefficients. Solution: Since A contains the identity matrix A is a nonempty subset. Also, the product of two matrices with integer coefficients is again a matrix with integer coefficients. Hence A is closed under multiplication. However, we notice that −1 1 2 0 2 0 0 ∈A but = 2 ∈ / A. 0 1 0 1 0 1 Hence A is not closed under taking inverses so it is not a subgroup of G. Exercise: 14 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = SR , i.e., the set of bijections from R to R with the operation of composition, and A is the subset of functions f (x) = xp/q , where p and q are odd integers. Solution: Let f, g ∈ A. Then f , g are given by f (x) = xa/b , g(x) = xc/d , where a, b, c, d are odd integers. We will use the One-Step Subgroup Criterion. Clearly A is nonempty; for example, f given by f (x) = x3/5 is in A. Notice that the inverse of g in A is given by g −1 (x) = xd/c . Then (f ◦ g −1 )(x) = f (g −1 (x)) = f (xd/c ) = (xd/c )a/b = xad/bc . Since the product of two odd integers is odd, ad and bc are also odd integers. Therefore (f ◦ g −1 )(x) ∈ A and A is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 15 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = SR and let A = {f ∈ G | f (Z) ⊆ Z}. Solution: We will show that A is not a subgroup by a counterexample. Consider the element f given by f (x) = x2 . The image of f restricted to Z is f (Z) = {z 2 | z ∈ Z}, which is a subset of Z; therefore f ∈ A. But

3.5. SUBGROUPS

107

√ the inverse of f is given by f −1 (x) = x1/2 . But then f −1 (3) = 3, which is not in Z. Therefore the image of −1 f√ restricted to Z, f −1 (Z), cannot be a subset of Z, since at least one integer (3) is mapped to a non-integer ( 3). Therefore A is not even closed under taking inverses, so it cannot be a subgroup. Exercise: 16 Section 3.5 Question: Prove or disprove that the given subset A of the given group G is a subgroup, where G = SR and let A = {f ∈ G | f (Z) ⊆ Z}. Solution: We will use the definition of a subgroup. Clearly A is nonempty; for example, f : Z → Z given by f (x) = x is in A. Since Z ⊆ R, the set A is the symmetric group SZ , the group of all bijections Z → Z, and it is a subset of G = SR . Moreover, since A is a group itself, it is closed under the operation and under taking inverses. Therefore, A is a subgroup of G by the definition of a subgroup. Exercise: 17 Section 3.5 Question: Find an example to illustrate that the union of two subgroups is not necessarily another subgroup. Solution: Let G = D4 , the group of dihedral symmetries of a square. Let A be generated by sr2 , so 2 A = hsr i = {1, sr2 }. Let B be generated by r, so B = hri = {ι, r, r2 , r3 }. Because they are generated by elements in G, A and B are subgroups of G. The union of A and B is A ∪ B = {ι, r, r2 , r3 , r4 , sr2 }. But A ∪ B is not a subgroup because sr2 r = sr3 , which is not in A ∪ B, so A ∪ B is not closed under composition. Therefore, A ∪ B is the union of two subgroups but is not itself a subgroup. Exercise: 18 Section 3.5 Question: Let G be an abelian group that is not necessarily finite. Define the subset Tor(G) to be the subset of elements that have finite order. Prove that Tor(G) is a subgroup. [Tor(G) is called the torsion subgroup of G.] Solution: Let x, y ∈ Tor G. Then the orders of x and y are finite and are given by |x| = n, |y| = m. We will use the One-Step Subgroup Criterion. Clearly Tor G is nonempty; for example, e ∈ Tor G because its order is 1. We know that y m = e, so (y m )−1 = e−1 , y −m = e, and (y −1 )m = e. Therefore the order of y −1 is bounded above by m: |y −1 | ≤ m. We consider |xy −1 |. Since xn = e and y −m = e, surely (xy −1 )nm = xnm (y −1 )nm = (xn )m ((y −1 )m )n = em en = e, so |xy −1 | ≤ nm. Since the order of xy −1 is bounded by the product of two finite numbers, it is itself finite. Therefore xy −1 ∈ Tor G and Tor G is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 19 Section 3.5 Question: Prove that a finite group G with order greater than 2 cannot have a subgroup H with |H| = |G| − 1. Solution: Suppose that G has a subgroup H with |H| = |G| − 1. Let g be the unique element in G − H. We will prove this is impossible by cases. Suppose first that g is not its own inverse. Then g −1 ∈ H. But because H is closed under taking inverses, −1 −1 (g ) = g ∈ H also. This contradicts the possibility of a subgroup H of order |G| − 1. Suppose alternatively that g is its own inverse. Let x ∈ H. By associativity, (g · g) · x = g · (g · x). But g · g = e and g · x ∈ H, since x 6= g −1 = g. Call g · x = y, an element of H. Now we have e · x = x = g · y, which implies that g −1 = x−1 · y, or g = x−1 · y. But H is closed under taking inverses, so x−1 ∈ H, and H is closed under the operation, so x−1 · y ∈ H. But then g ∈ H. This also contradicts the possibility of a subgroup H of order |G| − 1. Exercise: 20 Section 3.5 Question: Let G1 and G2 be two groups. Prove that {(x, e2 ) | x ∈ G1 } and that {(e1 , y) | y ∈ G2 } are subgroups of G1 ⊕ G2 . Solution: We will use the One-Step Subgroup Criterion. Call the first set A and the second set B. Clearly both are nonempty; for example, (e1 , e2 ) is in both. −1 −1 Let (x1 , e2 ), (x2 , e2 ) ∈ A. The inverse of (x2 , e2 ) is (x−1 2 , e2 ), because (x2 , e2 ) · (x2 , e2 ) = (x2 x2 , e2 e2 ) = −1 −1 (e1 , e2 ). Now (x1 , e2 ) · (x2 , e2 ) = (x1 x2 , e2 ) is an element of A because G1 is closed under the operation and under taking inverses, so x1 x−1 2 ∈ G1 . Therefore A is a subgroup of G1 ⊕G2 by the One-Step Subgroup Criterion. Similarly, let (e1 , y1 ), (e1 , y2 ) ∈ B. The inverse of (e1 , y2 ) is (e1 , y2−1 ), because (e1 , y2 )·(e1 , y2−1 ) = (e1 e1 , y2 y2−1 ) = (e1 , e2 ). Now (e1 , y1 ) · (e1 , y2 −1) = (e1 , y1 y2−1 ) is an element of B because G2 is closed under the operation and

108

CHAPTER 3. GROUPS

under taking inverses, so y1 y2−1 ∈ G2 . Therefore B is a subgroup of G1 ⊕G2 by the One-Step Subgroup Criterion.

Exercise: 21 Section 3.5 Question: Determine all the finite subgroups of (R∗ , ×). Solution: The only finite subgroup of (R∗ , ×) is {1}. We will show this by contradiction. Suppose there exists a subgroup A of (R∗ , ×) that contains a nonzero real number r that is not 1. Then all powers of r — r, r2 , r3 , . . . — are also in A because A is closed under multiplication. But there are an infinite number of powers of r in R∗ ; therefore A is infinite. We reject the possibility of a finite subgroup containing an element other than 1 and conclude that the only finite subgroup of (R∗ , ×) is {1}. Exercise: 22 Section 3.5 Question: Let F be Q, R, C, or Fp (or more generally any field). Prove that the subset in GLn (F ) of upper triangular matrices is a subgroup. [Recall that a matrix is upper triangular if all the entries below the main diagonal are 0.] Solution: Let H be the set of matrices in GLn (F ) that are upper triangular. We can characterize an upper triangular matrix A = (aij ), where aij = 0 whenever i > j. The identity matrix is upper triangular so H 6= ∅. Let A, B ∈ H. Then A = (aij ) and B = (bij ) where aij = bij = 0 when i > j. The product C = AB has entries cij . Suppose that i > j. Then cij =

n X

aik bkj =

k=1

i−1 X

aik bkj +

k=1

n X

aik bkj

k=i

Now, since A is upper triangular, aik = 0 for all k < i. Also, in the second summation, since i > j, we have k ≥ i > j and since B is upper triangular, bkj = 0 (since k > j). Consequently, when i > j, cij =

i−1 X

0bkj +

k=1

n X

aik 0 = 0.

k=i

Thus C is also upper triangular. Finally, suppose that A ∈ H. According to a theorem in linear algebra, (which follows from Cramer’s Rule) the ijth entry of A−1 is det Ai (~ej ) , det A where Ai (~ej ) is the matrix A where the column i is replaced by the the vector ~ej , the vector of all 0s except for a 1 in the j’th coordinate. Suppose that i > j. Then Ai (e~j ) is upper triangular but with a 0 on the diagonal. The determinant of an upper diagonal matrix is the product of the diagonal elements. Hence det Ai (~ej ) = 0 whenever i > j. Thus A−1 is an upper triangular matrix. Hence H is closed under inverses. Thus H is a subgroup of GLn (F ). Exercise: 23 Section 3.5 Question: Let F be as in the previous exercise. Prove that the subset of GLn (F ) of diagonal matrices is an abelian subgroup. Explain why this abelian subgroup is strictly larger than Z(GLn (F )) Solution: Let A and B be diagonal matrices in GLn (F ). Then     a1 0 · · · 0 b1 0 · · · 0  0 a2 · · · 0   0 b2 · · · 0      A= . and B =  .  . . .. . . . , . .. . . ..   ..  .. . ..  . 0

0

···

0

···

bn

b1 a1   0    and BA =  ..   .

0 b2 a2 .. .

··· ··· .. .

0 0 .. .

0

0

···

bn an

an

0

so their products are 

a1 b1  0  AB =  .  ..

0 a2 b2 .. .

··· ··· .. .

0 0 .. .

0

0

···

an bn





    

3.5. SUBGROUPS

109

by the definition of matrix multiplication. We observe that the diagonal matrices in GLn (F ) are closed under multiplication. Moreover, since  −1  a1 0 ··· 0  0 a−1 ··· 0  2   A−1 =  . .. ..  , ..  .. . . .  0

0

···

a−1 n

the diagonal matrices in GLn (F ) are closed under taking inverses. With these two facts, the diagonal matrices in GLn (F ) satisfy the definition of a subgroup. But ai , bi ∈ F , and F is a field, so the ai and bi commute. Therefore ai bi = bi ai for all i, making AB = BA. Therefore the diagonal matrices in GLn (F ) are not only a subgroup but are also an abelian subgroup. Finally, the set of diagonal matrices in GLn (F ) is strictly larger than the center Z(GLn (F )) because the center consists of only the diagonal matrices that are uniform on the diagonal:   a 0 ··· 0  0 a ··· 0    Z(GLn (F )) = { . . . }. . . ...   .. ..  0

0

···

a

Exercise: 24 Section 3.5 Question: Let G be an abelian group. Prove that the following two subsets are subgroups a) {g n | g ∈ G}, where n is a fixed integer. b) {g ∈ G | g n = 1}, where n is a fixed integer. Solution: Let G be an abelian group. We can prove that the following subsets are subgroups using the One-Step Subgroup Criterion. a) Let n be a fixed integer and let H = {g n | g ∈ G}. Since 1n = 1 so 1 ∈ H. Hence H is nonempty. Let x, y ∈ H. Then there exist g, h ∈ G with x = g n and y = hn . Then xy −1 = g n (hn )−1 = g n (h−1 )n = gg · · · gh−1 h−1 · · · h−1 = gh−1 gh−1 · · · gh−1 , where that last equality holds because G is abelian. Thus xy −1 = (gh−1 )n , so xy −1 ∈ H. By the One-Step Subgroup Criterion, H ≤ G. b) let n be a fixed integer and let K = {g ∈ G | g n = 1}. Since 1n = 1, then 1 ∈ K so K is nonempty. Let x, y ∈ K. Then, xn = y n = 1. Hence, because G is abelian, (xy −1 )n = xn y −n = xn (y n )−1 = 1 · 1−1 = 1. Thus xy −1 ∈ K so K is a subgroup by the One-Step Subgroup Criterion. Exercise: 23 Section 3.5 Question: Let G be a group and let {Hi }i∈C be a collection of subgroups of G, not necessarily finite. Prove that the intersection \ Hi i∈C

is a subgroup of G. T Solution: Let x, y ∈ i∈C Hi . Then x, y ∈ Hi for all i ∈ C. We will use the One-Step Subgroup Criterion. Clearly A is nonempty; e must be in all Hi and therefore in the intersection. Consider xy −1 . Since y ∈ Hi for −1 all i ∈ C and the Hi are closed under taking inverses, yT ∈ Hi for all i ∈ T C. Moreover, since x is also in Hi for all i ∈ C, xy −1 ∈ Hi for all i ∈ C. But then xy −1 ∈ i∈C Hi . Therefore i∈C Hi is a subgroup of G by the One-Step Subgroup Criterion. Exercise: 26 Section 3.5 \ Question: Prove that Z(G) = CG (a). a∈G

110

CHAPTER 3. GROUPS

Solution: We will prove this equality by double set inclusion. Recall that the center of G, Z(G) is given by −1 Z(G) = {g ∈ G | gag −1 = x T for all a ∈ G} and the centralizer T of a in G is given by CG (a) = {g ∈ G | gag = a}. First we will show that a∈G CG (a) ⊆ Z(G). Let g ∈ a∈G C (a). Then g ∈ C (a) for all a ∈ G. But then G T G gag −1 = a for all a ∈ G, meaning thatTg ∈ Z(G). Therefore a∈G CG (a) ⊆ Z(G). −1 Now we will show that Z(G) ⊆ a∈G T CG (a). Let g ∈ Z(G). Then gag = a for all a ∈ G. But then T g ∈ CG (a)Tfor all a ∈ G, meaning that g ∈ a∈G C (a). Therefore Z(G) ⊆ a∈G CG (a). T G Since a∈G CG (a) ⊆ Z(G) and Z(G) ⊆ a∈G CG (a), we conclude that the two sets are equal: Z(G) = \ CG (a). a∈G

Exercise: 27 Section 3.5 Question: Prove that the center Z(Dn ) of the dihedral group is {ι} if n is odd and {ι, rn/2 } if n is even. Solution: First consider elements of the form ri as candidates for the center. Rotations commute with rotations, so consider the product of ri with srj in both directions. We have ri srj = srj−i and srj ri = srj+i . For which i is j − i = j + i modulo n? It is for those i such that 2i = 0 modulo n; that is, i = 0 or i = n/2. So the only elements of the form ri that are in the center are r0 = ι and rn/2 — however, rn/2 exists only if n is even. Next consider elements of the form sri as candidates for the center. These elements do not commute with r, since rsri = sri−1 but sri r = sri+1 . Clearly sri−1 6= sri+1 unless i − 1 = i + 1 modulo n, but this is only true modulo 2, and the dihedral group is defined only for n ≥ 3. So no elements of the form sri can be in the center. Therefore, the center Z(Dn ) of the dihedral group is {ι} if n is odd and {ι, rn/2 } if n is even. Exercise: 28 Section 3.5 Question: Prove that for all n ≥ 3, the center of the symmetric group is Z(Sn ) = {id}. Solution: We will show that the center Z(Sn ) = {id} by cases. First, suppose there is a permutation (a1 a2 · · · ak ) of length greater than 2 in the center. It is impossible for (a1 a2 · · · ak ) to be in the center because it does not commute with (a1 a2 ), and elements in the center must commute with every element in the group. We observe that (a1 a2 )(a1 a2 · · · ak ) = (a2 a3 a4 · · · ak ), but (a1 a2 · · · ak )(a1 a2 ) = (a1 a3 · · · ak a2 ). Therefore the center contains no permutations of length greater than 2. Second, suppose there is a permutation (a1 a2 ) of length 2 in the center. Since n ≥ 3, there is another 2cycle (a1 a3 ). These two transpositions do not commute because (a1 a2 )(a1 a3 ) = (a1 a3 a2 ) but (a1 a3 )(a1 a2 ) = (a1 a2 a3 ). Therefore the center contains no permutations of length 2. Since the center contains no permutations of length 2 nor of length greater than 2, all permutations in the center are of length 1. But the only 1-cycle is the identity. Therefore Z(Sn ) = {id}. Exercise: 29 Section 3.5 Question: In the group Dn , calculate the centralizer and the normalizer for each of the subsets a) {s},

b) {r},

and

c) {ι, r, r2 , . . . , rn−1 }.

Solution: (a) CDn (s) = {ι, rn/2 , s, srn/2 } if n is even and {ι, s} if n is odd; NDn (s) = {ι, rn/2 , s, srn/2 } if n is even and {ι, s} if n is odd (b) CDn (r) = {ι, r, r2 , r3 , . . . , rn−1 }; NDn (r) = {ι, r, r2 , r3 , . . . , rn−1 } (c) CDn ({ι, r, r2 , . . . , rn−1 }) = {ι, r, r2 , . . . , rn−1 }; NDn ({ι, r, r2 , r3 , . . . , rn−1 }) = Dn . Exercise: 30 Section 3.5 Question: For the given group G, find the CG (A) and NG (A) of the respective sets A. a) G = S3 and A = {(1 2 3)}. b) G = S5 and A = {(1 2 3)}. c) G = S4 and A = {(1 2)}. Solution: (a) CG (A) = {id, (1 2 3), (1 3 2)} and NG (A) = {id, (1 2 3), (1 3 2)}. (b) CG (A) = {id, (1 2 3), (1 3 2), (4 5), (1 2 3)(4 5), (1 3 2)(4 5)} and NG (A) = {id, (1 2 3), (1 3 2), (4 5), (1 2 3)(4 5), (1 3 2)(4 5)}. (c) CG (A) = {id, (1 2), (3 4), (1 2)(3 4) and NG (A) = {id, (1 2), (3 4), (1 2)(3 4). Exercise: 31 Section 3.5 Question: For the given group G, find the CG (A) and NG (A) of the respective sets A.

3.5. SUBGROUPS

111

a) G = D6 and A = {s, r2 }. b) G = Q8 and A = {i, j}. c) G = S4 and A = {(1 2), (3 4)}. Solution: Normalizers and centralizers: a) Let G = D6 and A = {s, r2 }. For the centralizer, CG (A), note that rk s = sr−k , so rk s = srk if and only if k ≡ −k (mod 6), hence only for k = 0 or 3. Also for any k, we have rk r2 = r2 rk so 1 and r3 ∈ CG (A). For the reflections, s(srk ) = (srk )s ⇐⇒ rk = r−k ⇐⇒ k = 0or3. and r2 (srk ) = (srk )r2 ⇐⇒ srk−2 = srk+2 , which is never true. Hence, this shows that CG (A) = {1, r3 }. For the normalizer, rk sr−k = sr−2k ∈ A if and only if 2k ≡ 0 (mod 6), so k = 0 or 3. Also, rk r2 r−k = r2 ∈ A so we have no new conditions. The only rotations in NG (A) are 1 and r3 . Also (srk )s(srk )−1 ∈ A ⇐⇒ srk sr−k s = sr2k ∈ A ⇐⇒ 2k ≡ 0

(mod 6)

and (srk )r2 (srk )−1 = srk r2 r−k s = r−2 ∈ / A. We find again that NG (A) = {1, r3 }. b) Let G = Q8 and A = {i, j}. Now 1 and −1 commute with everything so ±1 ∈ CG (A). On the other hand, ±i do not commute with j; ±j do not commute with i; and ±k do not commute with either i or j. Hence CG (A) = {1, −1}. For the normalizer, iji−1 = k(−i) = −j ∈ / A so i (and also −i) are not in NG (A). Similarly, ±j ∈ / NG (A). Finally kik −1 = j(−k) = −i so k ∈ / NG (A) and also −k ∈ / NG (A) so NG (A) = {1, −1}. c) Let G = S4 and A = {(1 2), (3 4)}. A permutation that commutes with (1 2) must be a disjoint permutation or a disjoint permutation composed with (1 2). Similarly for the permutations that commute with (3 4). Hence CG (A) = {1, (1 2), (3 4), (1 2)(3 4)}. For the normalizer, we know that CG (A) ≤ NG (A). However, there are more elements in the normalizer. For example, (1 3 2 4) satisfies (1 3 2 4)(1 2)(1 3 2 4)−1 = (1 3 2 4)(1 2)(1 4 2 3) = (3 4) and (1 3 2 4)(3 4)(1 3 2 4)−1 = (1 3 2 4)(3 4)(1 4 2 3) = (1 2) Thus (1 3 2 4) ∈ NG (A). Similarly (1 3 2 4)−1 = (1 4 2 3) ∈ NG (A). In NG (A), we also have (1 3 2 4)(1 2) = (1 4)(2 3)

and

(1 2)(1 3 2 4) = (1 3)(2 4).

(Note, we obtains the last three by virtue of NG (A) closed under composition.) So far, we have found 8 elements in NG (A). We can see that the normalizer is not all of G since (1 3)(1 2)(1 3)−1 = (2 3) ∈ / A. Without some theorems that will allows to answer tis question quickly in the future, it is simply time consuming to check all the possibilities but we have found them all. So in fact, NG (A) = h(1 3 2 4), (1 2)i. Exercise: 32 Section 3.5 Question: Prove the assertion in Example 3.5.14 that BEij is the matrix of zeros everywhere except for the jth row of B as its ith row and that Eij B is the matrix of zeros everywhere except for the ith column of B as its jth column. Solution: Recall that B, and Eij are n × n matrices as they are elements of GLn (F ). Let A = BEkl and consider the (k, l) entry in A, akl . Let bkl be the (k, l) entry in B and let ekl be the (k, l) entry in Eij . Observe that k and l range from 1 to n but i and j are fixed. We know that akl = Σnp=1 bkp epl , and epl is always 0 except when p = i, l = j. Therefore akl = 0 unless l = j, and akj = bki eij = bki · 1 = bki — that is to say, the jth column of A is the ith column of B. Now let A = Eij B, otherwise notating as before. The entries of A are akl = Σnp=1 ekp bpl , and similarly, ail = bjl — that is to say, the ith row of A is the jth row of B. Exercise: 33 Section 3.5 Question: Let V1 ⊆ V2 be subsets of a group G.

112

CHAPTER 3. GROUPS

a) Prove that CG (V2 ) ≤ CG (V1 ). b) Prove that NG (V2 ) ≤ NG (V1 ). Solution: Let g ∈ CG (V2 ). Then for all v ∈ V2 , gvg −1 = v. But V1 ⊆ V2 , so v ∈ V1 also. But then gvg −1 = v for all v ∈ V1 , meaning that g ∈ CG (V1 ). Since every element in CG (V2 ) is also in CG (V1 ), it follows that CG (V2 ) ⊆ CG (V1 ). Since centralizers are groups, we can strengthen this result to CG (V2 ) ≤ CG (V1 ). Now let g ∈ NG (V2 ). Then for all v ∈ V2 , there exists v 0 ∈ V2 such that gvg −1 = v 0 . But V1 ⊆ V2 , so v, v 0 ∈ V1 also. But then gvg −1 = v 0 for all v ∈ V1 , meaning that g ∈ NG (V1 ). Since every element in NG (V2 ) is also in NG (V1 ), it follows that NG (V2 ) ⊆ NG (V1 ). Since centralizers are groups, we can strengthen this result to NG (V2 ) ≤ NG (V1 ). Exercise: 34 Section 3.5 Question: Prove Proposition 3.8.18. Solution: Let G be a group and A be a subset of G; then NG (A) is the normalizer of A in G. We will prove that NG (A) is a subgroup of G by the definition of a subgroup. Let x, y ∈ NG (A). Then xax−1 ∈ A, yay −1 ∈ A for all a ∈ A. To show that NG (A) is closed under the group operation, consider xya(xy)−1 = xyay −1 x−1 = xa0 x−1 = a00 ∈ A for some a0 , a00 ∈ A. Therefore (xy)A(xy)−1 ⊆ A, so xy ∈ NG (A), so NG (A) is closed under the group operation. To show that NG (A) is closed under taking inverses, we will show that conjugation A → A is a bijection. suppose that xa1 x−1 = a0 and xa2 x−1 = a0 . Then xa1 x−1 = a0 = xa2 x−1 , and by inverse conjugation a1 = a2 . Therefore the conjugation map a 7→ xax−1 is injective from A to A. But then each a ∈ A maps to a unique a0 ∈ A, requiring that the range fill all of A. So the conjugation map is a bijection A → A. As such, it has an inverse A → A: a0 7→ x−1 ax. But this means that for all a ∈ A, x−1 ax = a0 , so x−1 Ax ⊆ A, so for any x ∈ NG (A), x−1 ∈ NG (A); this makes NG (A) closed under taking inverses. Since NG (A) is closed under the group operation and under taking inverses, NG (A) is a subgroup of G. Exercise: 35 Section 3.5 Question: In (Z, +), list all the elements in the subgroup h12, 20i. Solution: Recall that h12, 20i is the set of all words of the form n × 12 + m × 20 because (Z, +) is abelian. 1 α2 (Note also that the multiplicative notation sα 1 s2 is written as α1 × s1 + α2 × s2 under additive notation.) For each n ∈ Z, we can partition n into different equivalent sums. For example, 2 = 2 + 0 = 1 + 1 = 0 + 2. Each partition corresponds to an element in h12, 20i: 2 × 12 + 0 × 20, 1 × 12 + 1 × 20, 0 × 12 = 2 × 20. However, because n and m can be negative, we also need to consider any possible negation of each partition, leaving us with (2, 0), (−2, 0), (1, 1), (−1, 1), (1, −1), (−1, −1), (0, 2), (0, −2) as possible pairs of coefficients for 12 and 20. We can write all the elements in h12, 20i by writing all the elements corresponding to partitions of 1 (including negations), all the elements corresponding to partitions of 2 (including negations), and so on. At any rate we can write the pattern for as long as we desire without missing any elements. Therefore h12, 20i = {0, 12, −12, 20, −20, 24, −24, 32, 8, −8, −32, 40, −40, . . . }. Exercise: 36 Section 3.5 Question: In D8 , list all the element in the subgroup hsr2 , sr6 i. Solution: Since D8 is a finite group, it is not hard to calculate all words formed from sr2 and sr6 by hand. We find that hsr2 , sr6 i = {1, sr2 , sr6 , r4 }. Exercise: 37 Section 3.5

1 0

1 1 , 1 1

Question: In GL2 (F3 ), list all the elements in 1 1 1 0 Solution: The subgroup , is given by 0 1 1 1

0 . 1

1 0

1 1 , 1 1

0 1 , 1 0

2 1 , 1 1

1 1 , 2 1

2 2 , 0 2

0 2 , 2 2

2 0 , 1 2

1 , 2

0 2

1 2 , 1 1

1 2 , 1 1

0 2 , 2 2

1 1 , 0 0

0 2 , 1 1

2 1 , 0 2

2 1 , 2 2

0 , 1

3.5. SUBGROUPS 0 1

113 2 0 , 1 1

2 1 , 2 2

1 0 , 0 2

1 2 , 0 0

1 2 , 2 0

0 2 , 2 0

2 0 , 2 1

2 0

Observe that this is the special linear group SL2 (F3 ) – the group of 2 × 2 matrices with determinant 1. Exercise: 38 Section 3.5 Question: Let G be a group and let S ⊆ G be a subset. Let C be the collection of subgroups of G that contain S. Prove that \ hSi = H. H∈C

Solution: Let G be a group and let S ⊆ G be a subset. Let C be the collection of subgroups of G that contain S. WE call S the subset \ S= H. H∈C αn 1 α2 Now in every expression of the form sα 1 s2 · · · sn with si ∈ S and αi ∈ Z, the elements si ∈ H for all H ∈ C α1 α2 αn αn 1 α2 and since every H is a subgroup, s1 s2 · · · sn ∈ H. Consequently, sα 1 s2 · · · sn ∈ S so we have shown that

hSi ≤ S. Conversely, hSi is a subgroup of G that contains S and so hSi ∈ C. Thus S ≤ hSi. Together, this shows that S = hSi. Exercise: 39 Section 3.5 Question: Prove that if A ⊆ B are subsets in a group G, then hAi ≤ hBi. αn 1 α2 Solution: Recall that we define the subgroup generated by a subset S by hSi = {sα 1 s2 · · · sn | n ∈ N, si ∈ β β α1 α2 1 2 βm n S, αi ∈ Z}. Therefore hAi = {a1 a2 · · · aα n | n ∈ N, ai ∈ A, αi ∈ Z} and hBi = {b1 b2 · · · bn | m ∈ N, bi ∈ B, βi ∈ Z}. αn 1 α2 Let a ∈ hAi; then a = aα 1 a2 · · · an for particular ai and αi . Since A ⊆ B, all ai are in B. Therefore a ∈ B. But a is an arbitrary element in A, so in fact A ⊆ B. That A ≤ B follows because A and B are groups. Exercise: 40 Section 3.5 Question: Prove that in S4 , the subgroup h(1 2 3), (1 2)(3 4)i is A4 . Solution: According to Example 3.5.6, A4 = {id, (1 2 3), (1 2 4), (1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 3 4), (2 4 3), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. We immediately have id, (1 2 3), (1 2)(3 4). The inverse and square of (1 2 3) is (1 3 2), and the inverse of (1 2)(3 4) is itself, so its square is id. Multiplying (1 2)(3 4) on the left and right by (1 2 3) and (1 3 2) gives (1 3 4), (1 4 3), (2 3 4), (2 4 3). We further see that (1 2)(3 4)(1 3 4) = (1 4 2) and (1 2)(3 4)(2 3 4) = (1 2 4). Also, (1 2 3)(1 4 3) = (1 4)(2 3) and (1 2)(3 4)(1 4)(2 3) = (1 3)(2 4). This gives all 3-cycles and pairs of disjoint 2-cycles. It is not hard to see that only a 2-cycle can compose with a 3-cycle to yield a 4-cycle, only a 4-cycle can compose with a 4-cycle to yield a 2-cycle, and only a 4-cycle can compose with two disjoint 2-cycles to yield a 2-cycle. As we have only 3-cycles and pairs disjoint 2-cycles, we cannot generate 2-cycles or 4-cycles. Therefore we have exhibited the entire group, and it is equal to A4 . Exercise: 41 Section 3.5 Question: This exercise finds generating subsets of Sn . a) Prove that Sn is generated by {(1 2), (2 3), (3 4), . . . , (n − 1 n)}. b) Prove that Sn is generated by {(1 2), (1 3), . . . , (1 n)}. c) Prove that Sn is generated by {(1 2), (1 2 3 · · · n)}. d) Show that S4 is not generated by {(1 2), (1 3 2 4)}. Solution: Consider the following subsets of Sn . Recall that every permutation can be expressed by a product of transpositions. a) A = {(1 2), (2 3), (3 4), . . . , (n − 1 n)}. The transposition (a b) can be written as (a b) = (b − 1 b) · · · (a + 1 a + 2)(a a + 1)(a + 1 a + 2) · · · (b − 1 b). Therefore, every transposition can be obtain as a product of transpositions from A. Hence, every permutation can be obtained as a product of elements in A. Thus hAi = Sn .

114

CHAPTER 3. GROUPS

b) B = {(1 2), (1 3), . . . , (1 n)}. The transposition (a b) can be written as (1 a)(1 b)(1 a). Using a similar reasoning to the previous part, Sn = hBi. c) C = {(1 2), (1 2 3 · · · n)}. Notice that (1 2 3 · · · n)(1 2)(1 2 3 · · · n)−1 = (2 3) and more generally that (1 2 3 · · · n)i (1 2)(1 2 3 · · · n)−i = (1 + i 2 + i). Hence, hCi contains A from part (a), so hCi = Sn . d) If we label the vertices of a cube as follows 3

1

2

4

then the permutation (1 3 2 4) corresponds to a rotation of the square by π/2 and the permutation (1 2) gives a reflection through a line. Thus, hDi will be like a dihedral group and will contain only 8 elements as opposed to the 24 elements in S4 . Exercise: 42 Section 3.5 Question: Prove that for any prime number p, the symmetric group Sp is generated by any transposition and any p-cycle. Solution: Let p be a prime number and let σ, τ ∈ Sp such that |σ| = p and |τ | = 2. We call H = hσ, τ i. We will write τ = (a b) with a ≤ b. Note that since p is prime, then hσi = hσ k i for all k not divisible by p. Furthermore, σ k is another p-cycle if p - k. Without loss of generality, we can chose to write the p-cycles starting with a (instead of 1). Hence, there exists some integer k such that σ k = (a b c3 c4 · · · cp )ρ with c3 6= a, b. Furthermore, H = h(a b), ρi. We can call a = c1 and b = c2 . Then we find that for 0 ≤ k ≤ p − 1, ρk τ ρ−k = (c1+k c2+k ), where we replace 2 + k with 1 if k = p − 1. These transpositions are all in H. So if ρ = (c1 c2 · · · cp ), then H contains all transpositions (ci ci+1 ) for 1 ≤ i ≤ p − 1. Furthermore, suppose that i < j. Then (cj−1 cj ) · · · (ci+1 ci+2 )(ci ci+1 )(ci+1 ci+2 ) · · · (cj−1 , cj ) = (ci cj . Hence, H contains all transpositions of the form (ci , cj ) for any 1 ≤ i < j ≤ p. Since ρ contains all the integers in its cycle, these transpositions (ci , cj ) account for all transpositions in Sp . Thus, since the set of transpositions generate Sp , we deduce that H = Sp . Exercise: 43 Section 3.5 Question: Label the vertices of a tetrahedron with integers {1, 2, 3, 4}. Prove that the group of rigid motions of a tetrahedron is A4 . Solution: There are three kinds of rigid motions of a tetrahedron — the identity, rotations of a face around the opposite vertex, and three “twists” that swap pairs of vertices. By Example 3.5.6, the alternating group A4 consists of the identity, symmetries fixing one vertex, and symmetries swapping pairs of vertices. Therefore A4 is the group of rigid motions of a tetrahedron. Exercise: 44 Section 3.5 Question: Show that if p is prime, then in the alternating group Ap , we have Ap = h(1 2 3), (1 2 3 · · · p)i. Solution: Let p be an odd prime and consider the subgroup H = h(1 2 3), (1 2 3 · · · p)i in Ap . Note that (1 2 3 · · · p)(1 2 3)(1 2 3 · · · p)−1 = (2 3 4)

3.5. SUBGROUPS

115

and by taking other powers (1 2 3 · · · p)k (1 2 3)(1 2 3 · · · p)−k , we get all the 3-cycles (1 2 3), (2 3 4), (3 4 5), up to (p 1 2). The subgroup H also contains (3 4 5)(1 2 3)(3 4 5)−1 = (1 2 4) and (4 5 6)(1 2 4)(4 5 6)−1 = (1 2 5) and all 3-cycles of the form (1 2 a) for 3 ≤ a ≤ p. By doing (2 3 4)(1 2 3)(2 3 4)−1 = (1 3 4) and by the same reasoning as above, we can obtain all the 3-cycles of the form (1 3 a) where 4 ≤ a ≤ p. (Note that we already have (1 3 2) by virtue of (1 3 2) = (1 2 3)2 . Continuing in a similar fashion, we can obtain all 3-cycles. Now every even permutation sσ can be written as a finite product of an even number of transpositions, σ = τ1 τ2 · · · τ2k . Now if τ2i−1 τ2i consists of nondisjoint transpositions, we have τ2i−1 τ2i = (a b)(b c) = (a b c). If τ2i−1 τ2i consists of disjoint transpositions, say (a b)(c d), then τ2i−1 τ2i = (a c b)(a c d). Hence, every even permutation can be written as a finite product of 3-cycles. Thus, since H contains all 3-cycles, we deduce that H = Ap . Exercise: 45 Section 3.5 Question: Show that (R, +) is not finitely generated. Solution: If (R, +) were finitely generated, we could write any r ∈ R as r = c1 · r1 + c2 · r2 + · · · + cn · rn for a finite set {r1 , r2 , . . . , rn } ⊆ R and a finite set {c1 , c2 , . . . , cn } ⊆ Z. (This simplified linear combination form of Definition 3.5.19 is possible because (R, +) is abelian.) Suppose that there is such a finite set {r1 , r2 , . . . , rn } that generates (R, +). We will show that this set cannot even generate (Q, +), a subset of (R, +). First, the sum of an irrational number and a rational number is irrational, because otherwise the irrational number would be the difference of two rational numbers, which is rational. Second, the product of an integer and an irrational number remains irrational for similar reasons. These two facts mean that only the rational elements in {r1 , r2 , . . . , rn } can combine linearly to generate a rational number. Therefore consider the subset {q1 , q2 , . . . , qm } of rational numbers in {r1 , r2 , . . . , rn }. By Example 3.5.23, this set cannot generate Q. Therefore, no finite set {r1 , r2 , . . . , rn } can even generate (Q, +), let alone (R, +). Exercise: 46 Section 3.5 Question: Describe the elements in Tor(C∗ ), the torsion subgroup of C∗ . (See Exercise 3.5.18.) Show that Tor(C∗ ) is not finitely generated. Solution: The torsion subgroup Tor(C∗ ) is the set of all c ∈ C∗ with finite order. Therefore if c ∈ Tor(C∗ ), then cn = 1 for some n ∈ Z>0 , so the torsion subgroup of C∗ contains all the nth roots of unity for all n, as well as any product of roots of unity by closure. This means that Tor(C∗ ) is generated by the principal nth roots of unity for all n. Can any finite set of principal nth roots of unity generate all the principal nth roots of unity? Suppose some finite set of principal nth roots G = {ζn1 , ζn2 , . . . , ζnm } does generate Tor(C∗ ); we will show that G cannot generate ζp , where p is the first prime not in G. The argument of each ζni ∈ G is 2πki /ni for some 0 ≤ ki ≤ ni . Although G contains primitive roots of unity for which their is a particular ki , by closure we can choose any ki in the range. Consider a product of powers of elements of G in polar form: ei2πk1 /n1 ei2πk2 /n2 · · · ei2πkm /nm = ei(2πk1 /n1 +2πk2 /n2 +···+2πkm /nm ) = z. This can only be a pth root of unity if z p = 1. But if z p = 1, then ep·i(2πk1 /n1 +2πk2 /n2 +···+2πkm /nm ) = ei2πk , which is only possible if p clears all the denominators. (Note that ki ≤ ni for all i, so the denominators cannot be cleared by canceling with the respective numerators.) But then n1 , n2 , . . . , nm divide p — but p is prime, so this is impossible. Therefore we cannot generate a pth root of unity by any product of powers of elements in G. This proves that Tor(C∗ ) is not finitely generated. Exercise: 47 Section 3.5 Question: Let H and K be two subgroups of G. Prove that H ∪ K is a subgroup of G if and only if H ⊆ K or K ⊆ H. Solution: First, suppose that H ∪ K is a subgroup of G. Then for all x ∈ H ∪ K, x ∈ H and x ∈ K. Suppose H 6⊆ K and K 6⊆ K. Then there is h ∈ H such that h 6∈ K and there is k ∈ K such that k 6∈ H. Consider the product hk, which is in the union H ∪ K, meaning that either hk ∈ H or hk ∈ K — without loss of generality, take hk ∈ H. But h−1 ∈ H, and H is closed, so h−1 hk = k ∈ H, which contradicts our original assumption. Therefore we reject that H 6⊆ K and K 6⊆ K to find that either H ⊆ K or K ⊆ K.

116

CHAPTER 3. GROUPS

Second, suppose that H ⊆ K or K ⊆ H. Without loss of generality, let H ⊆ K. Then H ∪ K = K, and we know K is a subgroup of G, so H ∪ K is a subgroup of G. Exercise: 48 Section 3.5 Question: Let H be a subgroup of a group G and let g ∈ G. Prove that if n is the smallest positive integer such that g n ∈ H, then n divides |g|. Solution: Let H be a subgroup of a group G and let g ∈ G. Let n be the smallest positive integer such that g n ∈ H. Call g n = h ∈ H. Suppose that |g| = d so in particular g d = e ∈ H. We perform the integer division of d by n to get d = nq + r with 0 ≤ r < n. So g r = g d−nq = h−q ∈ H. Since r < n and n is the least positive integer such that g n is in H, we must conclude that r = 0. Thus d = nq so n divides the order of g. Exercise: 49 Section 3.5 Question: Let H be a subgroup of a group G. a) Show that H ≤ NG (H). b) Show that if A is a subset of G, then A is not necessarily a subset of NG (A). Solution: Recall that NG (H) = {g ∈ G | gHg −1 ⊆ H. For (a), note that if g ∈ H ⊆ G, gHg −1 is certainly a subset of H again; it is exactly H. So all h ∈ H are in the normalizer, making H a subset of the normalizer, but H is a group, so it is a subgroup of the normalizer. For (b), consider G = D6 and A = {r, sr}. Consider the conjugation of A by r, which is in A. In particular, rsrr−1 = rs = sr−1 , which is not in A; therefore r is not in the normalizer of A, making it impossible for A to be in the normalizer of A. This counterexample shows that a subset of a group need not also be a subset of its own normalizer in that group. Exercise: 50 Section 3.5 Question: Consider the group (Q, +). a) Prove that if H and K are any two non-trivial subgroups of Q then H ∩ K is nontrivial. b) Prove that the above result does not necessarily hold if Q is replaced with R. Solution: For (a), since H is nontrivial, let ab ∈ H, where a and b are integers with no common factors, a is not 0, and b is not negative. Since H is closed under addition, add ab to itself b times to find that a is an integer element in H. Since H is closed under taking inverses, |a| ∈ H, where |a| is the absolute value of a. Now consider K, which contains some element dc , where c and d are integers with no common factors, c is not 0, and d is not negative. As before, |c| is in K. But as |a| and |c| are integers, it makes sense to add |a| to itself |c| times and to add |c| to itself |a| times. Then |a| · |c| ∈ H and |c| · |a| ∈ K, but |a| · |c| = |c| · |a|, so we have constructed an element that is in both H and K. √ For √ (b), consider the subgroups H = h1i and K = h 2i. The subgroup H is equivalent to (Z, +), while K = {n 2 | n ∈ Z}, so their intersection is trivial: H ∩ K = {0}.

3.6 – Lattice of Subgroups Exercise: 1 Section 3.6 Question: Determine the subgroup lattice of Z20 . Solution: We write Z20 = {1, x, x2 , · · · , x19 } with x20 = 1. Then the subgroup lattice of Z20 is: Z20 hz 2 i hz 4 i hz 5 i hz 10 i {1}

3.6. LATTICE OF SUBGROUPS

117

Exercise: 2 Section 3.6 Question: Determine the subgroup lattice of Z/105Z. Solution: The subgroup lattice of Z/105Z is: h1̄i

h7i

h35i

h5i

h21i

h3i

h15i

{0̄}

Exercise: 3 Section 3.6 Question: Determine the subgroup lattice of Z100 . Solution: We write Z100 = {1, x, x2 , · · · , x99 } with x100 = 1. Then the subgroup lattice of Z100 is: Z100 hz 5 i hz 25 i

hz 2 i hz 10 i

hz 50 i

hz 4 i hz 20 i

{1}

Exercise: 4 Section 3.6 Question: Determine the subgroup lattice of U (20). Solution: We first need to identify the elements in U (20). They are U (20) = {1, 3, 7, 9, 11, 13, 17, 19}. The lattice of subgroup is: U (20)

h3i

h11, 19i

h17i

h11i

h9i

h19i

{0̄} The group U (20) is not cyclic, since no element generates the whole group. Indeed, the subgroup h11, 19i is not cyclic so U (20) is not cyclic. We can generate U (20) with two elements 3 and 19. Exercise: 5 Section 3.6 Question: Determine the subgroup lattice of U (35). Underline in the lattice the subgroups of U (35) that are cyclic. Solution: We consider U (35). A quick calculation shows that |2| = 12. Furthermore, 34, which has order 2, is not in h2i. Then it is an easy check to see that U (35) = h2, 34i.

118

CHAPTER 3. GROUPS h2, 34i h4, 34i h16, 34i

h8, 34i

h29, 34i

h2i

{34}

h4i h16i

h8i h29i

{1}

Exercise: 6 Section 3.6 Question: Determine the subgroup lattice for Zp2 q where p and q are prime numbers. Solution: Consider the group Zp2 q where p and q are distinct prime numbers. Suppose that Zp2 q is generated by z. Then the subgroup lattice is the following. Zp2 q hz p i

hz q i

2

hz pq i

hz p i {e}

Exercise: 7 Section 3.6 Question: Determine the subgroup lattice for Zpn where p is a prime number and n is a positive integer. Solution: The only subgroups of Zpn are cyclic subgroups of order d with d | pn . If z is a generator of Zpn , then the lattice of subgroups is Z pn hz p i .. . hz p

n−2

hz p

n−1

{e}

Exercise: 8 Section 3.6 Question: Determine the subgroup lattice of Z3 ⊕ Z9 .

i i

3.6. LATTICE OF SUBGROUPS

119

Solution: We write Z3 = {1, x, x2 } with x3 = 1 and Z9 = {1, z, z 2 , . . . , z 8 } with z 9 = 1. Then the subgroup lattice of Z3 ⊕ Z9 is: Z3 ⊕ Z9

h(1, z)i

h(x, z)i

h(x2 , z)i h(x, 1), (1, z 3 )i

h(1, z 3 )i

h(x, 1)i

h(x, z 3 )i

h(x2 , z 3 )i

{1}

Exercise: 9 Section 3.6 Question: Determine the subgroup lattice of D4 . Solution: The subgroup lattice of D4 . D4 hs, r2 i hsi

hsr, r2 i

hri

hsr2 i hr2 i

hsri hsr3 i

{ι}

Exercise: 10 Section 3.6 Question: Determine the subgroup lattice of Dp where p is a prime number. Solution: Dp hri

hsi {ι}

Exercise: 11 Section 3.6 Question: Determine the subgroup lattice of D8 . Solution:

hsri

· · · hsrp−1 i

120

CHAPTER 3. GROUPS D8

hs, r2 i

hs, r4 i

hsi

hsr2 , r4 i

hri

hsr, r2 i

hr2 i

hsr3 , r4 i hsr, r4 i

hsr4 i hsr2 i hsr6 i hr4 i hsr7 i hsr3 i hsr5 i hsri

{ι}

Exercise: 12 Section 3.6 Question: Determine the subgroup lattice of S3 . Solution: S3 h(1 2 3)i h(1 2)i h(1 3)i h(2 3)i {1}

Exercise: 13 Section 3.6 Question: Determine the subgroup lattice of Z2 ⊕ Z2 ⊕ Z2 . Solution: We will assume that Z2 is generated by an element z, so that Z2 = {1, z}, where z 2 = 1. Define the subgroups H1 = h(z, 1, 1), (1, z, 1)i, H2 = h(z, 1, 1), (1, z, z)i, H3 = h(z, 1, 1), (1, 1, z)i, H4 = h(z, z, 1), (1, z, z)i H5 = h(1, z, 1), (z, 1, z)i, H6 = h(1, z, 1), (1, 1, z)i, H7 = h(1, 1, z), (z, z, 1)i. Then the subgroup lattice of Z2 ⊕ Z2 ⊕ Z2 is: Z2 ⊕ Z2 ⊕ Z2

H1

H2

h(z, 1, 1)i

H3

h(z, z, 1)i

h(1, z, 1)i

H4

h(z, z, z)i

{(1, 1, 1)}

H5

h(z, 1, z)i

H6

h(1, 1, z)i

H7

h(1, z, z)i

3.6. LATTICE OF SUBGROUPS

121

Exercise: 14 Section 3.6 Question: Determine the subgroup lattice of Zp ⊕ Zp where p is a prime number. Also show that all nontrivial strict subgroups are cyclic. Solution: Suppose that Zp is generated by some element z so that Zp = {1, z, z 2 , . . . , z p−1 }, and z p = 1. Zp ⊕ Zp

h(z, 1)i

h(z, z)i

h(z, z 2 )i

···

h(z, z p−1 )i

h(1, z)i

{(1, 1)}

Exercise: 15 Section 3.6 Question: Show that there is no group whose subgroup lattice has the shape of a 3-cube, i.e., has the same lattice as the poset (P({1, 2, 3}), ⊆). Solution: Assume that a group does have a subgroup lattice that has the same shape as the 3-cube. Then the first two rows would be as follows. hxi

hyi

hxyi

{e} Furthermore, since there are no subgroups between hxi and {e}, then |x| is prime. Similarly, |y| is prime. We know that the third subgroup also must be generated by a single element and this element has to be xy since xy ∈ / hxi and xy ∈ / hyi. Note that the assumption of this starting portion of the lattice tells us that hxi ∩ hyi = {e} and similarly hxi ∩ hxyi = {e} and hyi ∩ hxyi = {e}. This implies that x and y must have order 2. Assume without loss of generality that |x| > 2. Then x2 y ∈ / hxi since y ∈ / hxi; also x2 y ∈ / hyi, since x2 ∈ / hyi; and x2 y ∈ / hxyi since 2 k k−1 x y = (xy) implies that x = (xy) , a contradiction. Hence x and y have order 2. Furthermore, we also can conclude that xy has order 2. Assume otherwise. Then xyxy 6= e. Consider the element x(xyxy) = yxy, which is not equal to x. This element has order 2 because (yxy)(yxy) = yxxy = yy = e. But then yxy 6= x and it cannot be y and it cannot be in hxyi. Hence, we would need to have another subgroup of prime order hyxyi, but we do not. Thus, we know that |x| = |y| = |xy| = 2. Furthermore, yx cannot be x or y so yx = xy. We can conclude so far that G must have the following sublattice. hx, yi hxi

hyi

hxyi

{e} The lattice of the 3-cube does not have this as a sublattice. No element on the third row (going up) of the Hasse diagram of (P({1, 2, 3}), ⊆) has three edges coming into it. Exercise: 16 Section 3.6 Question: Prove that the lattice of A4 is that which is given in Example 3.6.7. Solution: The subgroups of A4 of order 2 are h(1 2)(3 4)i, h(1 3)(2 4)i, and h(1 4)(2 3)i. The subgroups of order 3 are generated by single elements of order 3. The only element of order 3 in A4 are 3-cycles so these subgroups

122

CHAPTER 3. GROUPS

are h(1 2 3)i, h(1 2 4)i, h(1 3 4)i, and h(2 3 4)i. These, along with {id} give all the subgroups generated by single elements. Now (1 2)(3 4)(1 3)(2 4) = (1 4)(2 3) and (1 2)(3 4)(1 4)(2 3) = (1 3)(2 4) so (1 3)(2 4)(1 4)(2 3) = (1 2)(3 4) and ultimately the subgroup K = h(1 2)(3 4), (1 3)(2 4)i is a subgroup of order 4. Now consider a subgroup generated by a 3 cycle and a (2, 2)-cycle. Without loss of generality, consider H = h(1 2 3), (1 2)(3 4)i. We observe that (1 2 3)(1 2)(3 4)(1 3 2) = (1 4)(2 3) and (1 3 2)(1 2)(3 4)(1 2 3) = (1 3)(2 4). So H contains K. Now we also calculate that (1 2)(3 4)(1 2 3)(1 2)(3 4) = (1 4 2) (1 3)(2 4)(1 2 3)(1 3)(2 4) = (1 3 4) (1 4)(2 3)(1 2 3)(1 4)(2 3) = (2 4 3) But the H must also contain (1 2 4), (1 4 3), and (2 3 4). This shows that any subgroup that contains a 3 cycle and a (2, 2) is all of A4 . Now consider a subgroup generated by two 3-cycles. Without loss of generality, consider the specific subgroup Q = h(1 2 3), (1 2 4)i. We immediately have (1 2 3)(1 2 4) = (1 3)(2 4). In A4 , the product of two 3-cycles that are not inverses of each other gives a (2, 2)-cycle. Hence, by the above reasoning, Q = A4 . We have shown that the lattice in Example 3.6.7 has all the subgroups of A4 and it’s easy to check that the edges properly describe subgroup containment. Exercise: 17 Section 3.6 Question: Let m, n ∈ Z where we consider Z as a group equipped with addition. a) Find a generator of hmi ∩ hni. b) Find a generator of the join of hmi and hni. Solution: Let m, n ∈ Z where we consider Z as a group equipped with addition. Note that the subgroup hni consists of all multiples of n. a) hmi ∩ hni = hlcm(m, n)i. b) The join of (i.e., the smallest subgroup that contains both) hmi and hni is hgcd(m, n)i. Exercise: 18 Section 3.6 Question: Determine (without creating the subgroup lattice) the number and order of all cyclic subgroups of Z5 ⊕ Z15 . Solution: Suppose that Z5 is generated by z and Z15 is generated by an element y. There are cyclic subgroups of order 3, 5, and 15. • There is only 1 subgroup of order 3, namely h(1, y 5 )i. • There are 6 subgroups of order 5, namely h(1, y 3 )i, h(z, y 3 )i, h(z 2 , y 3 )i, h(z 3 , y 3 )i, h(z 4 , y 3 )i, and h(z, 1)i. Note that for example (z, y 9 ), which is an element of order 5 is in h(z 2 , y 3 )i. • In order to get a cyclic group of order 15, it must be generated by an element of the form (a, y k ), where gcd(k, 15) = 1. However, since k is relatively prime to 15, there exists ` such that k` ≡ 1 (mod 15). Thus (a` , y) ∈ h(a, y k )i and in fact, (a` , y) generates h(a, y k )i. Hence, there are 5 distinct cyclic subgroups of order 15, namely h(z i , y)i for i = 0, 1, 2, 3, 4.

Exercise: 19 Section 3.6 Question: Determine (without creating the subgroup lattice) the number and order of all cyclic subgroups of Z15 ⊕ Z15 . Solution: Suppose that Z15 is generated by z. There are cyclic subgroups of order 3, 5, and 15. • There are 4 subgroups of order 3, namely h(z 5 , 1)i, h(z 5 , z 5 )i, h(z 5 , z 10 )i, and h(1, z 5 )i. • There are 6 subgroups of order 5, namely h(z 3 , 1)i, h(z 3 , z 3 )i, h(z 3 , z 6 )i, h(z 3 , z 9 )i, h(z 3 , z 12 )i, and h(1, z 3 )i.

3.6. LATTICE OF SUBGROUPS

123

• In order to get a cyclic group of order 15, it must be generated by an element of the form (z k , z m ), where gcd(k, 15) = 1 or gcd(m, 15) = 1. If gcd(k, 15) = 1, then there exists ` that is relatively prime to 15 such that k` ≡ 1 (mod 15). Thus h(z k , z m )i is generated by (z k , z m )` = (z, z m` ). There are 15 subgroups of the form h(z, z a )i. Similarly, there are 15 subgroups of the form h(z b , z)i and these are all cyclic of order 15 as well. Now 15+15 double counts subgroups that can be generated by (z k , z m ), where k and m are both relatively prime to 15. But again h(z k , z m )i = h(z, z m` )i, where m and ` are both relatively prime to 15 so m` is as well. If a is the remainder of m` when divided by 15, there are φ(15) possibilities for a. Hence, the total number of cyclic groups of order 15 is 15 + 15 − φ(15) = 22. Exercise: 20 Section 3.6 Question: Let G = D6 . Use the lattice in Example 3.6.8 to determine: a) CD6 (s) b) ND6 (sr) c) ND6 (s, r3 ) Solution: Let G = D6 . a) If A = {s}, then CD6 (s) = CD6 ({1, s}), so CD6 (s) is a subgroup of D6 that contains the subgroup hsi. The subgroup lattice of D6 shows us 4 options, namely hsi, hs, r2 i, hs, r3 i, D6 . We notice that r2 s(r2 )−1 = sr−4 6= s so r2 ∈ / CD6 (s). This rules out hs, r2 i and D6 . On the other hand sss−1 = s

r3 s(r3 )−1 = sr−3 r−3 = s.

and

Hence r3 ∈ CD6 (s). Since CD6 (s) is a subgroup, then hs, r3 i ⊆ CD6 (s) and in fact these are equal since CD6 (s) cannot be a bigger group. b) Since {sr} is a single element subset, than ND6 (sr) = CD6 (sr) = hsr, r3 i, where we follow the pattern of part (a). c) Consider ND6 (s, r3 ). It is easy to check that the center of D6 is Z(D6 ) = hr3 i so r3 ∈ ND6 (s, r3 ). Hence ND6 (s, r3 ) must be one of the 5 subgroups containing hr3 i. Notice that rsr−1 = sr−2 = sr4 ∈ / {s, r3 } so r∈ / ND6 (s, r3 ). On the other hand sss−1 = s ∈ {s, r3 } and r3 sr−3 = s ∈ {s, r3 } so s ∈ ND6 (s, r3 ). We conclude that N (s, r3 ) = hs, r3 i. Exercise: 21 Section 3.6 Question: Let g, h ∈ G such that |g| = 12 and |h| = 5. Prove that hgi ∩ hhi = {e}. Solution: Any element g k in hgi has order 12/ gcd(12, k). In particular, the order of g k divides 12. Similarly, the order of any element in hhi divides 5. Hence, any element in hgi ∩ hhi must have an order that divides both 12 and 5. Such an element must have order 1, which means that it is the group identity. Exercise: 22 Section 3.6 Question: Let x and y be elements in a group G such that |x| and |y| are relatively prime. Prove that hxi ∩ hyi = {e}. Solution: Any element xk in hxi has order |x|/ gcd(|x|, k). In particular, the order of xk divides |x|. Similarly, the order of any element in hyi divides |y|. Hence, any element in hxi ∩ hyi must have an order that divides both |x| and |y|. Since gcd(|x|, |y|), the only positive number that divides both |x| and |y| is 1. Hence, the only element in hxi ∩ hyi is the group identity. Exercise: 23 Section 3.6 Question: Let G be a group that has exactly one non-trivial proper subgroup. Prove that G is a cyclic group of order p2 where p is a prime number. Solution: Call H the non-trivial proper subgroup of G. The lattice of G is: G H {e}

124

CHAPTER 3. GROUPS

Let x ∈ H − {e}. Then hxi is a nontrivial subgroup of H. However, H has no proper nontrivial subgroups so H = hxi and is cyclic. Assume |x| = |H| = n is a composite integer with n = ab and a, b > 1. Then hxa i is a proper nontrivial subgroup of H. This is a contradiction so the order of x is a prime number p. Let y ∈ G − H. Since y ∈ / H, the subgroup hyi is a distinct nontrivial subgroup of H. Hence, because of the conditions on G, we deduce that hyi = G and in particular that G is a cyclic group. Let |y| = n. Since x ∈ hyi, then x = y a for some a. By Proposition 3.3.7, p = n/ gcd(n, a) and in particular, p | n with n = pm. By Proposition 3.3.7, the element y p has order m. Assume that m 6= p. Then the subgroup hy p i is a nontrivial proper subgroup of G of order m, distinct from H. This is a contradiction. Hence m = p and we deduce that G is a cyclic group of order p2 . Exercise: 24 Section 3.6 Question: Let G be an abelian group that has exactly two non-trivial proper subgroups, neither of which is contained in the other. Prove that G is a cyclic group of order pq where p and q are distinct primes. Solution: Call H and K the two non-trivial proper subgroups of G. The lattice of G is: G K H {e} Let x ∈ H − {e}. Then hxi is a nontrivial subgroup of H. However, H has no proper nontrivial subgroups so H = hxi and is cyclic. Assume |x| = |H| = n is a composite integer with n = ab and a, b > 1. Then hxa i is a proper nontrivial subgroup of H. This is a contradiction so the order of x is a prime number p. The same reasoning for K holds in that K is a cyclic subgroup generated by some element y 6= e of prime order q. Now consider the element xy ∈ G. We prove by contradiction that xy ∈ / H. Assuming xy ∈ H, then xy = xa a−1 for some integer a. Then y = x , which means that y ∈ H ∩ K. However, H ∩ K = {e} but y 6= e. This is a contradiction. Thus xy ∈ / H. By an identical reasoning, xy ∈ / K. Consequently, the group hxyi is a subgroup of G that is neither a subgroup of H or of K. Hence, hxyi = G. This shows that G is cyclic of order pq. To show that p 6= q, observe that the cyclic group of order p2 (say generated by z) has only one nontrivial proper subgroup, namely the subgroup hz p i. Hence, the provided conditions on G imply that it is cyclic of order pq, with p and q distinct primes. Exercise: 25 Section 3.6 Question: Prove that if G is a group whose entire lattice of subgroups consists of one chain, then G is cyclic and of order pn where p is prime and n is a positive integer. Solution: Suppose that G is a group whose subgroup lattice consists of one chain. We have Hn = G .. . H2 H1 {e} The subgroup H1 cannot require more than two (or more) generating element say x and y since otherwise both hxi and hyi would be subgroups. Hence, H1 = hzi is cyclic. Furthermore, the order of z is a prime number p since if the order were composite with |z| = nm, then hz m i would be a nontrivial subgroup.

3.7. GROUP HOMOMORPHISMS

125

Consider the subgroup Hi . Assume that Hi requires more than one generating element. Then Hi contains x and y such that hxi and hyi are subgroups such that the join of hxi and hyi strictly contains hxi and hyi. However, since the lattice of G is a chain we must have hxi = Hj and hyi = Hk for some 1 ≤ j, k < i. But then the join of these two subgroups is Hmax(j,k) . so this contradicts the fact that Hi requires more than one generating element. In particular, G itself is cyclic with G = hwi. Now suppose that |w| = n is not a prime power. Then n = pqn0 fro some distinct primes p and q. Then, G contains a cyclic subgroup K1 of order p and another K2 of order q. Neither an be contained inside the other, which contradicts the hypothesis that the subgroup lattice of G is a chain. We conclude that G is cyclic and |G| = pn where p is prime and n is a positive integer.

3.7 – Group Homomorphisms Exercise: 1 Section 3.7 Question: Let ϕ : G → H be a homomorphism between groups. Prove that Im ϕ ≤ H. Solution: For any h1 , h2 ∈ Im ϕ, there exist some g1 , g2 ∈ G where ϕ(g1 ) = h1 and ϕ(g2 ) = h2 . By properties of homomorphisms then ϕ(g1 g2−1 ) = ϕ(g1 )ϕ(g2−1 ) = ϕ(g1 )ϕ(g2 )−1 = h1 (h2 )−1 = h1 h−1 ∈ Im ϕ. By the One 2 Step Subgroup Criterion, Im ϕ ≤ H. Exercise: 2 Section 3.7 Question: Find all homomorphisms from Z15 to Z20 . Solution: We will let Z15 = {xi x15 = 1} and Z20 = {y j y 20 = 1}. We will examine the properties of a homomorphism ϕ : Z15 → Z20 . Since our homomorphism will be completely determined by where we send the generator, consider where ϕ could send x. Since we know that x15 = 1, we need a homomorphism where 1 = ϕ(1) = ϕ(x15 ) = ϕ(x)15 . In other words, ϕ(x) has an order that divides 15. Examining the numbers that divide 15 that also correspond to orders of elements in Z20 we have ϕ(x) must have order 1 or 5. So ϕ(x) must be sent to one of the following elements, {1, y 4 , y 8 , y 12 , y 16 }. We now prove that all of these work by considering the function fi (x) = x4i for some non-negative integer i. Consider than fi (xa xb ) = (xa xb )4i = x4ia x4ib = (xa )4i (xb )4i = f (xa )f (xb ). So then we have an map that preserves the relations of the generator and all of the elements’ relations follow. So ϕ is a homomorphism. Exercise: 3 Section 3.7 Question: Find all homomorphisms from Z4 to Z2 ⊕ Z2 . Solution: We will let Z4 = {xi x4 = 1} and Z2 ⊕ Z2 = {(y i , z j ) y 2 = 1 and z 2 = 1}. Since our homomorphism will be completely determined by where we send x, and since |x| = 4, we need |ϕ(x)| = 4, 2 or 1. In Z2 ⊕ Z2 we only have elements of order 2 and 1. Since we will always send x to an element of order 2 or 1, we notice a couple things, ϕ(x2 ) = ϕ(x)2 = 1 and ϕ(x3 ) = ϕ(x)3 = ϕ(x)ϕ(x)2 = ϕ(x) for any homomorphism ϕ. It is easy to see that because of these relations, sending x to any element of Z2 ⊕ Z2 will give us a different homomorphism. So we have ϕ(x) = (1, 1), ϕ(x) = (1, z), ϕ(x) = (y, 1), and ϕ(x) = (y, z). Exercise: 4 Section 3.7 Question: Prove that the function ϕ : G → G defined by ϕ(g) = g −1 is a homomorphism if and only if G is abelian. Solution: (=⇒) Suppose that the function ϕ : G → G defined by ϕ(g) = g −1 is a homomorphism. Then for any a, b ∈ G consider ϕ(ab) = (ab)−1 = b−1 a−1 . However, since ϕ is a homomorphism we see that ϕ(ab) = ϕ(a)ϕ(b) = a−1 b−1 . This implies that b−1 a−1 = a−1 b−1 for any a, b ∈ G. Taking the inverse of both sides we see that ab = ba for any a, b ∈ G. Which shows that G is abelian. (⇐=) Suppose that G is abelian. For any a, b ∈ G we have ϕ(ab) = (ab)−1 = b−1 a−1 . Since G is abelian then, b−1 a−1 = a−1 b−1 = ϕ(a)ϕ(b). Which shows that ϕ is a homomorphism. This completes the proof of the if and only if statement.

126

CHAPTER 3. GROUPS

Exercise: 5 Section 3.7 Question: Let Z33 be generated by an element y and Z12 by an element z. Suppose that ϕ : Z33 → Z12 is a homomorphism satisfying ϕ(y 7 ) = z 8 . Find ϕ(x); determine Ker ϕ; and determine Im ϕ. Solution: Since 7 is relatively prime to 33, y 7 is also a generator of Z33 . Now the multiplicative inverse of 7 in Z/33Z is 19 since 7 × 19 = 133 ≡ 1 (mod 33). Thus (y 7 )19 = y. Hence, since ϕ(y 7 ) = z 8 , then ϕ(y) = (z 8 )19 = z 152 = z 8 . Thus, for all a, ϕ(y a ) = z 8a . Then Ker ϕ = {y a | 8a ≡ 0

(mod 12)} = {y a | a ≡ 0

(mod 3)} = hy 3 i

Im ϕ = hz 8 i = hz 4 i.

Exercise: 6 Section 3.7 Question: Consider the function f : U (12)⊕(Z/16Z) → Z/4Z defined by f (a, b) = ab with elements considered now modulo 4. Prove that f is a well-defined function. Determine whether or not f is a homomorphism. If it is, determine Ker f ; if it is not, give a counter-example. Solution: To check whether a function is well-defined, we must determine if f (a, b) = f (c, d) when (a, b) ∼ (c, d). In our case, (a, b) ∼ (c, d) when (a mod 12, b mod 16) = (c mod 12, d mod 16). If c − a = 12k ⇒ c − a = 4l and if b − d = 16m ⇒ b − d = 4n for some integers k, l, m and n. This shows that ā = c̄ mod 4 and b̄ = d¯ ¯ mod 4 = cd mod 4 = f (c, d) which mod 4. This implies that f (a, b) = ab mod 4 = (ā)(b̄) mod 4 = (c̄)(d) shows that the function is well-defined. Now we will provide a counter example showing that f is not a homomorphism. A counter example will appear where ϕ(a, b) + ϕ(c, d) 6= ϕ(ac, b + d) or ab + cd 6= ac(b + d) mod 4. We let (a, b) = (5̄, 1̄) and (c, d) = (7̄, 3̄), then ϕ(5̄, 1̄) + ϕ(7̄, 3̄) = 5 · 1 + 7 · 3 mod 4 = 5 + 21 mod 4 = 2̄ mod 4. However, ϕ(5 · 7, 1 + 3) = ϕ(3̄, 4̄) = 3 ∗ 4 mod 4 = 0̄ mod 4. So f is not a homomorphism. Exercise: 7 Section 3.7 Question: What common fallacy in elementary algebra is addressed by the statement that “the function f : (R, +) → (R, +) with f (x) = x2 is not a homomorphism.” Solution: The common fallacy is how many students do the following, (x + y)2 = x2 + y 2 . In terms of f it would be, f (x + y) = (x + y)2 = x2 + y 2 = f (x) + f (y). However, this is not true since f is not a homomorphism.

Exercise: 8 Section 3.7 Question: Let G be a group. Determine whether or not the function ϕ : G ⊕ G → G defined by ϕ(x, y) = x−1 y is a homomorphism. Solution: Let (a, b), (c, d) ∈ G ⊕ G, now consider ϕ(ac, bd) = (ac)−1 bd = c−1 a−1 bd but ϕ(a, b)ϕ(c, d) = a−1 bc−1 d. Now we will set d = 1 and b = 1. Then the equations reduce to c−1 a−1 and a−1 c−1 . These two equations are equal if and only if G is an abelian group. So ϕ is not a homomorphism in general. Exercise: 9 Section 3.7 Question: Find all homomorphisms Z5 → S5 . Also find all homomorphisms S5 → Z5 . Solution: Let Z5 = {xi x5 = 1}. First we find all homomorphism from Z5 → S5 . Now, every element of Z5 (except the identity) has order 5. This implies that we must find a σ ∈ S5 where |σ| = 5. Since 5 is a prime number, the only permutations that have order 5 in S5 are 5-cycles (c.f. Exercise 3.4.22). Suppose that σ is any 5-cycle and we set ϕ(x) = σ. We will show that this is a homomorphism. Consider ϕ(a + b mod 5) = σ a+b mod 5 = σ ā mod 5 σ b̄ mod 5 = ϕ(ā mod 5)ϕ(b̄ mod 5). Then we have a different homomorphism for each 5-cycle and there are 4! = 24 different 5-cycles. Then we also include the trivial homomorphism (ϕ(x) = 1) and we end up with 25 different homomorphisms from Z5 → S5 . Now we will find all homomorphisms from S5 → Z5 . Now, by Exercise 3.4.9, every σ ∈ S5 can be decomposed into the composition of transpositions. Every transposition has order 2 which implies that for any transposition τ , |ϕ(τ )| = 2 or 1. However, Z5 has no elements of order 2 so this implies that for every transposition τ we have ϕ(τ ) = 1. For any homomorphism ϕ and any σ ∈ S5 we can write σ as σ1 σ2 · · · σn where each σi is a transposition. Then ϕ(σ) = ϕ(σ1 σ2 · · · σn ) = ϕ(σ1 )ϕ(σ2 ) . . . ϕ(σn ) = (1)(1) . . . (1) = 1. Which shows that for

3.7. GROUP HOMOMORPHISMS

127

any homomorphism, all permutations get mapped to 1. So the only homomorphism from Z5 → S5 is the trivial homomorphism. Exercise: 10 Section 3.7 Question: Let ϕ : G → H be a homomorphism. Prove that for all g ∈ G, the order |ϕ(g)| divides the order |g|. Solution: Let |g| = n. In particular, this means that g n = 1G . By properties of homomorphisms, ϕ(g n ) = ϕ(g)n , but also ϕ(g n ) = ϕ(1) = 1H . Hence, ϕ(g)n = 1H so by properties of homomorphisms (Corollary 3.3.6) |ϕ(g)| divides n = |g|. Exercise: 11 Section 3.7 Question: Prove Proposition 3.7.15: Let ϕ : G → H be a homomorphism of groups. Then 1. ϕ is injective if and only if Ker ϕ = {1G }. 2. ϕ is surjective if and only if H = Im ϕ. Solution: 1. We will prove the statement that ϕ is injective if and only if Ker ϕ = {1G }. (=⇒) Suppose that ϕ is injective. By Proposition 3.7.9, we know that ϕ(1) = 1 and since ϕ is injective no other element can go to 1. This implies that Ker ϕ = {1G }. (⇐=) Suppose that Ker ϕ = {1G }. Then for any a, b ∈ G, ϕ(a) = ϕ(b) ⇒ 1 = ϕ(a)ϕ(b)−1 = ϕ(a)ϕ(b−1 ) = ϕ(ab−1 ). Since Ker ϕ = {1G } we know that ab−1 = 1 which implies that a = b. This shows that ϕ is injective. This proves both directions of the if and only if statement. 2. We will prove the statement that ϕ is surjective if and only if H = Im ϕ. (=⇒) Suppose that ϕ is surjective. Then for any h ∈ H, we know there exists some g ∈ G such that ϕ(g) = h ⇒ h ∈ Im ϕ. Since the Im ϕ cannot be any bigger than H, this implies that H = Im ϕ. (⇐=) Suppose that H = Im ϕ. Then for every h ∈ H, there is some g ∈ G where ϕ(g) = h. Since H is the codomain, this implies that ϕ is surjective. So ϕ is surjective if and only if H = Im ϕ.

Exercise: 12 Section 3.7 Question: Prove that ϕ : Z ⊕ Z → Z defined by ϕ(x, y) = 2x + 3y is a homomorphism. Determine Ker ϕ. Describe the fiber ϕ−1 (6). Solution: Let (a1 , b1 ), (a2 , b2 ) ∈ Z ⊕ Z and consider ϕ(a1 + a2 , b1 + b2 ) = 2(a1 + a2 ) + 3(b1 + b2 ) = 2a1 + 3b1 + 2a2 + 3b2 = ϕ(a1 , b1 ) + ϕ(a2 , b2 ). So we can conclude that ϕ is a homomorphism. Next, we will determine Ker ϕ. Suppose (a, b) ∈ Z ⊕ Z and ϕ(a, b) = 0. Then ϕ(a, b) = 2a + 3b = 0 ⇒ 2a = −3b ⇒ a = −3/2b.

128

CHAPTER 3. GROUPS

So Ker ϕ = {(x, y) ∈ Z ⊕ Z x = −3/2y}. Consider the fiber ϕ−1 (6) or all pairs (a, b) where 2a + 3b = 6. Solving for b we find, 3b = 6 − 2a ⇒ b = 6−2a 3 . ) a ≡ 0 mod 3}. Since b needs to be an integer, we must have, a ≡ 0 mod 3. So ϕ−1 (6) = {(a, 6−2a 3 Exercise: 13 Section 3.7 Question: Let ϕ and ψ be homomorphisms between two groups G and H. Prove or disprove that {g ∈ G | ϕ(g) = ψ(g)} is a subgroup of G. Solution: Call H = {g ∈ G | ϕ(g) = ψ(g)}. Let x, y ∈ H. Then by the property of homomorphisms, ϕ(xy −1 ) = ϕ(x)ϕ(y)−1 and ψ(xy) = ψ(x)ψ(y)−1 . Now since x, y ∈ H, we have ϕ(xy −1 ) = ϕ(x)ϕ(y)−1 = ψ(x)ψ(y)−1 = ψ(xy −1 ). Thus xy −1 ∈ H. By the One-Step Subgroup Criterion, H is a subgroup of G. Exercise: 14 Section 3.7 Question: Consider the function f : U (33) → U (33) defined by f (x) = x2 . Show that f is a homomorphism and find the kernel and the image of f . Solution: Let x, y ∈ U (33) and consider f (xy) = (xy)2 = x2 y 2 = x̄2 ȳ 2 = f (x̄)f (ȳ). Since the previous equation holds for all x, y ∈ U (33), f is a homomorphism. If x ∈ Ker f , this implies that x2 = 33k + 1 for some integer k. Solving for 33k we find, x2 − 1 = 33k ⇒ (x + 1)(x − 1) = 33k. So we are looking for numbers that are surrounded by two numbers, which cumulatively have both 3 and 11 as prime factors (for instance, 10 since 3|(10 − 1) and 11|(10 + 1)). The only exception to this requirement is the identity, 1. Using this strategy we find that Ker f = {1, 10, 23, 32}. We can find the image by listing out the squares of each element of U (33), Im f = {12 ≡ 1, 22 ≡ 4, 42 ≡ 16, 52 ≡ 25, 72 ≡ 16, 82 ≡ 31, 102 ≡ 1, 132 ≡ 4, 142 ≡ 31, 162 ≡ 25, 172 ≡ 25, 192 ≡ 31, 202 ≡ 4, 232 ≡ 1, 252 ≡ 31, 262 ≡ 16, 282 ≡ 25, 292 ≡ 16, 312 ≡ 4, 322 ≡ 1} = {1, 4, 16, 25, 31}.

Exercise: 15 Section 3.7 Question: Prove that the function f : GL2 (R) → GL3 (R) with  2 a 2ab a b f = ac ad + bc c d c2 2cd

 b2 bd d2

is a homomorphism and find the kernel of f . Solution: Let

a c

b d

and

e g

f h

be two matrices in GL2 (R). Then a b e f f f c d g h  2  2  a 2ab b2 e 2ef f2 = ac ad + bc bd eg eh + f g f h c2 2cd d2 g2 2gh h2  a2 e2 + 2abeg + b2 g 2 2a2 ef + 2abeh + 2abf g + 2b2 gh 2 2 = ace + adeg + bceg + bdg 2acef + (ad + bc)(eh + f g) + 2bdgh c2 e2 + 2cdeg + d2 g 2 2c2 ef + 2cdeh + 2cdf g + 2d2 gh

 a2 f 2 + 2abf h + b2 h2 acf 2 + adf h + bcf h + bdh2  c2 f 2 + 2cdf h + d2 h2

3.7. GROUP HOMOMORPHISMS

129

and also a b e f f c d g h ae + bg af + bh =f ce + dg cf + dh   (ae + bg)2 2(ae + bg)(af + bh) (af + bh)2 = (ae + bg)(ce + dg) (ae + bg)(cf + dh) + (af + bh)(ce + dg) (af + bh)(cf + dh) (ce + dg)2 2(ce + dg)(cf + dh) (cf + dh)2   a2 e2 + 2abeg + b2 g 2 2a2 ef + 2abeh + 2abf g + 2b2 gh a2 f 2 + 2abf h + b2 h2 = ace2 + adeg + bceg + bdg 2 2acef + (ad + bc)(eh + f g) + 2bdgh acf 2 + adf h + bcf h + bdh2  . c2 e2 + 2cdeg + d2 g 2 2c2 ef + 2cdeh + 2cdf g + 2d2 gh c2 f 2 + 2cdf h + d2 h2 By inspection, we see that f (X)f (Y ) = f (XY ) for any two invertible matrices x, y ∈ GL2 (R). Thus f is a homomorphism. a b The kernel of f is all matrices such that c d  2 a ac c2

  2ab b2 1 ad + bc bd = 0 2cd d2 0

0 1 0

 0 0 . 1

From a2 = 1, b2 = 0, c2 = 0 and d2 = 1, we deduce that a = ±1, b=c=0, d = ±1. All the other equations are satisfied except for the center of the matrix ad + bc = 1. Since bc = 0, this last equation implies that a = d = ±1. Hence the kernel of f is Ker f = {I, −I}, where I is the 2 × 2 identity matrix. Exercise: 16 Section 3.7 Question: Consider the multiplicative group R× and consider the function f : R× → R>0 × {1, −1} given by f (x) = (|x|, sign(x)). We assume that all groups have multiplication as their operation. Show that f is an isomorphism. Solution: First we show that f is a homomorphism. Let a, b ∈ R× and we will consider f (ab) = (|ab|, sign(ab)) and since sign and absolute value are both multiplicative homomorphisms we have (|ab|, sign(ab)) = (|a|, sign(a))(|b|, sign(b)) = f (a)f (b). This shows that f is a homomorphism. Now we will prove that it is indeed an isomorphism by providing an inverse function, f −1 : R>0 × {1, −1} → R× where f −1 (a, −1n ) = a(−1n ). To prove it is the inverse consider f −1 (f (a)) = f −1 (|a|, sign(a)) = |a|(sign(a)) = a and f (f −1 (a, −1n )) = f (a(−1)n ) and since a > 0 we know that this equals, (a, −1n ). Since we have an inverse function, f must be a bijection and therefore an isomorphism. Exercise: 17 Section 3.7 Question: Construct both the Cayley tables for both S3 and GL2 (F2 ) to show that the function given in Example 3.7.20 establishes an isomorphism between S3 and GL2 (F2 ). Solution: The Cayley table for S3 is

1 (1 2) (1 3) (2 3) (1 2 3) (1 3 2)

1 1 (1 2) (1 3) (2 3) (1 2 3) (1 3 2)

(1 2) (1 2) 1 (1 2 3) (1 3 2) (1 3) (2 3)

(1 3) (2 3) (1 2 3) (1 3 2) (1 3) (2 3) (1 2 3) (1 3 2) (1 3 2) (1 2 3) (2 3) (1 3) 1 (1 3 2) (1 2) (2 3) (1 2 3) 1 (1 3) (1 2) (2 3) (1 2) (1 3 2) 1 (1 2) (1 3) 1 (1 2 3)

130

CHAPTER 3. GROUPS

The Cayley table for GL2 (F2 ) is × 1 0 1 1 1 1 0 1 1 1 0 1

0 1 0 1 0 1 1 0 1 0 1 1

1 0 1 0 1 1 1 1 0 1 1 1 0 1

0 1 0 1 0 1 0 1 1 0 1 0 1 1

1 1 1 1 1 0 1 1 0 1 1 1 0 1

0 1 0 1 0 1 1 0 1 1 0 1 1 0

1 1 1 1 0 1 1 0 1 1 0 1 1 1

0 1 0 1 1 1 0 1 1 0 1 0 0 1

0 1 0 1 1 1 0 1 1 0 1 1 1 1

1 0 1 0 1 0 1 1 0 1 0 1 0 1

1 1 1 1 0 1 1 1 1 1 0 1 1 0

1 0 1 0 1 0 0 1 0 1 1 1 0 1

0 1 0 1 1 1 0 1 1 1 1 0 1 1

1 1 1 1 0 1 1 0 0 1 0 1 1 0

Since these two Cayley tables show that corresponding elements operate in corresponding ways, then the function described in Example 3.7.20 is an isomorphism. Exercise: 18 Section 3.7 Question: Find all homomorphisms Z → Z. Determine which ones are isomorphisms. Solution: We will construct a general homomorphism ϕ : Z → Z. We know that ϕ(0) = 0. Now consider what happens when we fix ϕ(1) = c. For any positive e ∈ Z we should have ϕ(e) = ϕ(1 + 1 + 1 . . . + 1(e times)) = ϕ(1) + ϕ(1) + . . . + ϕ(1)(e times) = c + c + . . . + c (e times) = e · c. Similarily if e were negative, ϕ(e) = ϕ(−1) + ϕ(−1) + . . . + ϕ(−1) (|e| times) = (−ϕ(1)) + (−ϕ(1)) + . . . + (−ϕ(1))(|e| times) = −c + −c + . . . + −c = −c · |e| = ce. So that every element is completely determined by where ϕ(1) goes. In fact, we just established the result that ϕ(e) = e(ϕ(1)) for all e ∈ Z. So a different homomorphism is determined for every c we choose to send 1 to and we can send 1 to every single one of the integers. Letting ϕ(1) = 0 corresponds to the trivial homomorphism, letting ϕ(1) = 1 corresponds to the identity isomorphism, and letting ϕ(1) = −1 corresponds to another isomorphism (this is easily seen to be an isomorphism by observing the inverse function, ϕ−1 (1) = −1). However, letting ϕ(1) = c for when |c| > 1 does not give an isomorphism, since for any e ∈ Z, ϕ(e) = ce 6= 1 when |c| > 1 showing that ϕ cannot be injective. Exercise: 19 Section 3.7 Question: Prove that Zm ⊕ Zn ∼ = Zmn if m and n are relatively prime. i j Solution: Let Zm ⊕ Zn = {(x , y ) xm = 1 and y n = 1} and consider the element (x, y). If m and n are relatively prime then |(x, y)| = mn/ gcd(m, n) = mn. Since |(x, y)| = |Zm ⊕ Zn | = mn, then Zm ⊕ Zn must be cyclic. By Proposition 3.7.25, since |Zm ⊕ Zn | = |Zmn | and both are cyclic, it follows that Zm ⊕ Zn ∼ = Zmn . Exercise: 20 Section 3.7 Question: Prove that Z9 ⊕ Z3 ∼ 6 Z27 . = i j Solution: Let Z9 ⊕ Z3 = {(x , y ) x9 = 1 and y 3 = 1}. We will show they are not isomorphic by showing that Z9 ⊕ Z3 does not have an element of order 27. Every element in Z9 is of order 1, 3 or 9 and every element in Z3 is of order 1 or 3. Any element (a, b) ∈ Z9 ⊕ Z3 has order |a||b|/ gcd(|a|, |b|). Considering all possible combinations of orders, we see that any element in Z9 ⊕ Z3 has order 1, 3, or 9. Since no element has order 27, we conclude that Z9 ⊕ Z3 ∼ 6 Z27 . = Exercise: 21 Section 3.7 Question: Prove that Zm ⊕ Zn ∼ = Zlcm(m,n) ⊕ Zgcd(m,n) .

3.7. GROUP HOMOMORPHISMS

131

Solution: Write Zm = {1, x, x2 , . . . , xm−1 } and Zn = {1, y, y 2 , . . . , y n−1 }. Since in a direct sum G ⊕ H, |(g, h)| = lcm(|g|, |h|), we know that in Zm ⊕ Zn , the element (x, y) has order lcm(m, n). Define a, b by the positive integers satisfying an = lcm(m, n) = bm. Notice that a gcd(m, n) =

lcm(m, n) lcm(m, n) gcd(m, n) mn gcd(m, n) = = = m, n n n

and similarly that b gcd(m, n) = n. We also write Zlcm(m,n) = {1, z, z 2 , . . . , z lcm(m,n)−1 } and Zgcd(m,n) = {1, u, u2 , . . . , ugcd(m,n)−1 }. Consider the function f : Zlcm(m,n) ⊕ Zgcd(m,n) → Zm ⊕ Zn determined by f (z k , u` ) = (xk+` , y k ). We can see that f is a homomorphism from f ((z k , u` )(z α , uβ )) = f (z k+α , u`+β ) = (xk+α+`+β , y `+β ) = (xk+` , y ` )(xα+β , y β ) = f (z k , u` )f (z α , uβ ). The kernel of f is ker f = {(z k , u` ) | k + ` ≡ 0

(mod m) and k ≡ 0

(mod n)}.

If k | n, then there exists some k 0 with k = nk 0 . Then k + ` ≡ 0 (mod m) implies that nk 0 + ` = mh or in other words nk 0 − mh + ` = 0. However, for all k 0 , h ∈ Z, the linear combination nk 0 − mh is a multiple of gcd(m, n). Hence, since 0 ≤ ` ≤ gcd(m, n), we deduce that we must have ` = 0. This implies that k is a multiple of m and a multiple of n, so k must be a multiple of lcm(m, n). However, 0 ≤ k ≤ lcm(m, n) − 1, so we deduce that k = 0. Hence, we have shown that ker f = {(1, 1)}. This implies that the homomorphism is injective. Furthermore, an injective function between sets of the same size (mn = lcm(m, n) gcd(m, n)) is also a surjection, so we have shown that f is a bijection. Since f is a bijection and a homomorphism, it is an isomorphism and thus Zm ⊕ Zn ∼ = Zlcm(m,n) ⊕ Zgcd(m,n) .

Exercise: 22 Section 3.7 Question: Let G1 and G2 be two arbitrary groups. Prove that G1 ⊕ G2 ∼ = G2 ⊕ G1 . Solution: We will define a map f : G1 ⊕ G2 → G2 ⊕ G1 where f (g1 , g2 ) = (g2 , g1 ). Since we can define an inverse function, f −1 (g2 , g1 ) = (g1 , g2 ), it is easy to see that f is a bijection. So we just need f to be a homomorphism. Consider f (g1 h1 , g2 h2 ) = (g2 h2 , g1 h1 ) = (g2 , g1 )(h2 , h1 ) = f (g1 , g2 )f (h1 , h2 ) so that f is indeed a homomorphism and so is an isomorphism. This shows that G1 ⊕ G2 ∼ = G2 ⊕ G1 . Exercise: 23 Section 3.7 Question: Let G1 and G2 be two arbitrary groups. Prove that the ‘projection’ maps p1 : G1 ⊕ G2 → G1 and p2 : G1 ⊕ G2 → G2 defined by p1 (x, y) = x and p2 (x, y) = y are homomorphisms. Find the kernels and images of p1 and p2 . Solution: Since our proof will be identical for both projections, we will choose to prove that p1 : G1 ⊕ G2 → G1 is a homomorphism. Let (g1 , h1 ), (g2 , h2 ) ∈ G1 ⊕ G2 and consider p1 (g1 g2 , h1 h2 ) = g1 g2 = g1 ∗ g2 = p1 (g1 , h1 )p2 (g2 , h2 ) which shows that p1 is a homomorphism. By the same reasoning, p2 is also a homomorphism. For p1 , any element of the form (1, h) ∈ G1 ⊕G2 will be mapped to the identity of G1 so that Ker p1 = {(1, h) ∀h ∈ G2 }. Similarily, Ker p2 = {(g, 1) ∀g ∈ G1 }. For any element g ∈ G1 , we know that p1 (g, 1) = g and likewise for any h ∈ G2 we know that p2 (1, h) = h so that both p1 and p2 are surjective. This implies that Im p1 = G1 and Im p2 = G2 .

132

CHAPTER 3. GROUPS

Exercise: 24 Section 3.7 Question: Prove that if a homomorphism ϕ : G → H is injective, then G ∼ = ϕ(G). Solution: Define f : G → ϕ(G) as f (g) = ϕ(g). Because ϕ is injective, we know that f is also injective. And since every element in our codomain is defined to be in the image of ϕ, we know that it is also in the image of f so that f is surjective. So f is a bijection. Let g, h ∈ G and consider f (gh) = ϕ(gh) = ϕ(g)ϕ(h) (since ϕ is a homomorphism) = f (g)f (h). This shows that f is a homomorphism. Therefore f is an isomorphism and G ∼ = ϕ(G). Exercise: 25 Section 3.7 Question: Prove that for every homomorphism f : (R, +) → (R>0 , ×) there exists a positive real number a such that satisfies f (q) = aq for all q ∈ Q. Solution: Let ϕ be a homomorphism such that ϕ : (R, +) → (R>0 , ×). Let ϕ(1) = a such that a ∈ R>0 . By Prop 3.7.9 (3), ϕ(1 × m) = (a)m for m ∈ R. Thus there exists a positive real number a such that ϕ(m) = am for all m ∈ R. Exercise: 26 Section 3.7 Question: Prove Proposition 3.7.17. Solution: Let ϕ : G → H be an isomorphism. Suppose h1 , h2 ∈ H. There exist g1 , g2 ∈ G such that ϕ(g1 ) = h1 and ϕ(g2 ) = h2 because ϕ is an isomorphism. It follows that ϕ(g1 ∗ g2 ) = ϕ(g1 ) ∗ ϕ(g2 ) = h1 ∗ h2 . Note that ϕ−1 (h1 ) = g1 and ϕ−1 (h2 ) = g2 . Additionally, ϕ−1 (h1 ∗ h2 ) = g1 ∗ g2 = ϕ−1 (h1 ) ∗ ϕ−1 (h2 ). Thus ϕ−1 is a homomorphism. Exercise: 27 Section 3.7 Question: Prove that if two groups G and H are infinite and cyclic, then G ∼ = H. Solution: Let G = hxi and H = hyi be two infinite cyclic groups. Then G = {1, x, x2 , ϕx3 , . . . } and 2 3 H = {1, y, y , y , . . . }, so elements in G and H are uniquely given by xn and y m for n, m ∈ Z≥0 . Define a function ϕ : G → H by the mapping xn 7→ y n . We claim this function is an isomorphism of groups. First, let xn , xm ∈ G and consider that ϕ(xn y m ) = ϕ(xn+m ) = y n+m = y n y m , meaning that ϕ is a homomorphism. Next, to show that ϕ is surjective, let y n ∈ H. Then xn satisfies ϕ(xn ) = y n , making ϕ a surjection.. Finally, to show that ϕ is injective, let xn , xm ∈ G and suppose that ϕ(xn ) = ϕ(y m ). Then y n = y m , so n = m, and xn = xm , making ϕ and injective. Therefore ϕ is a bijective homomorphism or an isomorphism between arbitrary infinite cyclic groups G and H. Exercise: 28 Section 3.7 Question: Prove that Z2 ⊕ Z2 ⊕ Z2 and Q8 are not isomorphic. Solution: The group Z2 ⊕ Z2 ⊕ Z2 is abelian and Q8 is not, so the groups cannot be isomorphic. Notice also that in Z2 ⊕ Z2 ⊕ Z2 every nonidentity element has order 2, whereas in Q8 there are six elements of order 4. Exercise: 29 Section 3.7 Question: Prove that D6 and A4 are not isomorphic. Solution: Consider r ∈ D6 where |r| = 6. An element of order 6 in A4 would have to be a 6-cycle or a disjoint 3-cycle and 2-cycle. However, the longest possible cycle is of length 4 in A4 and since there are only 4 numbers, any 3 cycle and 2 cycle will share at least 1 element and cannot be disjoint. So there is no element of order 6 in A4 and therefore there is no isomorphism between D6 and A4 . Exercise: 30 Section 3.7 Question: Prove that D12 and S4 are not isomorphic. Solution: Consider r ∈ D12 where |r| = 12. In S4 , all the element have orders 1, 2, 3 or 4. Since no element has order 12, an isomorphism between D12 and S4 cannot exist. Exercise: 31 Section 3.7 Question: Prove that (R∗ , ×) and (C∗ , ×) are not isomorphic. Solution: Consider i ∈ (C∗ , ×) where |i| = 4. The only elements of finite order in (R∗ , ×) are 1 and −1 which have orders 1 and 2 respectively. Since there is no element of order 4 in (R∗ , ×) we know that the two groups cannot be isomorphic.

3.7. GROUP HOMOMORPHISMS

133

Exercise: 32 Section 3.7 Question: Prove or disprove that U (7) and U (9) are isomorphic. Solution: By examination, we find that |U (7)| = |U (9)| = 6. With hopes that both are cyclic, we test some elements and find that 3̄ generates U (7) and 2̄ generates U (9). Since both are cyclic and of order 6, by Proposition 3.7.25, U (7) ∼ = U (9). Exercise: 33 Section 3.7 Question: Prove or disprove that U (44) and U (25) are isomorphic. Solution: Using Euler’s totient function we know that |U (44)| = (4−1)(11−1) = 30 and |U (25)| = (52 −5) = 20. Since the groups have different orders, they cannot be isomorphic. Exercise: 34 Section 3.7 Question: Let d be a divisor of n. Consider the function ϕ : U (n) → U (d) defined by ϕ(a) = a mod d. Determine the kernel and the image of ϕ. Solution: The kernel of ϕ is all multiples of d less than n, or Ker ϕ = {k · d | k = 0, 1, . . . , nd }. The image of ϕ is the congruence classes of all integers from 0 to d − 1, or Im ϕ = {0, 1, . . . , d − 1}. Exercise: 35 Section 3.7 Question: Prove that U (21) ∼ = Z2 ⊕ Z6 . Solution: First, write U (21) = {1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20}. It is easy to observe that 5 generates a cyclic subgroup in this order — h5i = {1, 5, 4, 20, 16, 17} — and that the remaining elements in U (21) are these elements multiplied by 2 and reduced modulo 21. Write Z2 = hx | x2 = 1i and Z6 = hy | y 6 = 1i. Then define a homomorphism ϕ : (Z2 ⊕ Z6 ) → U (21) by the mapping (xi , y j ) 7→ 2i 5j mod 21). We saw above that this method generates all of U (21), so this is a surjective function; as these are finite groups of the same cardinality, this makes it also injective. It remains to show that ϕ is a group homomorphism. Let (xi , y m ), (xj , y n ) ∈ Z2 ⊕ Z6 . Consider ϕ((xi , y m ) · j (x , y n )) = ϕ(xi xj , y m y n ) = ϕ(xi+j , y m+n ) = 2i+j 5m+n = 2i 2j 5m 5n = 2i 5m · 2j 5n . This makes ϕ a group homomorphism between U (21) and Z2 ⊕ Z6 . As ϕ is also a bijection, it follows that U (21) ∼ = Z2 ⊕ Z6 . Exercise: 36 Section 3.7 Question: Let G be the group of rigid motions of a cube. In Exercise 3.3.27 we saw that G has order 24. Prove that each rigid motion of the cube corresponds uniquely to a permutation of the 4 main diagonals through the cube. Conclude that G ∼ = S4 . Solution: Consider the 4 main diagonals of the cube. Suppose two of them stay fixed and the other two diagonals interchange. Call these diagonals AB and CD such that AC and BD are edges of the cube. This interchange of just the two diagonals occurs as the rigid motion of rotation by 180◦ around the line passing through the midpoints of AC and BD. There are six pairs of opposite cube edges, which correspond to the six transpositions in S4 . We know that the set of transpositions generate Sn . Hence, the set of rigid motions of the cube is isomorphic to S4 . More precisely, 4-cycles in S4 correspond to a rotation by 90◦ around an axis connecting midpoints of opposite faces. (2, 2)-cycles correspond to a rotation by 180◦ around an axis connecting midpoints of opposite faces. 3cycles correspond to rotations by 120◦ around a main diagonal. Exercise: 37 Section 3.7 Question: Let G be a group, let H be a subgroup, and let g ∈ G. a) Prove that gHg −1 = {ghg −1 | h ∈ H} is a subgroup of G. b) Prove that the function ϕg : H → gHg −1 by ϕg (h) = ghg −1 is an isomorphism between H and gHg −1 . [The subgroup gHg −1 is said to be a conjugate to H.] Solution: a) We will prove that gHg −1 is a subgroup using the one step subgroup criteria. We know that gHg −1 is nonempty since 1 ∈ H so that g1g −1 = 1 ∈ gHg −1 . Let x, y ∈ gHg −1 , then x = ghg −1 for some h ∈ H and y = gkg −1 for some k ∈ H. Consider xy −1 = (ghg −1 )(gkg −1 )−1 = ghg −1 gk −1 g −1 = gh(1)k −1 g −1 = g(hk −1 )g −1 and since hk −1 ∈ H we know that xy −1 = g(hk −1 )g −1 ∈ gHg −1 . So by the one step subgroup criteria, gHg −1 is a subgroup of G.

134

CHAPTER 3. GROUPS

b) First we will show that ϕg is a homomorphism. Let h, k ∈ H and consider ϕg (hk) = g(hk)g −1 = g(h(g −1 g)k)g −1 = (ghg −1 )(gkg −1 ) = ϕg (h)ϕg (k). So ϕg is a homomorphism. Suppose that ϕg (h) = ϕg (k) ⇒ ghg −1 = gkg −1 . Now operating on the right by g and the left by g −1 we get, (g −1 g)h(g −1 g) = (g −1 g)k(g −1 g) ⇒ h = k. This shows that ϕg is injective. For any x ∈ gHg −1 , x = ghg −1 for some h ∈ H, so x = ϕg (h) which shows that ϕg is surjective. This proves that ϕg is an isomorphism between H and gHg −1 . Exercise: 38 Section 3.7 Question: Let G be a group and let g ∈ G be any group element. Define the function ψg : G → G by ψg (x) = gxg −1 . a) Prove that ψg is an automorphism of G with inverse (ψg )−1 = ψg−1 . b) Prove that the function Ψ : G → Aut(G) defined by Ψ(g) = ψg is a homomorphism. [An automorphism of the form ψg is called an inner automorphism. The image of Ψ in Aut(G) is called the group of inner automorphisms and is denoted Inn(G).] Solution: a) First we show that ψg is a homomorphism. Let x, y ∈ G and consider ψg (xy) = g(xy)g −1 = g(x(g −1 g)y)g −1 = (gxg −1 )(gyg −1 ) = ψg (x)ψg (y). This shows that ψg is a homomorphism. Now for injectivity, suppose that ψg (x) = ψg (y) ⇒ gxg −1 = gyg −1 . Then, operating on the right by g and the left by g −1 shows that, g −1 gxg −1 g = g −1 gyg −1 g ⇒ x = y. So ψg is injective. Since |G| = |G| and ψg is injective, this implies that ψg is also surjective. So ψg is an isomorphism from G to G which, by definition, means ψg is an automorphism. Now we will prove that ψg−1 is the inverse function by showing that ψg−1 (ψg (x)) = ψg (ψg−1 (x)) = x. First, ψg−1 (ψg (x)) = ψg−1 (gxg −1 ) = g −1 (gxg −1 )g = (g −1 g)x(g −1 g) = x. Now the other way, ψg (ψg−1 (x)) = ψg (g −1 xg) = g(g −1 xg)g −1 = (gg −1 )x(gg −1 ) = x. So ψg−1 is the inverse of ψg . b) Let g, h, x ∈ G and consider Ψ(gh)(x) = ψgh (x) = (gh)x(gh)−1 = ghxh−1 g −1 = g(hxh−1 )g −1 = ψg (hxh−1 ) = ψg (ψh (x)) = ψg ◦ ψh (x) = Ψ(g)Ψ(h)(x). So Ψ is a homomorphism. Exercise: 39 Section 3.7 Question: Prove that Aut((Q, +)) ∼ = (Q∗ , ×).

3.7. GROUP HOMOMORPHISMS

135

Solution: Suppose that ϕ is an automorphism of (Q, +). Then for all ab , dc ∈ Q, ϕ( ab + dc ) = ϕ( ab ) + dc (c/d) = a a a ∗ ϕ( ad+bc bd ). But since ϕ is a map Q → Q, for each b ∈ Q, there exists λ ∈ Q such that b = λϕ( b ), since the a c ∗ rationals are closed under division. Therefore there exist λ1 , λ2 , λ3 ∈ Q such that ϕ( b + d ) = ϕ( ab ) + dc = ad+bc = λ3 ab + λ3 dc . Therefore λ1 = λ2 = λ3 . Since ab and dc were λ1 ab + λ2 dc , which is is equal to ϕ( ad+bc bd ) = λ3 bd arbitrary in Q, this means that the map ϕ is defined by a single nonzero rational number λ such that ϕ( ab ) = λ ab . For suggestive notation, we will write ϕλ in place of ϕ. Now that we have classified the isomorphisms of Aut((Q, +)), we will show the isomorphism with (Q∗ , ×). Let Ψ : Aut((Q, +)) → (Q∗ , ×) be defined by ϕλ 7→ λ. The map Ψ is clearly bijective, since each nonzero rational number corresponds exactly to one automorphism. To see that Ψ is a homomorphism, we observe that Ψ(ϕλ1 ϕλ2 ) = Ψ(ϕλ1 )Ψ(ϕλ2 ) = λ1 λ2 = Ψ(ϕλ1 λ2 ). Therefore there is an automorphism between Aut((Q, +)) and (Q∗ , ×).

Exercise: 40 Section 3.7 Question: This exercise determines the automorphism group Aut(Zn ). Suppose that Zn is generated by the element z. a) Prove that every homomorphism ψ : Zn → Zn is completely determined by where it sends the generator z. b) Prove that every homomorphism ψ : Zn → Zn is of the form ψ(g) = g a for some integer a with 0 ≤ a ≤ n−1. For the scope of this exercise, denote by ψa the homomorphism such that ψa (g) = g a . c) Prove that ψa ∈ Aut(Zn ) if and only if gcd(a, n) = 1. d) Show that the function Ψ : U (n) → Aut(Zn ) with Ψ(a) = ψa is an isomorphism to conclude that U (n) ∼ = Aut(Zn ). Solution: Throughout this Exercise we will let Zn = {z i z n = 1}. a) Consider a homomorphism ψ : Zn → Zn where ψ(z) = z i . Then, for any element z k ∈ Zn , consider ψ(z k ) = ψ(z)ψ(z) . . . ψ(z) (k times) = z a z a . . . z a (k times) = z ka . Which shows that the homomorphism is completely determined by where z is sent. b) Let ψ be an arbitrary homomorphism that is determined by letting ψ(z) = z a for 0 ≤ a ≤ n − 1. Now consider any element z i ∈ Zn , then ψ(z i ) = ψ(z)i = (z a )i = z ai = (z i )a . This shows that every homomorphism is of the form ψ(g) = g a where 0 ≤ a ≤ n − 1 and ψ(z) = z a . c) Since this is an if and only if statement, we will prove both directions. (=⇒) Suppose that ψa ∈ Aut(Zn ), this implies that ψa is an isomorphism which implies that ψa preserves the order of elements. Since |z| = n, we must have |ψa (z)| = |z a | = |z|/ gcd(a, |z|) = n/ gcd(a, n) = n. This implies gcd(a, n) = 1. (⇐=) Suppose that gcd(a, n) = 1. Then |ψa (z)| = |z a | = |z|/ gcd(a, |z|) = n/ gcd(a, n) = n. Which implies that ψa (z) ∈ Im ψa is a generator for Zn which implies that Im ψa = Zn and ψa is surjective. Since ψa is surjective it is also injective since we are mapping back into the same group which has finite order. Therefore ψa is an automorphism of Zn . This completes the proof of the if and only if statement. d) Let ā, b̄ ∈ U (n) and z ∈ Zn and consider Ψ(ab)(z) = ψab (z) = z ab = (z b )a = (ψb (z))a = ψa (ψb (z)) = (ψa ◦ ψb )(z) = (Ψ(ā)Ψ(b̄))(z) which shows that Ψ is a homomorphism. To prove injectivity, consider ā, b̄ ∈ U (n) where Ψ(ā) = Ψ(b̄), then Ψ(ā)(z) = Ψ(b̄)(z) ⇒ ψa (z) = ψb (z) ⇒ z a = z b ⇒ a ≡ b mod n, which shows that Ψ is injective. By part c we know that if ψa ∈ Aut(Zn ) then gcd(a, n) = 1, which implies that ā ∈ U (n) so that Ψ(ā) = ψa . for every ψa ∈ Aut(Zn ). So Ψ is surjective. We have shown that Ψ is an isomorphism and we conclude that U (n) ∼ = Aut(Zn ).

136

CHAPTER 3. GROUPS

3.8 – Group Presentations Exercise: 1 Section 3.8 Question: Find a set of generators and relations for S3 . Prove that D3 ∼ = S3 . Solution: It is easiest to consider a triangle and label the vertices 1, 2, 3 moving counter-clockwise. Then the permutation (1 2 3) corresponds to a counterclockwise rotation of the triangle and the permutation (1 2) corresponds to a reflection across the line from vertex 3 to the midpoint of side between 1 and 2. The powers of (1 2 3) are id, (1 2 3), (1 3 2) and the powers of (1 2) are itself and id. The compositions (1 2)(1 2 3) = (2 3) and (1 2)(1 3 2) = (1 3) complete the generation of the whole group S3 by words. We now prove D3 ∼ = S3 , which will establish relations for our presentation of S3 . By Theorem 3.8.8, we need only map the generators of D3 to the generators of S3 and show that the relations in D3 are satisfied by the generators of S3 . We define a map ϕ to send r 7→ (1 2 3) and s 7→ (1 2). Then, (1 2 3)3 = (1 2 3)(1 2 3)(1 2 3) = (1 2 3)(1 3 2) = id, so ϕ(r)3 = id = ϕ(e), satisfying the first relation in D3 . Next, (1 2 3)(1 2) = (1 3) = (1 2)(1 3 2), so ϕ(rs) = ϕ(sr−1 ), satisfying the second relation in D3 . By Theorem 3.8.8, the function ϕ extends to a homomorphism between D3 and S3 . It is easy to see that ϕ is injective and surjective because the groups are of small cardinality; therefore ϕ is an isomorphism from D3 to S3 . Finally, because we have established an isomorphism between D3 and S3 , we know that the relations used to define D3 can be used under isomorphism to define S3 by a presentation. Therefore, a presentation for S3 is h(1 2 3), (1 2) | (1 2 3)3 = (1 2)2 = id, (1 2 3)(1 2) = (1 2)(1 2 3)−1 i. Exercise: 2 Section 3.8 Question: Find a set of generators and relations for the group Z2 ⊕ Z2 ⊕ Z2 . Solution: Consider the symbols x, y, z, where x signifies (z, e, e), y signifies (e, z, e), xy signifies (z, z, e), and so on. It is easy to see that the three elements represented by x, y, z generate the whole group Z2 ⊕ Z2 ⊕ Z2 . We need relations to assure that any generator has order 2 and that the generators commute with one another, since certainly xy and yx should refer to the same element. Therefore, a presentation is hx, y, z | x2 = y 2 = z 2 , xy = yx, xz = zx, yz = zyi. Exercise: 3 Section 3.8 Question: Prove that in any group, given any two elements x, y ∈ G, the subgroups hx, yi and hx, xyi are isomorphic. Solution: Notice that hx, xyi contains x−1 xy = y, a word made from x and xy. Similarly, hx, yi contains xy, a word made from x and y. Then, by the closure of groups, hx, xyi contains all elements of hx, yi, and vice versa. Define a map ϕ by x 7→ x, y 7→ y. Clearly x and y in hx, xyi satisfy the same relations as x and y in hx, yi because these are subgroups of the same group. Therefore by Theorem 3.8.8, ϕ extends to a homomorphism from hx, yi to hx, xyi. It is clear that ϕ is injective and surjective since every element in hx, yi is also in hx, xyi, making ϕ an isomorphism between the two subgroups. Since G was arbitrary, this result holds in any group. Exercise: 4 Section 3.8 Question: Consider the group presentation G = hx, y | x4 = y 3 = 1, xy = y 2 x2 i. Prove that G is the trivial group G = {1}. Solution: Notice first from xy = y 2 x2 we get by multiplying on the left by y the identity yxy = x2 . Then multiplying on the right by y 2 gives yx = x2 y 2 . Consider the expression xyxy. On the one hand we have xyxy = (xy)xy = (y 2 x2 )xy = y 2 x3 y and on the other we have xyxy = x(yx)y = x(x2 y 2 )y = x3 y 3 = x3 . Thus y 2 x3 y = x3 . Multiplying on the left by y gives us x3 y = yx3 . From x4 = 1, we have x9 = x. Then yx = yx9 = x3 yx6 = x6 yx3 = x9 y = xy. Hence y commutes with x. Since y and x commute, the relation xy = y 2 x2 gives yx = y 2 x2 and then by multiplying on the left by y we get y 2 x = x2 . Multiplying on the right by x−1 gives y 2 = x. Then x3 = y 6 = 1 and we already have x4 = 1. Thus x = xgcd(4,3) = 1. We then deduce that G = {1}.

3.8. GROUP PRESENTATIONS

137

Exercise: 5 Section 3.8 Question: Prove that hi, j | i4 = j 4 = 1, i2 = j 2 , ij = j 3 ii is a presentation of the quaternion group Q8 . In particular, show how to write −1, −i, −j, k, and −k as operations on the generators i and j. Solution: First, we know k = ij must be a relation in order for this to be the quaternion group, so we define the element k in this way. Immediately we observe, using the given relations, that k 2 = (ij)2 = ijij = ij 4 i = i2 = j 2 . It must now be the case that −1 = i2 = j 2 , and since we have shown that k 2 = i2 = j 2 , we have −1 = i2 = j 2 = k 2 as desired. It remains to write −i, −j, −k in terms of these generators. We know (−1)i must equal −i, so −i = (−1)i = i2 i = i3 , and by the same token −j = j 3 , −k = k 3 . We have now exhibited all the relations that determine the quaternion group except jk = i, ki = j, ji = −k, kj = −i, ik = −j. (See Example 3.3.12.) From the relation ij = j 3 i, we have k = −ji or ji = −k. From the definition of k, we have jk = jij = j 4 i = i, ki = iji = j 3 i2 = j 5 = j, kj = ijj = j 3 ij = j 6 i = j 2 i = i2 i = −i, and ik = iij = j 2 j = −j. This satisfies all the relations and establishes that the given group is Q8 . Exercise: 6 Section 3.8 Question: Show that the group G = hx, y | x5 = 1, y 2 = 1, x2 yxy = 1i is isomorphic to Z2 . Solution: x2 yxy = 1 ⇒ x2 = yx4 y ⇒ x2 = yx2 x2 y ⇒ x2 = y(yx4 y)(yx4 y)y ⇒ x2 = x3 ⇒ x = 1. Thus as the element x is trivial in G, G is characterized solely by the element y and is clearly isomorphic to Z2 . Exercise: 7 Section 3.8 Question: Prove that D7 and hx, y | x7 = y 3 = 1, yx = x2 yi have the same lattice structure. Solution: We need to find the subgroups of G = hx, y | x7 = y 3 = 1, yx = x2 yi. Obviously, hxi and hyi are subgroups of order 7 and 3 respectively. (Using the relation yx = x2 y, we put all the powers of x to the right of the word.) The powers of xy are xy = xy; (xy)2 = x3 y 2 ; and (xy)3 = xyx3 y 2 = xx3 yy 2 = 1. Hence, hxyi = {1, xy, x3 y 2 }. Similarly, we have 5 more subgroups of order 3: hx2 yi = {1, x2 y, x6 y 2 } hx3 yi = {1, x3 y, x2 y 2 } hx4 yi = {1, x4 y, x5 y 2 } hx5 yi = {1, x5 y, xy 2 } hx6 yi = {1, x6 y, x4 y 2 }. A subgroup that contains any power of x and any element of order 3 will also contain y and therefore be the whole group. Furthermore, if we take any any two elements of order 3 from different subgroups above, we can use them to obtain a power of x alone, and then by our previous comment we can deduce that any subgroup generated by such elements of order 3 is the whole group. Hence, the subgroup lattice of G is the following. G hxi hyi

hxyi

hx2 yi

hx3 yi

hx4 yi

hx5 yi

hx6 yi

{1} From the solution of Exercise 3.6.10, we see that this has the same lattice as D7 . Exercise: 8 Section 3.8 Question: Let G = ha, b | a6 = b7 = 1, ab = b3 ai. 1. Prove that every element in G can be written uniquely as bm an for 0 ≤ m ≤ 6 and 0 ≤ n ≤ 5. 2. Write the element a4 b2 a−2 b5 in the form bm an . 3. Find the order of ab and list all the powers (ab)i .

138

CHAPTER 3. GROUPS

Solution: 1. Let 0 ≤ i < 6 and 0 ≤ j < 7. Clearly ai can be written as b0 ai and bj can be written as bj a0 . All elements in the form bj ai are trivially in the correct form. Elements in the form ai bj can be written as b3j(mod7) ai . However, because 3 and 7 are relatively prime, for 0 < j < 7, {3j(mod7)} = {j}. Thus any element in the form ai bj is equal to a unique element in the form bj ai , making all unique elements in G be defined simply by bj ai . 2. a4 b2 a−2 b5 = a4 b2 a4 b5 = a4 b6 a4 = b3 a2 3. (ab)1 = ab ⇒ (ab)2 = abab = ab4 a = b5 a2 ⇒ (ab)3 = b5 a2 ab = b4 a3 ⇒ (ab)4 = b4 a3 ab = ba4 ⇒ (ab)5 = ba4 ab = b6 a5 ⇒ (ab)6 = b6 a5 ab = 1. Thus the order of ab = 6.

Exercise: 9 Section 3.8 Question: The quasidihedral group of order 16 is defined by QD16 = ha, b | a8 = b2 = 1, ab = ba3 i. Show that QD16 has 16 elements and then draw the subgroup lattice of QD16 . Solution: From the relation ab = ba3 , we see that in any word involving powers of a and b, we can move the b element to the left, though changing the powers on a. For example, ak b = ak−1 ba3 = ak−2 ba6 = · · · = ba3k . Hence every element in the group can be written as b` ak for ` = 0, 1, and k = 0, 1, . . . , 7. By associativity, it does not matter in what order we move the b generator through a word via the relation ab = ba3 . Hence QD16 has exactly 16 elements. In order to find the subgroup lattice, we do need to do quite a bit of work. We list some of the subgroups: hbi = {1, b} ha4 i = {1, a4 } ha2 i = {1, a2 , a4 , a6 } hbai = {1, ba, a4 , ba5 } k

and more generally

k

hba i = {1, ba } if k is even hbak i = {1, bak , a4 , bak+4 } if k is odd hb, a4 i = {1, a4 , b, ba4 } hba2 , a4 i = {1, a4 , ba2 , ba6 } we’ve found all the subgroups of order 4, 2 cyclic and 2 ∼ = Z2 ⊕ Z2 hai = {1, a, a2 , . . . , a7 } this is the only cyclic subgroup of order 8 hb, a2 i = {1, a2 , a4 , a6 , b, ba2 , ba4 , ba6 } hba, a2 i = {1, a2 , a4 , a6 , ba, ba3 , ba5 , ba7 } This is all the subgroups. The subgroup lattice looks like:

3.8. GROUP PRESENTATIONS

139 QD16

hb, a2 i

hba, a2 i

hai

hb, a4 i

hbai

ha2 i

hba3 i

hba2 , a4 i

hba4 i

hbi

ha4 i

hba2 i

hba6 i

{1}

Exercise: 10 Section 3.8 Question: Prove that if all nonidentity elements in a finite group G have order 2, then there exists a nonnegative integer n such that G = ha1 , a2 , . . . , an | a2i = 1 ai aj = aj ai for all i, j = 1, 2, . . . , ni. n times

}| { z Prove also that G ∼ = Z2n = Z2 ⊕ Z2 ⊕ · · · ⊕ Z2 . Solution: Suppose that G is a finite group such that all nonidentity elements have order 2. If G = {1}, then the conclusion of the exercise is satisfied trivially. Now, let n be the minimum number of generators of the group G. This number n exists by the well-ordering of the integers. We prove that G has the form G = ha1 , a2 , . . . , an | a2i = 1 ai aj = aj ai for all i, j = 1, 2, . . . , ni by induction on n. If n = 1, the G is cyclic, generated by some element a and we must have a2 = 1. Hence, G has the form described in the conclusion of the exercise. Now suppose that for some positive integer n, every finite group minimally generated by n elements such that all nonidentity elements have order 2 has the desired form. Consider a finite group G0 generated by n+1 elements such that all nonidentity elements have order 2. Suppose that G0 is generated by elements b1 , b2 , . . . , bn+1 . Then subgroup H = hb1 , b2 , . . . , bn i is generated by n elements and is such that all nonidentity elements have order 2. Then by the induction hypothesis, H has the form H = hc1 , c2 , . . . , cm | c2i = 1 ci cj = cj ci for all i, j = 1, 2, . . . , mi. A priori, we do not necessarily have m = n or ci = bi for 1 ≤ i ≤ n. However, {c1 , c2 , . . . , cm , bn+1 } then generate G0 so by the minimality of generators, we conclude that m + 1 = n + 1, so that m = n. Thus, G0 = hc1 , c2 , . . . , cn , bn+1 i and these generators satisfy c2i = b2n+1 = 1 and ci cj = cj ci for all 1 ≤ i, j, ≤ n. Since all nonidentity elements have order 2, (ci bn+1 )2 = 1 for all i. Thus, ci bn+1 ci bn+1 = 1 =⇒ bn+1 ci bn+1 = ci =⇒ ci bn+1 = bn+1 ci . Setting cn+1 = bn+1 , we see that G0 = hc1 , c2 , . . . , cn | c2i = 1 ci cj = cj ci for all i, j = 1, 2, . . . , n + 1i. By induction, the conclusion holds for all finite groups. Because of the relations in G as described above, every element can be written as ap11 ap22 · · · apnn , where pi = {0, 1}. Set Z2 = hz | z 2 = 1i. Note that the function ϕ : G → Zn2 defined by ϕ(ap11 ap22 · · · apnn ) = (z p1 , z p2 , . . . , z pn )

140

CHAPTER 3. GROUPS

is a surjective homomorphism. It is also easy to see that it is injective as well. Hence, G ∼ = Z2n . Exercise: 11 Section 3.8 Question: Prove that the subgroup h(1 2 3 4), (2 4)i in S4 is isomorphic to D4 . Solution: By labeling the vertices of the square transformed by D4 as 1, 2, 3, 4, it is clear that the permutation (1 2 3 4) is a counterclockwise rotation and the permutation (2 4) is a reflection across the diagonal from 2 to 4. Therefore we define a map ϕ such that ϕ(r) = (1 2 3 4) and ϕ(s) = (2 4). We will show that ϕ satisfies the conditions of Theorem 3.8.8. We observe that ϕ(r)4 = (1 2 3 4)(1 2 3 4)(1 2 3 4)(1 2 3 4) = (1 3)(2 4)(1 3)(2 4) = id = ϕ(r4 ) = ϕ(1) and ϕ(r)ϕ(s) = (1 2 3 4)(2 4) = (1 2)(3 4) = (2 4)(1 4 3 2) = (2 4)(1 2 3 4)−1 = ϕ(sr−1 ). Therefore, by the theorem, ϕ extends to a homomorphism between h(1 2 3 4), (2 4)i and D4 . It is not hard to see that h(1 2 3 4), (2 4)i has 8 elements, the same as D4 , making ϕ an isomorphism between h(1 2 3 4), (2 4)i and D4 .

Exercise: 12 Section 3.8 Question: We work in the group S5 . a) Find a subgroup of S5 isomorphic to D5 . b) Find a subgroup of S5 isomorphic to D6 . c) Prove that S5 does not have any subgroups isomorphic to Dn for n ≥ 7. Solution: Working in S5 . a) In to have a subgroup isomorphic to D5 , we need to find an element σ of order 5 and an element τ of order 2 such that στ = τ σ −1 . To be motivated, we cna look at what r and s do to the vertices of a pentagon for inspiration. Here is a subgroup that works: h(1 2 3 4 5), (2 5)(3 4)i. b) In to have a subgroup isomorphic to D6 , we need to find an element σ of order 6 and an element τ of order 2 such that στ = τ σ −1 . The motivation we used in part (a) is not available here. The only elements of order 6 in S5 are products of a 3-cycle and a 2-cycle. Without loss of generality (because we could always relable numbers), we can take σ = (1 2 3)(4 5). Looking for a 2-cycle τ with the desired properties we find that the following subgroup has generators with the desired properties: h(1 2 3)(4 5), (1 2)i. c) S5 cannot contain a subgroup isomorphic to Dn for n ≥ 7 because in Dn , the rotation has order n, whereas in S5 , the highest order of any element is 6. Exercise: 13 Section 3.8 Question: Consider homomorphisms Q8 → Z4 . 1. Prove that there exists no homomorphism ϕ : Q8 → Z4 = hz | z 4 = 1i such that ϕ(i) = z and ϕ(j) = 1. 2. Prove that there exists a homomorphism ψ : Q8 → Z4 = hz | z 4 = 1i such that ψ(i) = z 2 and ψ(j) = 1. Solution: 1. Suppose there exists a homomorphism ϕ : Q8 → Z4 = hz | z 4 = 1i such that ϕ(i) = z and ϕ(j) = 1. ϕ(−1) = ϕ(i ∗ i) = ϕ(i) ∗ ϕ(i) = z ∗ z = z 2 . But ϕ(−1) = ϕ(j ∗ j) = ϕ(j) ∗ ϕ(j) = 1 ∗ 1 = 1. Contradiction. Therefore no homomorphism exists with these properties. 2. Let ψ : Q8 → Z4 = hz | z 4 = 1i such that ψ(i) = z 2 , ψ(j) = 1, ψ(k) = z 2 and ψ(1) = 1. It can be seen that for a, b ∈ Q8 that ψ(a ∗ b) = ψ(a) ∗ ψ(b). This defines a homomorphism with the requisite properties. Exercise: 14 Section 3.8 Question: Prove that there exists a homomorphism from ϕ : D6 → GL2 (F3 ) such that 1 1 2 0 ϕ(r) = and ϕ(s) = . 0 1 0 1

3.8. GROUP PRESENTATIONS

141

Solution: For this exercise, we simply need to check the generator extension theorem. We note that in GL2 (F3 ), 1 1 0 1 has order 3 so raised to the 6th power it does give the identity. Also 2 0 0 1 has order 2. We check the relation rs = sr2 . 1 0 and

2 0

0 1 1 0

1 2 1 0

0 2 = 1 0

2 1 2 = 1 0

0 1 1 0

1 1 2 2 = 1 0

1 1

The given matrices satisfy the hypothesis for the Extension Theorem on Generators so there does exist a desired homomorphism ϕ. (The homomorphism ϕ is not injective but the same homomorphism from D3 is, establishing that 1 1 2 0 , 0 1 0 1 is a subgroup of GL2 (F3 ) isomorphic to D3 .) Exercise: 15 Section 3.8 Question: Prove that the subgroup

0 1

2 1 , 0 1

1 2

in GL2 (F3 ) is isomorphic to Q8 . Solution: Call the subgroup in GL2 (F3 ) by G = hA, Bi. We will use Theorem 3.8.8 and define a map ϕ : Q8 → G that sends i 7→ A and j 7→ B. We must show that ϕ(i)4 = ϕ(j)4 = I, ϕ(i)2 = ϕ(j)2 = ϕ(−1), and ϕ(i)ϕ(j) = ϕ(j)3 ϕ(i). For the first relation, we calculate that ϕ(i)4 = A4 = I = B 4 = ϕ(j)4 . Next, we note that 2 0 2 2 ϕ(i) = A = = B 2 = ϕ(j)2 . 0 2 So this matrix corresponds to −1 in Q8 . What is important is that ϕ(i)2 = ϕ(j)2 ; by defining ϕ(−1) to be equal to this matrix, we satisfy ϕ(i)2 = ϕ(j)2 = ϕ(−1), and the second relation is met. Finally, we calculate 2 1 ϕ(i)ϕ(j) = AB = = B 3 A = ϕ(j 3 )ϕ(i). 1 1 This satisfies the third relation and makes ϕ a homomorphism between Q8 and G because of Theorem 3.8.8. We compute all the possible elements in G as words of A and B: 1 0 0 2 1 1 2 0 2 1 2 2 1 1 2 1 , , , , , , , . 0 1 1 0 1 2 0 2 1 1 2 1 1 2 2 0 Since there are eight elements in G, just as in Q8 , we know that ϕ is an isomorphism between G and Q8 , as desired. Exercise: 16 Section 3.8 Question: Fix a positive integer n. Show that a function Dn → GL2 (R) that maps cos(2π/n) − sin(2π/n) 1 0 r 7−→ and s 7−→ sin(2π/n) cos(2π/n) 0 −1

142

CHAPTER 3. GROUPS

extends uniquely to a homomorphism ϕ : Dn → GL2 (R). Show also that this ϕ is injective. Solution: It is a standard check in linear algebra that because of trigonometric addition rules, cos(α) − sin(α) cos(β) − sin(β) cos(α + β) − sin(α + β) = . sin(α) cos(α) sin(β) cos(β) sin(α + β) cos(α + β) Hence,

n cos(2π/n) − sin(2π/n) cos(2π) − sin(2π) 1 = = sin(2π/n) cos(2π/n) sin(2π) cos(2π) 0

0 . 1

Also,

1 0

0 −1

2

=

0 . 1

1 0

Finally,

cos(2π/n) − sin(2π/n) 1 sin(2π/n) cos(2π/n) 0

0 cos(2π/n) sin(2π/n) = −1 sin(2π/n) − cos(2π/n) 1 0 cos(2π/n) sin(2π/n) = 0 −1 − sin(2π/n) cos(2π/n) −1 1 0 cos(2π/n) − sin(2π/n) = . 0 −1 sin(2π/n) cos(2π/n)

By the Extension Theorem on Generators, ϕ extends uniquely to a homomorphism Dn → GL2 (R). It is not hard to see that the matrices cos(2πk/n) − sin(2πk/n) cos(2πk/n) sin(2πk/n) k k ϕ(r ) = and ϕ(r s) = sin(2πk/n) cos(2πk/n) sin(2πk/n) − cos(2πk/n) for 0 ≤ k ≤ n − 1 are all distinct matrices. Hence, ϕ injective. Exercise: 17 Section 3.8 Question: Prove that the subgroup 2 0

0 1 , 1 0

1 1

in GL2 (F7 ) is isomorphic to the group G2 in Example 3.8.6. Solution: The group G2 is defined in the example by G2 = ha, b | a3 = b7 = 1, ab = b2 ai. The obvious strategy is to define a map ϕ that sends 2 0 1 1 a 7→ , b 7→ 0 1 0 1 and to show that this map satisfies the conditions of Theorem 3.8.8. We see immediately that 3 2 0 8 0 1 0 = = 0 1 0 1 0 1 3 1 1 1 7 1 0 = = 0 1 0 1 0 1 2 2 0 1 1 2 2 1 1 2 0 = = , 0 1 0 1 0 1 0 1 0 1 meaning that the relations of G2 are satisfied, and ϕ extends to a homomorphism from the given subgroup to G2 . We are given in the example that G2 has 21 elements. By brute force, it is not difficult to see that the given matrix subgroup also has only 21 elements, making ϕ a bijection and an isomorphism of groups. Exercise: 18 Section 3.6 Question: Sketch the Cayley graph for Z6 ⊕ Z2 using directed edges and colors corresponding to generators. Solution: Let Z6 be generated by an element x and Z2 by an element y. Then the Cayley graph of Z6 ⊕ Z2 is as follows.

3.8. GROUP PRESENTATIONS

143 (x2 , y)

(x1 , y)

(x2 , 1)

(x3 , y)

(x1 , 1)

(x3 , 1)

(x0 , 1)

(x4 , 1)

(x0 , y)

(x5 , 1)

(x4 , y)

(x5 , y)

Exercise: 19 Section 3.8 Question: Sketch the Cayley graph of D4 using the generating set {s, sr}. Solution: r3 sr

s

3

s r

ι

2

sr sr

sr2 r

Exercise: 20 Section 3.8 Question: Show that A4 is generated by {(1 2 3), (1 2)(3 4)} and then sketch the Cayley graph of A4 using these generators. Solution: Example 3.6.7 (and Exercise 3.6.16) shows that A4 is generated by {(1 2 3), (1 2)(3 4)}. The nicest presentation of the Cayley diagram of A4 with the given generators is on a (semiregular) truncated tetrahedron. In the diagram below, the single arrow corresponds to multiplication on the left by (123) and the double arrow corresponds to multiplication on the left by (12)(34). (143) (132) (14)(23) (142)

(123) 1 (13)(24) (134)

(243) (12)(34)

(124) (234)

144

CHAPTER 3. GROUPS

Exercise: 21 Section 3.8 Question: Decide whether some group has a tetrahedron as its Cayley graph for some set of generators. Solution: Suppose that Z2 is generated by the element z. Then the group Z2 ⊕ Z2 , with the (redundant) set of generators {(z, 1), (1, z), (z, z)} has the tetrahedron as it Cayley graph. Exercise: 22 Section 3.8 αm 1 α2 Question: Let S be a set and let F (S) be the free group on S. Prove that two reduced words w1 = sα 1 s2 · sm β1 β2 and w2 = t1 t2 · · · tβnn are equal in F (S) if and only if m = n, si = ti for all 1 ≤ i ≤ n and αi = βi for all 1 ≤ i ≤ n. Solution: We prove this result by induction on max(m, n). As a base case, consider max(m, n) = 1. Then if m = n = 0, then w1 = 1 = w2 . If m = 0 and n = 1, then w1 = 0 and w2 = tβ1 1 and it is obvious that w1 6= w2 . β1 1 Similarly if m = 1 and n = 0. If m = n = 1, then w1 = sα 1 and w2 = t1 . Then, w1 = w2 is equivalent to α1 −β1 α1 −β1 1 −β1 . 1 = s1 t1 . But then s1 t1 is not a reduced word, which means that s1 = t1 . Hence, we have 1 = sα 1 Since the symbols have no relations on themselves, this implies that α1 = β1 . αm 1 α2 For the induction step, suppose that for some positive integer k, two reduced words sα 1 s2 · sm and β1 β2 t1 t2 · · · tβnn with max(m, n) ≤ k are equal in F (S) if and only if m = n, si = ti for all 1 ≤ i ≤ n and αi = βi for all 1 ≤ i ≤ n. Now consider two reduced words w1 and w2 for the same form but with max(m, n) ≤ k + 1. 2 −β1 t1 . Hence, we have We know that w1 = w2 if and only if w1 w2−1 = 1. By Theorem 3.8.4, w2−1 = tn−βn · · · t−β 2 αm −βn 2 −β1 1 α2 t1 . 1 = sα · · · t−β 2 1 s2 · sm tn

Because of this equality, the word on the right is not reduced. All adjacent symbols s1 , s2 , . . . , sm and tn , · · · , t2 , t1 are distinct by definition except possibly for sn and tm . Since the word is not reduced, we must have sm = tn . Furthermore, if αm 6= βn , then −β

αm −βn 2 −β1 1 α2 tn−1n−1 · · · t−β w1 w2−1 = sα t1 1 s2 · sm 2

is reduced since every symbol in the word is distinct from the one before. This is a contradiction since w1 w2−1 = 1. β1 β2 αm βn 1 α2 Thus, αm = βn . We have proved that if sα 1 s2 · sm = t1 t2 · · · tn as reduced words, then tn = sm and βn−1 αm−1 β1 β2 α1 α2 αm = βn . This implies that s1 s2 · sm−1 = t1 t2 · · · tn−1 . We can now apply the induction hypothesis since max(m − 1, n − 1) ≤ k. Thus m = n and si = ti for all 1 ≤ i ≤ k + 1 and αi = βi as well. By induction, the theorem holds. Exercise: 23 Section 3.8 Question: Prove that Aut(Z2 ⊕ Z2 ) ∼ = S3 . Solution: The group Z2 ⊕Z2 can be written as the presentation hx, y | x2 = y 2 = 1, xy = yxi. An automorphism ϕ on Z2 ⊕ Z2 must send the generator x to another element of order 2 and the generator y to a different element of order 2. Every nonidentity element in Z2 ⊕ Z2 has order 2. Also, every automorphism ϕ must satisfy ϕ(1) = 1. If we label x as 1, y as 2, and xy as 3, then we can describe an automorphism as a bijection from {x, y, xy} back into itself, and hence as a permutation on {1, 2, 3}. Hence Aut(Z2 ⊕ Z2 ) is a subgroup of S3 . We need to see if we get the whole group. It is easy to show that the bijection ϕ1 defined by ϕ1 (x) = y and ϕ1 (y) = x, will also satisfy ϕ1 (xy) = xy and is a homomorphism. (This automorphism corresponds to (1 2).) However, it is also easy to show that the function ϕ2 defined by ϕ2 (x) = xy, ϕ2 (y) = y and ϕ2 (xy) = x is a homomorphism and a bijection. This ϕ2 corresponds to the permutation (1 3) in S3 . However, S3 is generated by (1 2) and (1 3) so Aut(Z2 ⊕ Z2 ) is isomorphic to S3 . Exercise: 24 Section 3.8 Question: This exercise guides the reader to prove that Aut(Q8 ) ∼ = S4 . a) Let ϕ be an arbitrary automorphism of Aut(Q8 ). Prove that 1. ϕ(i) ∈ {i, −i, j, −j, k, −k}; 2. ϕ(−1) = −1; 3. ϕ(j) ∈ {i, −i, j, −j, k, −k} − {ϕ(i), ϕ(−i)}; 4. all values of ϕ(x) are determined by ϕ(i) and ϕ(j).

3.8. GROUP PRESENTATIONS

145

b) Prove that all functions that are allowed by the above restrictions are valid automorphisms. Conclude that | Aut(Q8 )| = 24. c) By labeling the faces of a cube with {i, −i, j, −j, k, −k} where x and −x are opposite faces, show that Aut(Q8 ) is isomorphic to the group of rigid motions of a cube. d) Use Exercise 3.7.36 to conclude that Aut(Q8 ) ∼ = S4 . Solution: A proof that Aut(Q8 ) ∼ = S4 . a) Let ϕ ∈ Aut(Q8 ). 1. By Proposition 3.7.21, any automorphism ϕQ8 → Q8 must map i into another element of order 4. Hence, ϕ(i) ∈ {i, −i, j, −j, k, −k}. 2. Similarly, any automorphism ϕQ8 → Q8 must map −1 into another element of order 2. However, there is only one element of order 2 so ϕ(−1) = −1. 3. By the homomorphism property, ϕ(−i) = (−1) · ϕ(i). So if ϕ(i) = ±i, ±j, or ±k, then, ϕ(−i) = ∓i, ∓j, or ∓k. Furthermore, ϕ(j) must be an element of order 4 as well so ϕ(j) ∈ {i, −i, j, −j, k, −k} − {ϕ(i), ϕ(−i)}. 4. We know that hi, ji = Q8 so all values of ϕ are determined by ϕ(i) and ϕ(j). b) Exercise 3.8.5 gives a presentation of Q8 . It is easy to check that any of the possibilities described with the restrictions above give a homomorphism by the Extension Theorem on Generators. Furthermore, Q8 is generated by any two elements of order 4 that are not (−1) multiples of each other. {ϕ(i), ϕ(j)} consists of elements of order 4 that are not (−1) multiples of each other so ϕ is surjective. A surjective function between sets of the same size is also injective. Hence, ϕ is an isomorphism, and in this case an automorphism. We can tell that | Aut(Q8 )| = 24 because our restrictions impose 6 possibilities for ϕ(i) and then 4 independent possibilities for ϕ(j). c) We label the faces of a cube as described. Then mapping i to the six possibilities for ϕ(i) corresponds to mapping the face i to any of the faces of the cube. The condition ϕ(−1) = −1 corresponds to the opposite face −i being mapped to (−1)ϕ(i). In a rigid motion of the cube, once we know how one face is moved, there are 4 possible rotations around the axis perpendicular to and through the center of that face. These correspond to the choice options of ϕ(j). Hence, Aut(Q8 ) is precisely the group of rigid motions of a cube. d) By Exercise 3.7.36, Aut(Q8 ) ∼ = S4 . Exercise: 25 Section 3.8 Question: Consider the sculpture entitled Quintrino depicted in Figure 3.13 and let G be its group of symmetry. a) Show that |G| = 60. b) Show that G can be generated by two elements σ and τ . c) Show that G can be viewed as a subgroup of S12 by writing σ and τ explicitly as elements in S12 . d) (*) Show that G is isomorphic to A5 . Solution: Consider the sculpture entitled Quintrino a) There are 12 pentagonal star shapes. The group of symmetry of Quintrino must map a star to another star. Furthermore, with the knowledge that a reference star is mapped to another star, there are 5 ways to rotate that mapped star. Also, the group of symmetry of Quintrino must preserve the structure of the sculpture so knowing where 3 linearly independent edges map to, we know where the entire structure is mapped to. Having knowledge of exactly the position and orientation of one face, determines the rest of the structure. Hence, there are 12 × 5 = 60 elements in the group of symmetry. b) Fix two adjacent stars (faces) F1 and F2 . Consider σ the rotation by 360/5 = 72◦ degrees around the axis perpendicular to F1 and through the center of F1 and consider τ the rotation by 360/5 = 72◦ degrees around the axis perpendicular to F2 and through the center of F2 . c) We may consider how the group of symmetry of Quintrino moves the faces, i.e., consider G as a function on the set of faces (the 12 stars). This way of viewing G will make it a subgroup of S12 . A natural way to number the faces is start with one face as 1; the five pentagonal faces around it as 2, 3, 4, 5, and 6; and label the remaining six faces as 13 − x, where x is number on the face opposite the face with number x. Then if F1 is the face with number 1 and F2 with number 2, we have σ = (2 3 4 5 6)(7 11 10 9 8) τ = (1 3 8 9 6)(4 7 12 10 5).

146

CHAPTER 3. GROUPS

3.9 – Groups in Geometry Exercise: 1 Section 3.9 Question: Calculate the equations of transformation for the rotation by π/3 about the point (3, 2). Solution: By Example 3.9.6, the equations will be given by x p1 cos α − sin α x − p1 7−→ + y p2 sin α cos α y − p2 where α = π/3 and (p1 , p2 ) = (3, 2). Substituting, we have x 3 cos π/3 − sin π/3 x−3 7−→ + y 2 sin π/3 cos π/3 y−2 √ x−3 3 1/2 − 3/2 = + √ y−2 2 3/2 1/2 √ 3 (1/2)(x − 3) − ( 3/2)(y − 2) √ = + 2 ( 3/2)(x − 3) + (1/2)(y − 2) √ √ 3 (1/2)x − 3/2√− ( 3/2)y + 3 √ = + 2 ( 3/2)x − 3 3/2 + (1/2)y + 1 √ √ 3 + (1/2)x − 3/2√− ( 3/2)y + 3 √ = 2 + ( 3/2)x − 3 3/2 + (1/2)y + 1 √ √ (1/2)x − ( 3/2)y + √3 + 3/2 √ = . ( 3/2)x + (1/2)y − 3 3/2 + 3 Exercise: 2 Section 3.9 Question: Determine the equations of transformation for the reflection through the line that makes an angle of π/6 with the x-axis and goes through the point (3, 4). √ Solution: A line making an angle of π/6 with the x-axis will have√slope sin(π/6)/ cos(π/6) = 3/3. Using slope-intercept form through the point (3, 4), the y-intercept is 4 − 3. Therefore if we translate the line by √ −4 + 3 in the y direction, we have a line through the origin making an angle of π/6 with the x-axis, allowing us to apply the theory of this section. Combining Example 3.9.5 and Example 3.9.6, the equations will be given by x p1 cos 2β − sin 2β x − p1 7−→ + y p2 sin 2β cos 2β y − p2 √ where β = π/6 and (p1 , p2 ) = (0, 4 − 3). Substituting, we have 0√ x − 0√ x cos π/3 − sin π/3 7−→ + y sin π/3 cos π/3 4− 3 y − (4 − 3) √ 0√ x √ 1/2 − 3/2 = + √ 4− 3 y−4+ 3 3/2 1/2 √ √ 0√ (1/2)x − ( 3/2)(y − 4 + √3) √ = + 4− 3 ( 3/2)x + (1/2)(y − 4 + 3) √ √ 0√ (1/2)x − ( 3/2)y + 3/2√− 3/2 √ = + 4− 3 ( 3/2)x + (1/2)y − 2 + 3/2 √ √ (1/2)x − ( 3/2)y + 3/2 √ − 3/2 √ = . ( 3/2)x + (1/2)y + 2 + 3 3/2 Exercise: 3 Section 3.9 Question: Determine the equations of transformation for the reflection through the line 2x + 3y = 6. Solution: This line has a y-intercept of 2, so if we translate it by (0, −2), it runs through the√origin. Its slope is −2/3, so it forms a triangle with horizontal leg 3 and vertical leg −2, and hence hypotenuse 11. This allows

3.9. GROUPS IN GEOMETRY

147

√ us to determine the angle the line makes at the origin by arcsin(2/ 11), for which we will write β. (Note that β is in the fourth quadrant.) Combining Example 3.9.5 and Example 3.9.6, the equations will be given by x p1 cos 2β − sin 2β x − p1 7−→ + y p2 sin 2β cos 2β y − p2 where β is as above and (p1 , p2 ) = (0, −2). Substituting, we have x 0 cos 2β − sin 2β x−0 7−→ + y 2 sin 2β cos 2β y−2 0 x cos 2β − (y − 2) sin 2β = + 2 x sin 2β + (y − 2) cos 2β x cos 2β − y sin 2β + 2 sin 2β = . x sin 2β + y cos 2β − 2 cos 2β + 2

Exercise: 4 Section 3.9 Question: Determine the equations of the isometry of reflection-translation through the line x = y with a translation distance of 1 (in the positive x-direction). Solution: To reflect across the line x = y, we only swap the x and y coordinates. To translate 1 unit in the positive x-direction, we merely add 1 to the new x-coordinate. Therefore the equations are x y+1 7−→ . y x

Exercise: 5 Section 3.9 Question: We work in R3 . Show by direct matrix calculations that rotation by θ about the x-axis, followed by rotation by α about the y-axis is not generally equal to the rotation by α about the y-axis followed by rotation by θ about the x-axis. Solution: The matrix for a rotation by θ around the x-axis is   1 0 0 Rx,θ = 0 cos θ − sin θ 0 sin θ cos θ and the matrix for a rotation by α around the y-axis is  cos θ Ry,α =  0 − sin θ Computing the two orders, we have   1 0 0 cos θ Rx,θ Ry,α = 0 cos θ − sin θ  0 0 sin θ cos θ − sin θ and



cos θ Ry,α Rx,θ =  0 − sin θ

 0 sin θ 1 1 0  0 0 cos θ 0

0 cos θ sin θ

0 1 0

0 1 0

 sin θ 0 . cos θ

  cos θ sin θ 0  =  sin2 θ cos θ − sin θ cos θ   0 cos θ − sin θ =  0 cos θ − sin θ

0 cos θ sin θ

 0 − sin θ cos θ cos2 θ

sin2 θ cos θ sin θ cos θ

 sin θ cos θ − sin θ  , cos2 θ

which are not the same in general. Exercise: 6 Section 3.9 Question: Use Theorem 3.9.5 to prove the claim that it suffices to know how an isometry f maps three non-collinear points to know f uniquely.

148

CHAPTER 3. GROUPS

Solution: Suppose we know that an isometry of the plane f maps three points p~, ~q and ~r to the three points ~u, ~v , and w. ~ By Theorem 3.9.5, we know that f (~x) = A~x + ~b for an orthogonal matrix A and a fixed point ~b. Then f (~q) − f (~ p) = A~q − A~ p = A(~q − p~) and similarly, f (~r) − f (~ p) = A(~r − p~). Furthermore, since p~, ~q and ~r are noncollinear, then ~q − p~ and ~r − p~ are linearly independent. Then we know that as equations with 2 × 2 matrices ~ − ~u . A ~q − p~ ~r − p~ = ~v − ~u w Thus, A = ~v − ~u

−1 w ~ − ~u ~q − p~ ~r − p~ ,

where the inverse exists because the columns are linearly independent. Knowing A, we can now find ~b uniquely by ~b = vu − A~ p. Once we know A and ~b, we know the isometry uniquely. Exercise: 7 Section 3.9 Question: Let L1 and L2 be two parallel lines in the Euclidean plane. Prove that reflection through L1 followed by reflection through L2 is a translation of vector ~v that is twice the perpendicular displace from L1 to L2 . Solution: We will use the result of Exercise 3.9.6 as an aide. Consider the diagram below that depicts two parallel lines as well as three chosen points that are not collinear. L1

L2

B 00

B

C0

A

C

A00

C 00

Reflection through L1 maps leaves A and B alone but maps C to C 0 . Then reflection through L2 maps A and B to A00 and B 00 and maps C 0 to C 00 as shown. On the set {A, B, C}, the resulting transformation acts as a translation by a vector ~v that is twice the displacement from L1 to L2 . By Exercise 3.9.6, the composition of the two reflections is this translation. Exercise: 8 Section 3.9 Question: Let L1 and L2 be two lines in the plane that are not parallel. Prove that reflection through L1 followed by reflection through L2 is a rotation about their point of intersection of an angle that is double the angle from L1 to L2 (in a counterclockwise direction). Solution: Suppose that L1 and L2 are lines that intersection at a point p~ = pp12 . Suppose that L1 makes an angle of β1 with the x-axis and that L2 makes an angle of β2 with the x-axis. By Example 3.9.7 and 3.9.8, reflection through Li has the function x p1 cos 2βi sin 2βi x − p1 fi = + . y p2 sin 2βi − cos 2βi y − p2 Thus, the composition has the functional expression x p1 cos 2β2 sin 2β2 cos 2β1 sin 2β1 x − p1 f2 ◦ f1 = + y p2 sin 2β2 − cos 2β2 sin 2β1 − cos 2β1 y − p2 p1 cos 2β2 cos 2β1 + sin 2β2 sin 2β1 cos 2β2 sin 2β1 − sin 2β2 cos 2β1 x − p1 = + p2 sin 2β2 cos 2β1 − cos 2β2 sin 2β1 cos 2β2 cos 2β1 + sin 2β2 sin 2β1 y − p2 p1 cos(2(β2 − β1 )) − sin(2(β2 − β1 )) x − p1 = + . p2 sin(2(β2 − β1 )) cos(2(β2 − β1 )) y − p2

3.9. GROUPS IN GEOMETRY

149

According to Examples 3.9.7 and 3.9.8, this isometry corresponds to rotation by 2(β2 − β1 ) about the point p~, which is the point of intersection of L1 and L2 . Exercise: 9 Section 3.9 Question: Let A = (0, 0), B = (1, 0), and C = (0, 1). Determine the isometry obtained by composing a rotation about A of angle 2π/3, followed by a rotation about B of angle 2π/3, followed by a rotation about C of angle 2π/3. Find the equations of transformation and describe it in simpler terms. Solution: The rotation by 2π/3 about A has the analytic expression of √ ! 3 1 − − x x 2 . R1 = √32 1 y y − 2

2

The rotation by 2π/3 about B has the analytic expression of √ ! 3 1 − x x−1 1 − 2 2 √ R2 = = + 3 y y 0 − 12 2

− 23 − 12

! 3 x 2√ . + y − 23

1 − √2

− 23 − 12

√

! √ 3 x + 23 . y 2

3 2

The rotation by 2π/3 about C has the analytic expression of √ ! 1 − 23 − x 0 x 2 √ R3 = + = 3 y−1 y 1 − 12 2 Then, x R2 ◦ R1 = y

1 − √2 3 2

√

− 23 − 12

√

1 − √2

3 2

!2 3 x 2√ + = y − 23

−√12 − 23

√ ! 3 3 x 2√ 2 + , y − 23 − 21

and so x R3 ◦ R2 ◦ R1 = y =

1 − √2 3 2

x + y

√ ! − 23 − 12 3 2 √ 3 3 2

−√12 − 23

√

3 2 − 12

! x + 23 y

√

− 23

!

+

3 2√

− 23

.

√ The resulting transformation is a translation by (3/2, 3 3/2). Exercise: 10 Section 3.9 Question: Let f be the isometry of rotation by α about the point A = (a1 , a2 ) and let g be the isometry of rotation by β about the point B = (b1 , b2 ). Show that f ◦ g may be described by a rotation by α + β about B followed by a translation and give this translation vector. Solution: The function f is described by x a1 cos α − sin α x − a1 cos α − sin α x 1 − cos α sin α a1 f = + = + . y a2 sin α cos α y − a2 sin α cos α y − sin α 1 − cos α a2 Similarly, x cos β g = y sin β

− sin β cos β

x 1 − cos β + y − sin β

sin β 1 − cos β

b1 . b2

The composition of these two rotations has the expression x cos β − sin β cos α − sin α x 1 − cos α sin α a1 g◦f = + + y sin β cos β sin α cos α y − sin α 1 − cos α a2 1 − cos β sin β b1 − sin β 1 − cos β b2 cos(α + β) − sin(α + β) x cos β − sin β 1 − cos α sin α a1 = + + sin(α + β) cos(α + β) y sin β cos β − sin α 1 − cos α a2 1 − cos β sin β b1 . − sin β 1 − cos β b2

150

CHAPTER 3. GROUPS

This is the expression of a rotation by α + β around the origin followed by a translation. [This composition will in fact turn out to be a rotation about a given point in the plane.] Exercise: 11 Section 3.9 Question: Let f be the plane Euclidean isometry of rotation by α about a point A and let t be a translation by a vector ~v . Prove that the conjugate f ◦ t ◦ f −1 is a translation and determine this translation explicitly. Solution: If A = (a1 , a2 ), the analytic expression of f is x cos α − sin α x − a1 a f = + 1 . y sin α cos α y − a2 a2 Thus, the analytic expression of t ◦ f −1 is x cos α t ◦ f −1 = y − sin α

sin α cos α

x − a1 a + v1 + 1 . y − a2 a2 + v2

Finally, the analytic expression of f ◦ t ◦ f −1 is cos α − sin α cos α sin α x − a1 a1 + v1 a a −1 x f ◦t◦f = + − 1 + 1 y sin α cos α − sin α cos α y − a2 a2 + v2 a2 a2 cos α − sin α cos α sin α x − a1 v a = + 1 + 1 sin α cos α − sin α cos α y − a2 v2 a2 x − a1 cos α − sin α v1 a = + + 1 y − a2 sin α cos α v2 a2 x cos α − sin α v1 = + . y sin α cos α v2 cos α − sin α −1 Thus, f ◦ t ◦ f is a translation by a vector ~v , which is the vector ~v rotated by an angle α. sin α cos α Exercise: 12 Section 3.9 Question: Let g be the plane Euclidean isometry of reflection through a line L and let t be a translation by a vector ~v . Prove that the conjugate g ◦ t ◦ g −1 is a translation and determine this translation explicitly. Solution: Suppose that L is a line through a point A = (a1 , a2 ) and with angle β with respect to the x-axis. Then by Examples 3.9.7 and 3.9.8, the analytic expression of g is x cos 2β sin 2β x − a1 a g = + 1 . y sin 2β − cos 2β y − a2 a2 Note that g = g −1 so the analytic expression of t ◦ f −1 is x cos 2β sin 2β x − a1 a + v1 t ◦ g −1 = + 1 . y sin 2β − cos 2β y − a2 a2 + v2 Finally, the analytic expression of g ◦ t ◦ g −1 is x cos 2β sin 2β cos 2β sin 2β x − a1 a + v1 a a g ◦ t ◦ g −1 = + 1 − 1 + 1 y sin 2β − cos 2β sin 2β − cos 2β y − a2 a2 + v2 a2 a2 cos 2β sin 2β cos 2β sin 2β x − a1 v a = + 1 + 1 sin 2β − cos 2β sin 2β − cos 2β y − a2 v2 a2 x − a1 cos 2β sin 2β v1 a + 1 = + y − a2 sin 2β − cos 2β v2 a2 x cos 2β sin 2β v1 = + . y sin 2β − cos 2β v2 cos 2β sin 2β −1 Thus, g ◦ t ◦ g is a translation by a vector ~v , which is the vector ~v reflected through a line sin 2β − cos 2β through the origin that makes an angle of β with respect to the x-axis.

3.9. GROUPS IN GEOMETRY

151

Exercise: 13 Section 3.9 Question: Prove that orthogonal matrices have determinant 1 or −1. Prove also that SO(n) = {A ∈ O(n) | det(A) = 1} is a subgroup of O(n) and hence of GLn (R). [The subgroup SO(n) is called the special orthogonal group.] Solution: An orthogonal matrix is an n × n matrix such that A> A = I. Taking the determinant of this condition gives det(A> A) = 1 =⇒ det(A> ) det(A) = 1 =⇒ det(A)2 = 1. Hence, the determinant of orthogonal matrices is 1 or −1. Furthermore, the determinant function is a homomorphism from O(n) to the group ({1, −1}, ×). By definition, SO(n) is the kernel of det : O(n) → ({1, −1}, ×) so SO(n) is a subgroup of O(n). Exercise: 14 Section 3.9 Question: Prove that SO(2) (see Exercise 3.9.13) consists of matrices cos θ − sin θ θ ∈ [0, 2π) . sin θ cos θ Solution: The set of matrices in SO(2) consists of 2 × 2 real matrices a11 a12 A= a21 a22 such that a211 + a221 = 1, a212 + a222 = 1, a11 a12 + a21 a22 = 0 and a11 a22 − a12 a21 = 1. For any pair of real numbers (x, y) ∈ R2 that satisfy x2 + y 2 = 1, there exists θ ∈ [0, 2π) such that x = cos θ and y = sin θ. So the first condition implies that there exists θ with a11 = cos θ and a21 = sin θ. The second condition is similar so there exists β such that a12 = cos β and a22 = sin β. The condition a11 a12 + a21 a22 = 0 is then equivalent to 0 = cos θ cos β + sin θ sin β = cos(β − θ), which means that β − θ = π2 + πk for some k ∈ Z. Finally the condition a11 a22 + a12 a21 = 1 implies that π π π + πk . 1 = cos θ sin θ + + πk − sin θ cos θ + + πk = sin 2 2 2 This implies that k is even. Hence, β = θ + π2 . Thus, the matrix A is cos θ − sin θ A= . sin θ cos θ

Exercise: 15 Section 3.9 Question: Prove that the function dt : R2 × R2 → R given by dt ((x1 , y1 ), (x2 , y2 )) = |x2 − x1 | + |y2 − y1 | satisfies the conditions of a metric on R2 as defined in Definition 3.9.2. Prove also that the set of surjective isometries for dt is a group and show that it is not equal to the group of Euclidean isometries. Solution: Consider the function dt : R2 × R2 → R given by dt ((x1 , y1 ), (x2 , y2 )) = |x2 − x1 | + |y2 − y1 |. Identity: It is obvious that for all P ∈ R2 , dt (P, P ) = 0. Conversely, suppose that dt (P, Q) = 0. Then |xQ − xP | + |yQ − yP | = 0. Since both |xQ − xP | and |yQ − yP | are nonnegative, they must both be 0 for their sum to be zero. Hence xQ = xP and yQ = yP . Thus Q = P . Symmetry: We have dt ((x2 , y2 ), (x1 , y1 )) = |x1 − x2 | + |y1 − y2 | = |x2 − x1 | + |y2 − y1 | = dt ((x1 , y1 ), (x2 , y2 )).

152

CHAPTER 3. GROUPS

Triangle Inequality: It is well known that for any two real numbers a and b, we have (*) |a + b| ≤ |a| + |b|. For any three points P , Q, and R in R2 , we have dt (P, R) = |xR − xP | + |yR − yP | = |(xR − xQ ) + (xQ − xP )| + |(yR − yQ ) + (yQ − yP )| ≤ |xR − xQ | + |xQ − xP | + |yR − yQ | + |yQ − yP | = dt (Q, R) + dt (P, Q). Let Gt be the set of surjective isometries on R2 equipped with composition, i.e., Gt is the set of functions f : R2 → R2 such that dt (f (P ), f (Q)) = dt (P, Q) for all P, Q ∈ R2 . We first prove that composition is a binary operation on Gt . If f, g ∈ Gt , then for all P, Q ∈ R2 , dt (f ◦ g(P ), f ◦ g(Q)) = dt (f (g(P )), f (g(Q))) = dt (g(P ), g(Q)) = dt (P, Q). Hence, f ◦ g ∈ Gt . It is easy to see that an isometry is injective since if f (P ) = f (Q), then dt (P, Q) = dt (f (P ), f (Q)) = 0, so by the identity axiom, P = Q. Since Gt consists of surjective functions, we conclude that all functions in Gt are bijections. Obviously, function composition is associative. Also, the identity function on R2 is a surjective isometry so is in Gt . Finally, for a bijective isometry f ∈ Gt , for all P, Q ∈ R2 , since f is an isometry, then applied to f −1 (P ) and f −1 (Q), we have dt (P, Q) = dt (f (f −1 (P )), f (f −1 (Q))) = dt (f −1 (P ), f −1 (Q)). Thus, the inverse function f −1 ∈ Gt . We conclude that Gt is a group. Exercise: 16 Section 3.9 Question: Find a presentation for the frieze group associated to the pattern in Figure 3.15. Solution: The pattern in Figure 3.15 is invariant under a unit translation t that maps, say one vertical opening flower to the next vertical opening flower to the right. The element t has infinite order in the frieze group G of this pattern. It is obvious that the pattern is not symmetric by reflection through the x-axis, and no glide-reflections either. We might wonder if there are reflections through vertical lines through the vertical opening flowers but that is not the case because of the hatch work pattern. We might expect that there are rotations of 180◦ around the center of the hatchwork but that is not the case either because of the vertical opening flowers. We conclude that G = hti. Exercise: 17 Section 3.9 Question: Find a presentation for the frieze group associated to the pattern in Figure 3.16. Solution: The chair rail pattern has no rotational symmetry and no reflection through the x-axis or glide reflections. Let s1 be the reflection through a vertical axis through the middle of a specific oval medallion and let s2 be the reflection through a vertical axis through the middle of the three lines immediately to the right of the specific medallion. The reflections s1 and s2 are symmetries of this frieze pattern. Note that s2 ◦ s1 = t corresponds to a unit translation to the right, i.e., mapping one oval medallion to the medallion immediately to the right. The frieze group for this pattern is G = hs1 , s2 | s21 = s22 = 1, i. Note that any reflection through a vertical axis can be expressed as tn s1 t−n of tn s2 t−n for an appropriate integer n. Exercise: 18 Section 3.9 Question: Find a presentation for the frieze group associated to the pattern in Figure 3.18. Show that it is the same as the frieze group associated to the pattern in Figure 3.17. Solution: This frieze group includes a number of symmetries: • reflections through vertical axes through the middle of gray or white “lamp” patterns; • glide reflections that maps one gray “lamp” into a white “lamp”;

3.9. GROUPS IN GEOMETRY

153

• rotations by 180◦ about any point along the x-axis (middle) that intersects a “lamp” boundary. Consider the white upsidedown lamp L in the middle of the part of the pattern shown. Call s the reflection through axis of symmetry of L. Call g the glide-reflection that maps the lamp L to the rightside up lamp immediately to its right. Note that t = g 2 is a unit translation symmetry to the right. Call r the rotation around the center of rotation immediately on the right of L. We claim that the frieze group G of this pattern is generated by s, g, and r. Every reflection through a vertical line through a white lamp has the form tn st−n for an appropriate integer n. Every reflection through a vertical line through a gray lamp has the form (tn g)s(tn g)−1 for an appropriate integer n. Powers of g give all glide-reflections and translations that keep the pattern invariant. Every rotation by 180◦ around some center of rotation of the symmetry pattern has the form tn rt−n or (tn g)r(tn g)−1 for an appropriate integer n. We also observe that r = gs. Hence, in fact, G is generated by r and s with G = hr, s | r2 = s2 = 1i.

Exercise: 19 Section 3.9 Question: Find a presentation for the frieze group associated to the following pattern and then sketch the fundamental pattern. Solution: This pattern has the same frieze group as the pattern in Exercise 3.9.18. Exercise: 20 Section 3.9 Question: Prove that there are only 7 nonisomorphic frieze groups. Solution: In order to prove that there are only 7 nonisomorphic frieze group, it is useful to consider what types of patterns can occur within one section such that the entire pattern is mapped out by applying the translation subgroup on that section. That “fundamental” section will be drawn between dashed lines below. We observe that if there is a rotation symmetry of 180◦ about a point A, then there is also a rotation −−→ symmetry of 180◦ about a point B, where 2AB is the unit translation. Furthermore, if the frieze group has a reflection symmetry through a vertical axis L, then it has reflection symmetry through a vertical axis L0 such that L0 is displaced from L by a translation 21 ~v , where ~v is the unit translation symmetry. Also, any group with a unit glide reflection symmetry g is such that g 2 is he unit translation symmetry. The following frieze pattern consists of just the translation group.

Frieze group with translation and horizontal reflection symmetry.

Frieze group with translation and vertical reflection symmetry.

Frieze group with translation and rotational symmetry.

154

CHAPTER 3. GROUPS

Frieze group with translation, horizontal reflection, and vertical reflection symmetry.

Frieze group with glide reflection symmetry.

Frieze group with glide reflection symmetry and vertical reflection symmetry.

All other locations for centers of rotation or vertical axes of symmetry lead to the same groups 2 through 5. Any frieze group with glide reflection symmetry such that the unit translation symmetry is still between the hashed lines is given in groups 6 and 7. Exercise: 21 Section 3.9 Question: Prove that the wallpaper group for the following pattern is not isomorphic to the wallpaper groups for either Figure 3.19a or 3.19b.

Solution: The wallpaper group for this pattern has elements of order 3, rotations of 120◦ around various centers (centers of the flowers; centers of the negative space of the flowers). However, the wallpaper group for 3.19a does no have any elements of order 3. Hence the wallpaper group of this exercise is not isomorphic to that of Figure 3.19a. On the other hand, the wallpaper group of Figure 3.19b has elements of order 6, symmetries of rotation by 60◦ around the centers of flowers. The wallpaper group in this exercise does not have elements of order 6 so is not isomorphic to that of Figure 3.19b. Exercise: 22 Section 3.9 Question: Prove that the wallpaper group for the following pattern is not isomorphic to the wallpaper groups for either Figure 3.19a or 3.19b.

3.9. GROUPS IN GEOMETRY

155

Solution: The wallpaper group for the pattern in this exercise has elements of order 4, e.g., rotations by 90◦ around the centers of the dark squares. Neither of the wallpaper groups in Figure 3.19a nor 3.19b have elements of order 4. Hence, the wallpaper groups cannot be isomorphic.

Exercise: 23 Section 3.9 Question: Sketch a reasonable portion of the pattern generated by the following fundamental pattern and the group indicated in each exercise. Assume the minimum distance for any translation is 2. The wallpaper group corresponding to Figure 3.19a.

Solution:

Exercise: 24 Section 3.9 Question: Sketch a reasonable portion of the pattern generated by the following fundamental pattern and the group indicated in each exercise. Assume the minimum distance for any translation is 2. The wallpaper group corresponding to Figure 3.19b.

156

CHAPTER 3. GROUPS

Solution:

Exercise: 25 Section 3.9 Question: sketch a reasonable portion of the pattern generated by the following fundamental pattern and the group indicated in each exercise. Assume the minimum distance for any translation is 2. The wallpaper group corresponding to the pattern in Exercise 3.9.29.

Solution:

3.9. GROUPS IN GEOMETRY

157

Exercise: 26 Section 3.9 Question: sketch a reasonable portion of the pattern generated by the following fundamental pattern and the group indicated in each exercise. Assume the minimum distance for any translation is 2. The wallpaper group corresponding to the pattern in Exercise 3.9.21.

Solution:

Exercise: 27 Section 3.9 Question: Find a presentation for the wallpaper group corresponding to Figure 3.19a. Solution: Call G the wallpaper group corresponding to Figure 3.19a. One element in G is the vertical glide reflection g1 as discussed in the text. We consider this glide reflection to be with respect to a specific vertical line and gliding upward a unit start fish. Call g2 the vertical glide reflection along the vertical line (that has glide reflection symmetry) immediately to the right of that for g2 . Note that g12 = g22 and both correspond to a unit symmetry of vertical translation ty . Furthermore, g2 g1 produces a translation of symmetry that is vertical by one unit and horizontal by one unit. Hence, if we call tx the unit symmetry translation in the horizontal direction, we have tx = g2−2 (g2 g1 ) = g2−1 g1 . Then every other glide reflection is of the form tnx g1 t−n or tnx g2 t−n x x for some integer n. So g1 and g2 generate G.

158

CHAPTER 3. GROUPS

We consider the relations on g1 and g2 . We already have g12 = g22 . Since translations commute with one another, we obtain a relation g2−1 g1 g2 g1 = g2 g1 g2−1 g1 ⇐⇒ g2−1 g1 g2 = g2 g1 g2−1 ⇐⇒ g2−1 g1 g22 = g2 g1 ⇐⇒ g2−1 g13 = g2 g1 ⇐⇒ g2−1 g22 g1 = g2 g1 . Hence, given the relation g12 = g22 , we automatically have the relation that translations commute. Hence, a presentation for the wallpaper group corresponding to Figure 3.19a is G = hg1 , g2 | g12 = g22 i. The subgroup hg2 g1 , g2−1 g1 i ∼ = Z ⊕ Z is the subgroup of translations. Exercise: 28 Section 3.9 Question: Find a presentation for the wallpaper group corresponding to the pattern in Exercise 3.9.22. Solution: The wallpaper group G for the pattern in Exercise 3.9.22 has many different symmetries. Around the center of each black square, the D4 dihedral symmetries are symmetries of this pattern. Fix one specific black square B and call r and s the rotation by 90◦ around its center and reflection through its horizontal axis. (Note: r and s generated the D4 symmetry of the wallpaper group only around this one specific black square. There are an infinite number of other black squares and similar rotations and reflections.) Furthermore, call t the unit translation to the right, i.e., the translation of the plane that moves one black square to the black square immediately to the right. We claim that r, s, and t generate G. Obviously, r has order 4, s has order 2, and t has infinite order. We also know that rs = sr−1 . Also, st = ts, which corresponds to a horizontal glide reflection. Note that rtr−1 corresponds to a unit vertical translation. Since translations commute, we have the relation rtr−1 t = trtr−1 . We also observe the additional relation that r2 tr−2 = t−1 , which can be rewritten as r2 t = t−1 r2 or equivalently tr2 = r2 t−1 . Every translation symmetry can be obtained by a composition of t and rtr−1 . For example, a translation by the vector (m, n), where we view the centers of black squares as integral units, is tm (rtr−1 )n = tm rtn r−1 . The rotation by 90◦ k around the center of the black square located at (m, n) is given by (tm rtn r−1 )rk (tm rtn r−1 )−1 . The reflection similar to srk but for a line through the center of the black square located at (m, n) is given by (tm rtn r−1 )srk (tm rtn r−1 )−1 . There are some other reflections and rotations in the wallpaper group that do not arise from the dihedral symmetries about the center of some black square. For example, consider the vertical reflection symmetry w of axis halfway between B and the black square immediate to its right. By previous exercises in this section, we know that w(sr2 ) = t. Thus w = tr2 s. Then sw = ws = tr2 gives rotation by 180◦ about the middle of the gray diamond between B and the next black square to the right. Also wrs = tr2 srs = tr gives the rotation by 90◦ about the point of intersection between the w axis and the rs axis. In this way, every symmetry of the wallpaper pattern is obtained by r, s, and t. We have the presentation G = hr, s, t | r4 = s2 = 1, rs = sr−1 , tr2 = r2 t−1 , rtr−1 t = trtr−1 i.

Exercise: 29 Section 3.9 Question: Find a presentation for the wallpaper group corresponding to the wallpaper pattern here below. Solution: Let x be a horizontal translation at a fixed distance d and y be a translation at an angle of π/3 at the same distance of d (we will refer to it as translating diagonally). Let s be a reflection over the vertical axis

3.10. DIFFIE-HELLMAN PUBLIC KEY

159

which splits one of the black flowered bulbs in half. We will read a series of instructions from the left to right, for instance sxy will be interpreted as ”First we reflect over the fixed axis, then shift horizontally and diagonally”. Then, G = hx, y, s | s2 = 1, xy = yx, x−1 s = sx, sy = x−1 ys i. Note that the subgroup H = hx, yi (or the set of translations) is isomorphic to Z ⊕ Z.

3.10 – Diffie-Hellman Public Key Exercise: 1 Section 3.10 Question: Show all your steps as you perform Fast Exponentiation to calculate by hand 5̄73 in the group U (153). Solution: We first need to calculate the binary expansion of 73. We get 73 = 64 + 8 + 1 = (1001001)2 . The Fast Exponentiation algorithm begins with x := 5, which corresponds to the stage i = 6. Then we have the following steps: i = 5: x := x2 = 25. i = 4: x := x2 = 625 = 13. i = 3: x := 5x2 = 845 = 80. i = 2: x := x2 = 6400 = 127. i = 1: x := x2 = 16129 = 64. i = 0: x := 5x2 = 20480 = 131. Thus, in the group U (153), we have 5̄73 = 131. Exercise: 2 Section 3.10 Question: Show all your steps as you perform Fast Exponentiation to calculate by hand 2̄111 in the group U (501). Solution: We first need to calculate the binary expansion of 111. We get 111 = 64 + 32 + 8 + 4 + 2 + 1 = (1101111)2 . The Fast Exponentiation algorithm begins with x := 2, which corresponds to the stage i = 6. Then we have the following steps: i = 5: x := 2x2 = 8. i = 4: x := x2 = 64. i = 3: x := 2x2 = 8192 = 176. i = 2: x := 2x2 = 61952 = 329. i = 1: x := 2x2 = 216482 = 50. i = 0: x := 2x2 = 5000 = 491. Thus, in the group U (501), we have 2̄111 = 491. Exercise: 3 Section 3.10 Question: Show all your steps as you perform Fast Exponentiation to calculate by hand 3̄795 in the group U (7703). Solution: We first need to calculate the binary expansion of 795. We get 795 = 512 + 256 + 16 + 8 + 2 + 1 = (1100011011)2 . The Fast Exponentiation algorithm begins with x := 3, which corresponds to the stage i = 9. Then we have the following steps:

160

CHAPTER 3. GROUPS

i = 8: x := 3x2 = 27. i = 7: x := x2 = 729. i = 6: x := x2 = 531441 = 7637. i = 5: x := x2 = 58323769 = 4356. i = 4: x := 3x2 = 56924208 = 6741. i = 3: x := 3x2 = 136323243 = 3252. i = 2: x := x2 = 10575504 = 6988. i = 1: x := 3x2 = 146496432 = 778. i = 0: x := 3x2 = 1815852 = 5647. Thus, in the group U (7703), we have 3̄795 = 5647. Exercise: 4 Section 3.10 Question: Consider the group G = GL2 (F199 ). Show all your steps as you perform Fast Exponentiation to calculate by hand 42 0̄ 23 . 10 3 Solution: We first need to calculate the binary expansion of 42. We get 42 = 32 + 8 + 2 = (101010)2 . The Fast 0̄ 23 Exponentiation algorithm begins with x := , which corresponds to the stage i = 5. Then we have the 10 3 following steps: 31 69 2 i = 4: x := x = . 30 40 0̄ 23 2 0̄ 23 46 123 36 34 i = 3: x := x = = . 10 3 10 3 140 88 84 101 172 81 i = 2: x := x2 = . 165 122 0̄ 23 164 133 136 191 0̄ 23 2 x = = . i = 1: x := 10 3 10 3 153 190 109 109 112 30 . i = 0: x := x2 = 39 64 Thus, in GL2 (F199 ), we have

0̄ 23 10 3

42

=

112 39

30 . 64

Exercise: 5 Section 3.10 Question: Show all your steps as you perform Fast Exponentiation to calculate by hand 9 3 2 1 1 in GL2 (R). Solution:

We first need to calculate the binary expansion of 9. We get 9 = 8 + 1 = (1001)2 . The Fast 3 2 Exponentiation algorithm begins with x := , which corresponds to the stage i = 3. Then we have the 1 1 following steps:

3.10. DIFFIE-HELLMAN PUBLIC KEY 2

i = 2: x := x =

11 4

8 . 3

201 112 i = 1: x := x = . 56 41 3 2 2 3 2 46673 i = 0: x := x = 1 1 1 1 13552 2

161

27104 7953

=

197123 97218 . 60225 35057

Thus, using 4 (instead of 9) matrix multiplications, we find that

3 1

9 2 197123 = 1 60225

97218 . 35057

Exercise: 6 Section 3.10 Question: Suppose that Alice and Bob decide to use the group U (3001) as their group G and the element 5̄ as the base. If Alice chooses a = 73 and Bob chooses b = 129, then what will be their common key using the Diffie-Hellman public key exchange algorithm? Solution: The common key that Alice and Bob will use is g ab = 5

73×129

9417

=5

= 1742.

Exercise: 7 Section 3.10 Question: Suppose that Alice and Bob decide to use the group U (4237) as their group G and the element 11 as the base. If Alice chooses a = 100 and Bob chooses b = 515, then what will be their common key using the Diffie-Hellman public key exchange algorithm? Solution: The common key that Alice and Bob will use is g ab = 11

100×515

= 11

51500

= 3484.

Exercise: 8 Section 3.10 Question: Play the role of Alice. Use the group G = U (3001) and the base g = 2̄. Change a and b to a = 437 and b = 1000. Send to Bob the ciphertext for the following message: “MEET ME AT DAWN” Solution: In numbers, the plaintext of Alice’s message is 1305,

520,

13,

500,

120,

4,

123,

1400.

The public key produced by Diffie-Hellman is g ab = 2066. Then the ciphertext that Alice will send Bob is 1232,

2963,

2850,

656,

1838,

2262,

2034,

2437.

Exercise: 9 Section 3.10 Question: Play the role of Alice. Use the group G = U (3001) but use the base g = 7̄. Bob sends you g b = 2442 and you decide to use a = 2319. Send to Bob the ciphertext for the following message: “SELL ENRON STOCKS NOW” a Solution: The Diffie-Helman public key that the algorithm requires is g ab = g b = 2977. The plaintext in numbers is 1905, 1212, 5, 1418, 1514, 19, 2015, 1119, 14, 1523. To get the ciphertext, each plaintext unit is multiplied by g ab . We get 2296,

922,

288,

1980,

2677,

2545,

2657,

153,

2665,

2461.

162

CHAPTER 3. GROUPS

Exercise: 10 Section 3.10 Question: Play the role of Bob. Use the group G = U (3517) and use the base g = 11. Alice sends you g a = 1651 and you tell Alice you will use b = 789. You receive the following ciphertext from Alice: 369,

665,

1860,

855,

3408,

1765,

1820,

1496.

Show the steps using fast exponentiation to recover the decryption key g −ab and use this to recover the plaintext for the message that Alice sent you. Solution: The decryption key that Bob needs to calculate it g −ab = g |G|a−ab = (g a )|G|−b . We have |G| − b = 3516 − 789 = 2727. We need to convert 2727 to binary: 2727 = (101010100111)2 . We use the fast exponentiation algorithm. It starts with x = g a = 1651, which corresponds to i = 11. i = 10: x := x2 = 126. i = 9: x := 1651x2 = 2592. i = 8: x := x2 = 994. i = 7: x := 1651x2 = 3047. i = 6: x := x2 = 2846. i = 5: x := 1651x2 = 1805. i = 4: x := x2 = 1283. i = 3: x := x2 = 133. i = 2: x := 1651x2 = 2888. i = 1: x := 1651x2 = 1432. i = 0: x := 1651x2 = 3480. We have calculated g −ab = 3480. To recover the plaintext, we take the ciphertext 369,

665,

1860,

855,

3408,

1765,

1820,

1496

14,

1520,

18,

516,

1518,

3000,

920.

and multiply by 3480. We get: 415,

Converting this mack into English, we recover the message: “DO NOT REPORT IT”. (This exercise had an error, the second to last plaintext block should have been 2000.) Exercise: 11 Section 3.10 Question: Play the role of Bob. Use the group G = U (7522) and use the base g = 3. Alice sends you g a = 2027 and you tell Alice you will use b = 2013. You receive the following ciphertext from Alice: 4433,

5198,

6996,

3275,

7067,

2568,

1894,

6037,

7208.

Show the steps using fast exponentiation to recover the decryption key g −ab and use this to recover the plaintext for the message that Alice sent you. Solution: The decryption key that Bob needs to calculate it g −ab = g |G|a−ab = (g a )|G|−b . We have |G| − b = φ(7522) − 2013 = 3760 − 2013 = 1747. We need to convert 1747 to binary: 1747 = (11011010011)2 . We use the fast exponentiation algorithm. It starts with x = g a = 2027, which corresponds to i = 10. i = 9: x := 2027x2 = 5195. i = 8: x := x2 = 6611. i = 7: x := 2027x2 = 7221. i = 6: x := 2027x2 = 6119.

3.10. DIFFIE-HELLMAN PUBLIC KEY

163

i = 5: x := x2 = 5167. i = 4: x := 2027x2 = 5713. i = 3: x := x2 = 411. i = 2: x := x2 = 3437. i = 1: x := 2027x2 = 255. i = 0: x := 2027x2 = 5191. We have calculated g −ab = 5191. Now we multiply the ciphertext by this decryption key and get the sequence 1905,

1404,

20,

805,

3,

1504,

500,

1415,

2300.

Converting this back to English, we get the phrase “SEND THE CODE NOW” Exercise: 12 Section 3.10 Question: We design the following Diffie-Hellman/ElGamal setup to encipher strings of bits (0 or 1). We will chose a prime number p with 210 < p < 211 and use the group G = U (p) and choose a base g ∈ G. We break up the string of bits into blocks of 10 bits and map a block of ten bits into a number between 0 and 1023 by using a binary expression, hence, (b0 , b1 , . . . , b9 ) −→ b0 + b1 · 2 + · · · + bk · 2k + · · · + b9 · 29 = mi . This is a plaintext unit mi and we view it as an element in U (p). With this setup to convert strings of bits to elements in U (p), we then apply the usual Diffie-Hellman key exchange and the ElGamal encryption. As one extra layer, when Alice sends the cipher text to Bob, she writes it as a bit string, but with the difference that since numbers c = mg ab can be expressed uniquely as an integer less than 211 , blocks of 10 bits become blocks of 11 bits. Here is the exercise. You play the role of Alice. You and Bob decide to use p = 1579 and the base of g = 7. Bob sends you g b = 993 and you decide to use a = 78. Show all the steps to create the Diffie-Hellman key g ab . Use this to create the ciphertext as described in the previous paragraph for the following string of bits: 1011010110 1011111001 1111100010. Show all the work. Solution: The first part of this exercise requires us to calculate the Diffie-Helman encryption key. We obtain it by (g b )a . The binary expansion of a = 78 is 78 = 64 + 8 + 4 + 2 = (1001110)2 . The fast exponentiation algorithm gives the following for calculating g ab . It starts with x = 993, which corresponds to i = 6. Then: i = 5: x := x2 = 753. i = 4: x := x2 = 148. i = 3: x := 993x2 = 1526. i = 2: x := 993x2 = 823. i = 1: x := 993x2 = 15. i = 0: x := x2 = 225. We have determined that g ab = 225. The plaintext message, converted into a sequence of elements in U (1579) is 726, 761,and 994. Multiplying by the key gives us the following sequence of integers of ciphertext: 713, 693, and 1011. Converting back to binary but with up to 10 bits, we have the ciphertext in binary of: 01011001001 01010110101 01111110011.

164

CHAPTER 3. GROUPS

Exercise: 13 Section 3.10 Question: Use the setup as described in Exercise 3.10.12 to encipher bit strings. This time, however, you play the role of Bob. You and Alice use the group U (1777) and the base g = 10, Alice sends you g a = 235, and you choose to use b = 1573. Alice has sent you the following string of bits: 00110110110 01110101001 10011011100. Show all the steps to create the Diffie-Hellman decryption key g −ab . Turn the ciphertext into strings of elements in U (1777). By multiplying by g −ab , recover the list of elements in U (1777) that correspond to plaintext. (These should be integers mi with 0 ≤ mi ≤ 1023.) Convert them to binary to recover the plaintext message in bit strings. 203

Solution: We first need to calculate the decryption key g −ab . We can get it as (g a )|G|−b = 235 . Converting 203 to binary, we have 203 = (11001011)2 . Implementing the fast exponentiation algorithm, we begin with x = 235, which corresponds to i = 7. Then we do the following. i = 6: x := 235x2 = 444. i = 5: x := x2 = 1666. i = 4: x := x2 = 1659. i = 3: x := 235x2 = 683. i = 2: x := x2 = 915. i = 1: x := 235x2 = 212. i = 0: x := 235x2 = 1129. This shows that the decryption key is g −ab = 1129. The ciphertext, as elements in U (1777) is 438, 937, and 1244. To recover the plaintext as a sequence of elements in U (1777), we simply multiply by 1129. We get the elements 496, 558, and 646, which converted back to binary using ten bits gives the plaintext of 0111110000 1000101110 1010000110.

Exercise: 14 Section 3.10 Question: We design the following application of Diffie-Hellman. We choose to encrypt 29 characters: 26 letters of the alphabet, space, the period “.”, and the comma “,”. We associate the number 0 to a space character, 1 through 26 for each of the letters, 27 to the period, and 28 to the comma. We use the group G = GL2 (F29 ), the general linear group on modular arithmetic base 29. Given a message in English, we write the numerical values of the characters in a 2 × n matrix, reading the characters of the alphabet by successive columns. Hence, “SAY FRIEND AND ENTER” would become the matrix 19 25 6 9 14 0 14 0 14 5 M= ∈ M2×10 (F29 ) 1 0 18 5 4 1 4 5 20 18 where we have refrained from putting the congruence bars over the top of the elements only for brevity. Then given a key K ∈ GL2 (F29 ), we encrypt the message into ciphertext by calculating the matrix C = KM . Hence, 3 4 with K = the ciphertext matrix becomes 5 9 3 17 3 18 0 4 0 20 6 0 C= 17 9 18 3 19 9 19 16 18 13 and “CQQICRRI SDI STPFR M” is the ciphertext message in characters. [Note that this enciphering scheme is not an ElGamal enciphering scheme as described in Section 3.10.4. Here, C = M and we do not use a function h so that the enciphering function Ek does not involve products in the group G.] Here is the exercise. You play the role of Alice. You and Bob decide on the above enciphering scheme. You will chose the key K in the usual Diffie-Hellman manner. You use the group G = GL2 (F29 ) and the base

3.10. DIFFIE-HELLMAN PUBLIC KEY

165

1 2 27 24 b g= . Bob sends you g = and you choose to use a = 17. Calculate the Diffie-Hellman key and 3 5 7 17 use this to determine the ciphertext corresponding to “COME HERE, NOW.” (Since there is an odd number of characters in the message, append a space on the end to make a message of even length.) Solution: We first create the message matrix. 3 13 0 5 M= 15 5 8 18

5 28

0 14

15 23

27 . 0

Now we need to calculate the public key K as created by the Diffie-Helman process. We need to calculate g ab = (g b )a . The power a = 17 has an expansion, namely a = (10001)2 . Following the fast easy binary 27 24 exponentiation, we start with K := g b = . Then we do 7 17 27 12 2 i = 3: K := K = . 18 22 17 8 i = 2: K := K 2 = . 12 4 8 23 2 . i = 1: K := K = 20 25 27 24 2 17 i = 0: K := K2 = . 7 17 11 7 This is the encryption key. Now according to the described encryption procedure, the ciphertext matrix is 0 24 20 26 22 6 15 25 C = KM = . 22 4 27 7 19 11 7 7 We can now put this back into English as the ciphertext message of “ VXDT.ZGVSFKOGYG” Exercise: 15 Section 3.10 Question: We use the communication protocol as described in Exercise 3.10.14. You use the same group 5 27 and the same base but this time you play the role of Bob. Alice sends you g a = and you chose 26 1 to use b = 12. After you send your g b to Alice, she then creates the public key and sends you the message “LPSKQIMBW.ECRBHL” in ciphertext. Show all the steps with fast exponentiation to calculate the deciphering key g −ab and recover the plaintext message. (Note that since we know how to take inverses of matrices in GL2 (F29 ), it suffices to calculate g ab and then find the inverse as opposed to calculating (g a )|G|−b .) Solution: The message that we received from Alice is written as a matrix as 12 19 17 13 23 5 18 8 C= . 16 11 9 2 27 3 2 12 a b To obtain the encryption key, we use fast exponentiation to calculate (g ) . Bob’s power b has the binary 5 27 expansion of b = 12 = (1100)2 . To get K, we start with K = , corresponding to i = 3. The remaining 26 1 steps of fast exponentiation are 5 27 17 13 2 i = 2: K := K = . 26 1 5 14 6 26 i = 1: K := K 2 = . 10 0 6 11 2 i = 0: K := K = . 2 28

166

CHAPTER 3. GROUPS

The decryption key is now K

−1

=

28 27

18 6

M = K −1 C =

We can now decipher the plaintext matrix as

15 14

5 28

0 20

23 15

28 0

20 8

18 5

5 . 27

Interpreting this as an English phrase, we get “ONE, TWO, THREE.” Exercise: 16 Section 3.10 Question: We design the following Diffie-Hellman/ElGamal setup. We choose to encrypt 30 characters: 26 letters of the alphabet, space, the period “.”, the comma “,” and the exclamation point “!”. We associate the number 0 to a space character, 1 through 26 for each of the letters, 27 to the period, 28 to the comma and 29 to “!”. We choose to compress triples of characters as follows: (b1 , b2 , b3 ), where each bi ∈ {0, 1, . . . , 29}, corresponds to the number b1 × 302 + b2 × 30 + b3 . The resulting possible numbers are between 0 and 303 − 1 = 26, 999. Now, the smallest prime bigger than 303 is p = 27011. We will work in the group U (27011) and we will view messages as sequences of elements in G encoded as described above. For example: “HI FRANK!” uses the compression of 8 × 302 + 9 × 30 + 0 = 7, 470

6 × 302 + 18 × 30 + 1 = 5, 941

14 × 30a = 2 + 11 × 30 + 29 = 12, 959

and hence corresponds to the sequence in G of 7470, 5941, 12959. Here’s the exercise. You are Bob. Alice and you use the system described above and we use the base of g = 2̄. Alice wants to send a message to you. She first sends her g a = 5, 738 ∈ G. You (Bob) chose the integer b = 10, 372 and send back g b = 255. You receive the following ciphertext from Alice: 11336

8377

17034 688

1031

13929.

Using Fast Exponentiation, determine the inverse of the public key, g −ab . Decipher the sequence of numbers corresponding to Alice’s plaintext message. From the message coding scheme, determine Alice’s original message (in English). 16638

. In binary, Solution: To get the decryption key, we need to calculate (g a )|G|−b = (g a )27010−10372 = 5738 b = 16638 = (100000011111110)2 . To implement fast exponentiation, we begin with x := g a = 5738, which corresponds to i = 14. Then we must do: i = 13: x := x2 = 25246. i = 12: x := x2 = 8960. i = 11: x := x2 = 4908. i = 10: x := x2 = 21663. i = 9: x := x2 = 23466. i = 8: x := x2 = 6910. i = 7: x := 5738x2 = 1347. i = 6: x := 5738x2 = 13024. i = 5: x := 5738x2 = 5575. i = 4: x := 5738x2 = 12552. i = 3: x := 5738x2 = 17011. i = 2: x := 5738x2 = 5833. i = 1: x := 5738x2 = 22711. i = 0: x := x2 = 14476.

3.10. DIFFIE-HELLMAN PUBLIC KEY

167

This shows that the decryption key is g −ab = 14476. To recover the plaintext as a sequence of elements in U (27011), we simply multiply the ciphertext by 14476 to get 8111

13073 765

19440 14684 26100.

Converting these to base 30, we get (9, 0, 11)30 , (14, 15, 23)30 , (0, 25, 15)30 , (21, 18, 0)30 , (16, 9, 14)30 , (29, 0, 0)30 . As an English phrase, this is: “I KNOW YOUR PIN!” Exercise: 17 Section 3.10 Question: Use the text to strings of groups elements as described in Exercise 3.10.16. Use the same group G = U (27011) but use the base g = 5. Play the role of Alice. Select a = 10, 000 while Bob sends you g b = 15128. Show all the steps to create the string of elements in G that are the ciphertext for the message “I WILL SURVIVE.” Solution: We will first compress the English string into a sequence of elements in G. This is (9, 0, 23)30 = 8123, (9, 12, 12)30 = 8472, (0, 19, 21)30 = 591, (18, 22, 9)30 = 16869, (22, 5, 27)30 = 19977. 10000

. In order to calculate the encryption key via the Diffie-Helman process, we need to calculate (g b )a = 15128 The binary expansion of a is 10000 = (10011100010000)2 . The fast exponentiation starts with x := 15128 and proceeds with the following steps: 1 = 12: x := x2 = 19192. 1 = 11: x := x2 = 10868. 1 = 10: x := 15128x2 = 10079. 1 = 9: x := 15128x2 = 1483. 1 = 8: x := 15128x2 = 17731. 1 = 7: x := x2 = 7332. 1 = 6: x := x2 = 6334. 1 = 5: x := x2 = 8221. 1 = 4: x := 15128x2 = 23394. 1 = 3: x := x2 = 9365. 1 = 2: x := x2 = 25519. 1 = 1: x := x2 = 11162. 1 = 0: x := x2 = 15512. We have calculated the encryption key of g ab = 15512. To implement the ElGamal with the above plaintext strings, the ciphertext is simply the product of the plaintext by g ab : 24672,

9149,

10863,

16371,

13032.

168

CHAPTER 3. GROUPS

3.11 – Semigroups and Monoids Exercise: 1 Section 3.11 Question: Prove that (N∗ , gcd) is a semigroup but not a monoid. Solution: First we show that gcd(a, b, c) = gcd(a, gcd(b, c)) for all natural numbers a, b, and c. Let d = gcd(b, c). And suppose that gcd(a, b, c) = e. Now, since e divides all three, it also divides b and c. This implies that e|d. Since e divides both a and d, this implies that e divides gcd(a, gcd(b, c)). Now, gcd(a, gcd(b, c)) divides a and d which implies that it divides a, b, and c so that gcd(a, gcd(b, c)) divides e. This proves the equality. We note that gcd(a, b, c) = gcd(c, a, b) = gcd(c, gcd(a, b)) = gcd(gcd(a, b), c) since order does not affect the operation. Then gcd(a, gcd(b, c)) = gcd(a, b, c) = gcd(gcd(a, b), c). This shows that (N∗ , gcd) is a semigroup. Suppose that there was a d so that for any a ∈ N∗ we have gcd(d, a) = a. Well this implies that d is a multiple of any a in the natural numbers. However this is impossible. So no such identity can exist. This shows that (N∗ , gcd) is not a monoid. Exercise: 2 Section 3.11 Question: Prove that (N∗ , lcm) is a semigroup and is a monoid. Solution: A semigroup is a set with an associative binary operation. Clearly lcm is a binary operation on the set N∗ ; the least common multiple of two natural numbers is again a natural number by Proposition 2.1.16. Recall that for a, b ∈ N∗ , lcm(a, b) = k if a|k, b|k, and a|k 0 and b|k 0 only if k|k 0 . We must show that lcm is associative on N∗ . Let a, b, c ∈ N∗ and consider lcm(a, lcm(b, c)). Let m1 = lcm(b, c); then b|m1 and c|m1 . Let k1 = lcm(a, m1 ), so a|k1 and m1 |k1 , so b|k1 and c|k1 also. Moreover, if there is k10 so that a|k10 and m1 |k10 (hence b|k10 and c|k10 ), then k1 |k10 . Consider now lcm(lcm(a, b), c). Let m2 = lcm(a, b); then a|m2 and b|m2 . Let k2 = lcm(m2 , c), so c|k2 and m2 |k2 , so a|k2 and b|k2 also. Moreover, if there is k20 so that a|k20 and m2 |k20 (hence a|k20 and b|k20 ), then k2 |k20 . We observe that k1 and k2 satisfy precisely the same conditions, so lcm(a, lcm(b, c)) = k1 = k2 = lcm(lcm(a, b), c), showing that lcm is associative on N∗ and making (N∗ , lcm) associative and hence a semigroup. To show that this structure is also a monoid, we must exhibit an identity element. Suppose that lcm(n, e) = n for all n ∈ N∗ and for some e ∈ N∗ . Then n|n, e|n, and if any m ∈ N∗ satisfies n|m and e|m then n|m. We claim e = 1 works for all n ∈ N∗ . Surely n|n, and surely 1|n, and since 1|m for all m, n|m obviously implies itself — that n|m. Therefore 1 is an identity for (N∗ , lcm), making this structure a monoid. Exercise: 3 Section 3.11 Question: Let p be a prime number. Recall the ordp function defined in (2.4). In Exercise 3.11.1 you showed that (N∗ , gcd) is a semigroup. Prove that ordp : (N∗ , gcd) → (N, min) is a semigroup homomorphism. Solution: In Exercise 2.1.20 we found that for all positive integers a and b, if αn 1 α2 a = pα 1 p2 · · · pn

and b = pβ1 1 pβ2 2 · · · pβnn

then min(α1 ,β1 ) min(α2 ,β2 ) n ,βn ) p2 · · · pmin(α . n

gcd(a, b) = p1

p∈ / {p1 , p2 , . . . , pn }: Then p does not divide gcd(a, b) either. Hence, ordp (gcd(a, b)) = 0 and ordp (a) = ordp (b) = 0 so min(ordp (a), ordp (b)) = 0, so these two are equal. p ∈ {p1 , p2 , . . . , pn }: Say p = pi . Then by the formula for the greatest common divisor above, we deduce that ordp (gcd(a, b)) = min(αi , βi ) = min(ordp (a), ordp (b)). In either case, ordp (gcd(a, b)) = min(ordp (a), ordp (b)). Thus, ordp is a semigroup homomorphism. Exercise: 4 Section 3.11 Question: Let (L, 4) be a lattice. a) Prove that (L, ◦), where a ◦ b = lub(a, b) is a semigroup. b) Prove that (L, ?), where a ? b = glb(a, b) is a semigroup.

3.11. SEMIGROUPS AND MONOIDS

169

Solution: Let (L, 4) be a lattice with partial order 4. a) Let a, b, c be elements in L. The element u = lub(a, lub(b, c)) obviously satisfies a 4 u, but by transitivity it also satisfies b 4 u and c 4 u. Hence, u is an upper bound for the set {a, b, c}. Let u0 = lub(lub(a, b), c). By a similar reasoning, u0 is an upper bound for the set {a, b, c}. We prove that both u and u0 are the least upper bound of {a, b, c}. Let U = lub(a, b, c). By definition of a least upper bound, U 4 u and U 4 u0 . On the other hand, b 4 U and c 4 U so lub(b, c) 4 U , again by the definition of least upper bound. And also, a 4 U and lub(b, c) 4 U , so again, u 4 U . Hence, we conclude that u = U by antisymmetry. By a similar reasoning, u0 = U as well. Thus lub(a, lub(b, c)) = lub(a, b, c) = lub(lub(a, b), c). We conclude that ◦ is an associative operation and hence that (L, ◦), where a ◦ b = lub(a, b) is a semigroup. b) The reasoning for this part is identical to that for the previous part. More precisely, if 4 is a partial order on L, then < is also a partial order on L and (L, <) is a lattice furthermore, the least upper bound of sets with respect to < is precisely the greatest lower bound of the set with respect to 4. Exercise: 5 Section 3.11 Question: Let (L, 4) be a lattice and let (Lop , 4op ) be the lattice defined by a 4op b ⇐⇒ b 4 a. Prove that the function f : L → Lop defined by f (a) = a is a semigroup isomorphism between (L, ◦), where a ◦ b gives the least upper bound between a and b in (L, 4) and (Lop , ?), where a ? b gives the greatest lower bound between a and b in (Lop , 4op ). Solution: For all a, b ∈ L, we have f (lub(a, b)) = lub(a, b). Now if u = lub(a, b), then a 4 u and b 4 u and if u0 is any other common upper bound, then u 4 u0 . These statements imply that u 4op a and u 4op b and if we also have u0 4op a and u0 4op b then u0 4op u. These conditions show that u is the greatest lower bound of {a, b} but with respect to 4op . The result follows. Exercise: 6 Section 3.11 Question: Let U be a set and P(U ) the power set of U . Prove that (P(U ), ∩) is a monoid but not a group. Solution: We know that ∩ is associative. Furthermore, for all A ∈ P(U ), we have A ∩ U = U ∩ A = A. Hence, U serves as the identity. This proves that (P(U ), ∩) is a monoid. However, if A 6= U , then for all B ∈ P(U ), we have A ∩ B ⊆ A ( U. Hence, there is no element B ∈ P(U ) such that A ∩ B = U . Thus, (P(U ), ∩) does not have inverses and therefore is not a group. Exercise: 7 Section 3.11 Question: Let U be a set and P(U ) the power set of U . Prove that (P(U ), ∪) is a monoid but not a group. Solution: We know that ∪ is associative. Furthermore, for all A ∈ P(U ), we have A ∪ ∅ = ∅ ∪ A = A. Hence, ∅ serves as the identity. This proves that (P(U ), ∪) is a monoid. However, if A 6= ∅, then for all B ∈ P(U ), we have ∅ ( A ⊆ A ∪ B. Hence, there is no element B ∈ P(U ) such that A ∪ B = ∅. Thus, (P(U ), ∪) does not have inverses and therefore is not a group. Exercise: 8 Section 3.11 Question: Let U be a set and consider the function f : P(U ) → P(U ) defined by f (A) = A, set complement. Prove that f is a monoid isomorphism from (P(U ), ∩) to (P(U ), ∪). Solution: It is obvious that f is a bijection since f is its own inverse function, namely f ◦ f is the identity on P(U ) fr all A ∈ P(U ). Also, by the DeMorgan laws f (A ∩ B) = A ∩ B = A ∪ B = f (A) ∪ f (B).

170

CHAPTER 3. GROUPS

Hence, f is a monoid homorphism. Consequently, we see that f is a monoid isomorphism. Exercise: 9 Section 3.11 Question: Let F be the set of functions from {1, 2, 3, 4} to itself and consider the monoid (F, ◦) with the operation of function composition. Let S be the smallest subsemigroup of F that contains the functions 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 f1 = , f2 = , f3 = , f4 = . 1 1 2 3 4 2 2 3 4 1 3 3 4 1 2 4 Prove that S is not a monoid. Prove that S contains no bijections. Solution: Let X be a set. Suppose that a function f : X → X is not surjective. Then, there exists x0 ∈ X such that there does not exist a ∈ X such that f (a) = x0 . Consequently, for any function g : X → X, there does not exist b ∈ X such that f (g(b)) = x0 . Hence, f ◦ g is not surjective. If S is the smallest subsemigroup of F that contains {f1 , f2 , f3 , f4 }, then all elements in S are obtained by a sequence of compositions of functions gi such that gi ∈ {f1 , f2 , f3 , f4 }. However, no function fi is surjective so by the result of the previous paragraph, no function in S is surjective either. Consequently, S does not contain any bijections and in particular, S does not contain the identity function so (S, ◦) is not a monoid. Exercise: 10 Section 3.11 Question: Prove that the power set of a monoid (M, ∗) is indeed a monoid as claimed in Example 3.11.17. Solution: We consider the power set monoid (P(M ), ∗) as defined in Example 3.11.17. Let R, S, and T be subsets of M . We have R ∗ (S ∗ T ) = {r ∗ (s ∗ t) | r ∈ R, s ∈ S, t ∈ T } = {(r ∗ s) ∗ t | r ∈ R, s ∈ S, t ∈ T }

by associativity in M

= (R ∗ S) ∗ T. Thus ∗ is associative on P(M ). Let e be the identity in M . Then for all S ∈ P(M ), we have S ∗{e} = {e}∗S = S. Thus, {e} is the identity in P(M ). Thus, (P(M ), ∗) is a monoid. Exercise: 11 Section 3.11 Question: Let U be a set and let Fun(U, U ) be the monoid of functions from U to U , with the operation of composition. Let A be a subset of U and let NA = {f ∈ Fun(U, U ) | f (A) ⊆ A}. 1. Show that NA is a submonoid of Fun(U, U ). 2. Give an example that shows that not every submonoid of Fun(U, U ) is NA for some A ⊆ U . Solution: Let U be a set and let Fun(U, U ) be the monoid of functions from U to U , with the operation of composition. Let A be a subset of U and let NA = {f ∈ Fun(U, U ) | f (A) ⊆ A}. a) It is obvious that the identity function idU maps idU (A) = A so idU ∈ NA . Furthermore, if f, g ∈ NA , then g(A) ⊆ A. Hence, f (g(A)) ⊆ f (A) ⊆ A and we conclude that f ◦ g ∈ NA . This shows that NA is a submonoid of Fun(U, U ). b) Let U = {1, 2, . . . , n} and consider the permutation group Sn as a subset of Fun(U, U ). Since Sn is a group, it is obviously a monoid and thus a submonoid of Fun(U, U ). The union of all the ranges of elements in Sn is {1, 2, . . . , n} so if Sn = NA for some A ⊆ U , then A = U . However, NU = Fun(U, U ) while Fun(U, U ) contains functions that are not bijections so Sn 6= Fun(U, U ). Hence, Sn is a submonoid that is not of the form NA . Exercise: 12 Section 3.11 Question: Call M the set of nonzero polynomials with real coefficients. The pair (M, ×) is a monoid with the polynomial 1 as the identity. Decide which of the following subsets of M are submonoids and justify your answer. 1. Nonconstant polynomials. 2. Polynomials whose constant coefficient it 1.

3.11. SEMIGROUPS AND MONOIDS

171

3. Palindromic polynomials. [A polynomial an xn + · · · + a1 x + a0 is called palindromic if an−i = ai for all 1 ≤ i ≤ n.] 4. Polynomials with odd coefficients. 5. Polynomials with an−1 = 0. 6. Polynomials with no real roots. Solution: Call M the set of nonzero polynomials with real coefficients. The pair (M, ×) is a monoid with the polynomial 1 as the identity. a) The set of nonconstant polynomials does not contain the identity 1. Hence, this set is not a submonoid. b) Let p(x) and q(x) be polynomials whose coefficient is 1. The constant term of p(x)q(x) is the product of their constant terms, i.e., 1 × 1 = 1. So p(x)q(x) also has a constant coefficient of 1. Furthermore, the identity polynomial has a constant coefficient of 1. c) The identity 1 is a palindromic polynomial. A palindromic polynomial has the alternative defining property that p(x) = xm p x1 , where m is the degree of p. Suppose that p(x), q(x) are palindromic polynomials of degree m and n respectively. Then the degree of p(x)q(x) is m + n. Also, 1 1 1 1 q = xm p xn q = p(x)q(x). xm+n p x x x x Thus p(x)q(x) is also palindromic. Hence, this subset is a submonoid. d) The set of polynomials with odd coefficients is not a submonoid. For example, (x + 1)(x + 1) = x2 + 2x + 1, so the product of two elements in this subset is not necessarily back in the subset. e) Consider polynomials a(x) with an−1 = 0, where n is the degree of the polynomial. The constant polynomial 1 has degree 0 so it satisfies the desired condition. Now let a(x) and b(x) be polynomials of degree m and n respectively. The degree of a(x)b(x) is m + n and the coefficient in front of the xm+n−1 term is am bn−1 + am−1 bn = am 0 + 0bn = 0. Thus, this set is closed under multiplication, so the subset is a submonoid. f) If p(x) and q(x) are two polynomials, then x0 is a root of p(x)q(x) if and only if x0 is a root of p(x) or q(x). Hence, if p(x)q(x) has a real root, the either p(x) or q(x) has a real root. The contrapositive is that if neither p(x) nor q(x) has a real root, then p(x)q(x) does not either. Hence, this subset of polynomials is closed under the operation. Furthermore, 1 does not have any real roots. Thus, the set of polynomials with no real roots is a submonoid. Exercise: 13 Section 3.11 Question: Let ϕ : M → N be a monoid homomorphism. We define the kernel of the monoid homomorphism as Ker ϕ = {m ∈ M | ϕ(m) = 1N }. Show that Ker ϕ is a submonoid of M . Solution: Let ϕ : M → N be a monoid homomorphism. By definition ϕ(1M ) = 1N . Hence, 1M ∈ Ker ϕ. Let m1 , m2 ∈ Ker ϕ. Then ϕ(m1 m2 ) = ϕ(m1 )ϕ(m2 ) = 1N · 1N = 1N . Thus Ker ϕ is closed under the operation in M . We have shown that Ker ϕ is a submonoid of M . Exercise: 14 Section 3.11 Question: Let ϕ : M → N be a monoid homomorphism. We define the image of the monoid homomorphism as Im ϕ = {n ∈ N | n = ϕ(m) for some m ∈ M }. Show that Im ϕ is a submonoid of N . Solution: Let ϕ : M → N be a monoid homomorphism. By definition, ϕ(1M ) = 1N . Hence, 1N ∈ Im ϕ. Let n1 , n2 ∈ Im ϕ. Then there exist m1 , m2 ∈ M such that ϕ(m1 ) = n1 and ϕ(m2 ) = n2 . But then ϕ(m1 m2 ) = ϕ(m1 )ϕ(m2 ) = n1 n2 .

172

CHAPTER 3. GROUPS

Since m1 m2 ∈ M , the operation n1 n2 ∈ Im ϕ. Hence, we conclude that Im ϕ is a submonoid of N . Exercise: 15 Section 3.11 Question: We denote by Z[x] the set of monic polynomials with coefficients in Z. Consider the monoids (Z[x] − {0}, ×) and (C, +). Define the function γm : Z[x] − {0} → C by γm (p(x)) =

d X

rim ,

i=1

where deg p(x) = d and r1 , r2 , . . . , rd are the roots of p(x), listed with multiplicity. Note that γm is only defined on nonzero polynomials and that if p(x) is a constant polynomial then γm (p(x)) = 0 because p(x) has no roots. a) Prove that γm is a monoid homomorphism. b) (*) Prove that the image of γm is the submonoid (Z, +). Solution: Define the function γm : Z[x] − {0} → C by γm (p(x)) =

d X

rim ,

i=1

where deg p(x) = d and r1 , r2 , . . . , rd are the roots of p(x), listed with multiplicity. a) Suppose that p(x), q(x) ∈ Z[x] − {0}. Let r1 , r2 , . . . , rd be the roots of p(x), not necessarily distinct, and let rd+1 , rd+2 , . . . , rd+k be the roots of q(x), also listed with multiplicity, where deg q(x) = k. Then the roots of p(x)q(x), listed with multiplicity, are r1 , r2 , . . . , rd+k . Hence, ! ! d+k d d+k X X X m m m γm (p(x)q(x)) = ri = ri + ri = γm (p(x)) + γm (q(x)). i=1

i=1

i=k+1

Furthermore, γm (1) = 0, the identity in the monoid (C, +). Thus, γm is a monoid homomorphism. b) By considering polynomials of the form p(x) = (x − 1)k , we see that γm (p(x)) = k for all nonnegative integers. If m is odd, then γm ((x + 1)k ) = −k, obtaining all the negative integers. This shows that Z ⊆ Im γm but we need to show the verse inclusion. The sum of roots of a polynomial is negative of the second to highest coefficient of the polynomial. Hence, in this case, if m = 1, then γm (p(x)) = −pd−1 , where pd−1 is coefficient of xd−1 , where deg p(x) = d. So if m = 1, then Im γm = Z. We claim that sums of powers of the roots of a polynomial can be expressed as a polynomial expression of the coefficients of the polynomial. For example, if p(x) = x2 + p1 x + p0 = (x − r1 )(x − r2 ), then γ1 (p(x)) = r1 + r2 = −p1 γ2 (p(x)) = r12 + r22 = (r1 + r2 )2 − 2r1 r2 = (−p1 )2 − 2p0 γ3 (p(x)) = r13 + r23 = (r1 + r2 )3 − 3r12 r2 − 3r1 r22 = (r1 + r2 )3 − 3r1 r2 (r1 + r + 2) = −p31 − 3p0 p1 . Similar results hold for polynomials with higher degree. (More details of this can be found in Chapter 11 of the textbook.) Because of this claim, the γm (p(x)) ∈ Z for all m and all p(x). We conclude that if m is odd, then Im γm = Z and if m is even, then Im γm = N, with (N, +) also a submonoid of (C, +). Exercise: 16 Section 3.11 Question: Prove that C 0 (R, R), the set of continuous real-valued function from R, is a submonoid of Fun(R, R) (equipped with composition). Solution: The identity function id(x) = x is continuous so id ∈ C 0 (R, R). Furthermore, by a theorem in analysis, usually encountered briefly at the beginning of the calculus sequence, the composition of two continuous functions is again continuous. Hence, C 0 (R, R) is closed under composition. By definition, C 0 (R, R) is a submonoid of Fun(R, R). Exercise: 17 Section 3.11 Question: Consider the monoid M = Fun(R, R) and consider the function  −x if x < 0    0 if 0 ≤ x < 1 f (x) =  1 − x if 1 ≤ x < 2    x if 2 ≤ x.

3.11. SEMIGROUPS AND MONOIDS

173

1. Prove that f 1 , f 2 , f 3 , f 4 are all distinct. 2. Prove that f n = f 4 for n ≥ 4. 3. This is an example of a sequence of the form (f k )k≥1 that is not constant but eventually constant. Explain why eventually constant sequences do not occur in groups. Solution: Consider the monoid M = Fun(R, R) and the function f (x) described in the exercise. a) We first note that if |x| ≥ 2, then f n (x) = |x| for all n ≥ 1. In order to study differences between the functions f k , we only need to consider what happens for x ∈ (−2, 2). Furthermore, because of the intervals over which f (x) is piecewise linear have integer lengths and because the linear pieces have slope 1, 0, or −1, differences in the linear pieces of f k only occur in the integer intervals between (−2, 2). We calculate that     |x| if |x| ≥ 2 |x| if |x| ≥ 2           1 + x if − 2 <≤ x ≤ −1 −1 − x if − 2 <≤ x ≤ −1   2 3 f (x) = 0 f (x) = 0 if − 1 ≤ 0 ≤ 0 if − 1 ≤ 0 ≤ 0     0 0 if 0 ≤ x ≤ 1 if 0 ≤ x ≤ 1       x − 1 if 1 ≤ x < 2, 0 if 1 ≤ x < 2,   |x| if |x| ≥ 2     0 if − 2 <≤ x ≤ −1  f 4 (x) = 0 if − 1 ≤ 0 ≤ 0    0 if 0 ≤ x ≤ 1    0 if 1 ≤ x < 2. We have calculated f 2 , f 3 , f 4 and can easily see that they are all distinct because the set over which they map to 0 continues to change. b) As noted earlier, if |x| ≥ 2, then f n (x) = |x|. For our calculations, we see that for all x with |x| < 2, we have f 4 (x) = 0. But f (0) = 0, so for all x with |x| < 2, we also have f n (x) = 0 for all n ≥ 4. c) In a group, if g a = g a+1 , then by operating on the left by g −a gives 1 = g. In this case, the sequence g k for k ≥ 1 simply gives the constant sequence of the identity element. Hence, in groups, there are no eventually constant but nonconstant sequences of the form g k . Exercise: 18 Section 3.11 Question: Let (S, ·) be a semigroup. Let e be some object not in S and consider the set S 0 = S ∪ {e}. Define the operation ∗ on S 0 by a ∗ b = a · b for all a, b ∈ S and a ∗ e = e ∗ a = a for all a ∈ S 0 . Prove that (S 0 , ∗) is a monoid. Solution: The exercise assumes that the semigroup is not a monoid. The definition of the operation ∗ is well-defined, though we need to add the operation e ∗ e = e. With these assumptions, ∗ is constructed so that e acts as an identity on S 0 . We know that ∗ is associative if restricted to S, but we must simply check that it is associative over all of S 0 . We have three cases with a, b ∈ S: Case 1: a ∗ (b ∗ e) = a ∗ b = (a ∗ b) ∗ e. Case 2: a ∗ (e ∗ b) = a ∗ b = (a ∗ e) ∗ b. Case 3: e ∗ (a ∗ b) = a ∗ b = (e ∗ a) ∗ b. Other cases for checking associativity in which two out of three of the variables are e, all lead to e ∗ (e ∗ a) = a, so they are all equal. Consequently, in all cases, we verify that associativity holds on S 0 . Exercise: 19 Section 3.11 Question: Prove that the relation ∼ defined (3.12) is indeed an equivalence relation. Solution: Let (M, +) be a commutative monoid with identity 0 and let ∼ be the relation described in (3.12). Reflexivity: Suppose that (m1 , m2 ) ∈ M ⊕ M . Then m1 + m2 = m2 + m1 so using k = 0, we have (m1 , m2 ) ∼ (m1 , m2 ).

174

CHAPTER 3. GROUPS

Symmetry: Suppose that (m1 , m2 ), (n1 , n2 ) ∈ M ⊕ M and suppose that (m1 , m2 ) ∼ (n1 , n2 ). Then there exists k ∈ M such that m1 +n2 +k = m2 +n1 +k. Since the operation + is commutative, n1 +m2 +k = n2 +m1 +k. Hence, (n1 , n2 ) ∼ (m1 , m2 ). Transitivity: Suppose that (m1 , m2 ), (n1 , n2 ), (p1 , p2 ) ∈ M ⊕ M with (m1 , m2 ) ∼ (n1 , n2 ) and (n1 , n2 ) ∼ (p1 , p2 ). Then there exists k and ` in M such that m1 + n2 + k = m2 + n1 + k and n1 + p2 + ` = n2 + p1 + `. Then adding these two, we get m1 + n1 + k + n2 + p2 + ` = m2 + n1 + k + n2 + p1 + ` ⇐⇒m1 + p2 + (n1 + k + n2 + `) = m2 + p1 + (n1 + k + n2 + `). Thus, (m1 , m2 ) ∼ (p1 , p2 ). We have proved that the ∼ relation described in (3.12) is an equivalence relation. Exercise: 20 Section 3.11 Question: Show that the Grothendieck construction applied to the monoid (N∗ , ×) gives the group (Q>0 , ×). Solution: Consider the monoid (N∗ , ×). If (a, b) ∼ (c, d) for pairs in N ∗ ⊕ N ∗ , then there exists k ∈ N∗ such that adk = bck. Since k 6= 0, we can use the cancellation property in the integers to conclude that (a, b) ∼ (c, d) if and only if ad = bc. This is the cross ratio equivalence relation whose equivalence classes define rational numbers. (See Example 1.3.10.) Since the denominators are positive, the equivalence classes give us Q>0 . The operation in the Grothendieck construction is (a, b) × (c, d) = (a × c, b × d), which is the usual operation of multiplication on Q>0 . Hence, the Grothendieck construction on (N∗ , ×) gives the group (Q>0 , ×). Exercise: 21 Section 3.11 Question: Consider the monoid (N, +). Let M = h4, 7i denote the smallest submonoid that contains the subset {4, 7}. [We say that M is the submonoid generated by {4, 7}.] List the first 15 elements in M . Prove that the group obtained from the Grothendieck construction applied to (M, +) is group-isomorphic to (Z, +). Solution: A submonoid of (N, +) must contain 0, so we know that M must contain 0. Otherwise, elements in the submonoid generated by {4, 7} are obtained from additions of these elements. For the first 15 elements of M we get M = {0, 4, 7, 8, 11, 12, 14, 15, 16, 18, 19, 20, 21, 22, 23, . . .}. Consider the equivalence relation ∼ on N ⊕ N required for the Grothendieck construction. Let (a, b) be any pair in N ⊕ N such that a − b = n ∈ Z. Suppose that n is positive. Then n(8 − 7) = a − b so 8n + b = 7n + a. Hence, f. If a = b so that a − b = 0, then (a, a) acts as the identity in M f. Finally, if n = a − b is n · (8, 7) = (a, b) in M negative, first note that n · (8, 7) = (−n) · (7, 8). Then (−n)(7 − 8) = a − b and hence, n · (8, 7) = (a, b). We have f is a group generated by a single element. Furthermore, M f is infinite so (M f, +) is isomorphic to shown that M (Z, +). Exercise: 22 Section 3.11 Question: Let A be an abelian group. Prove that the Grothendieck construction applied to A as a monoid is a group that is (group) isomorphic to A. e be the Grothendieck construction applied to A. Call e the Solution: Let A be an abelian group and let A e identity of A. We prove that the function ϕ : A → A defined by ϕ([(a, b)]) = ab−1 is a group isomorphism. Note that since A is a group, the ∼ equivalence relation becomes (a, b) ∼ (c, d) if and only if ad = bc. First, since ϕ is a function whose domain is a quotient set and ϕ is defined in reference to the elements of the original set, we must prove that ϕ is well-defined. Suppose that (a, b) ∼ (c, d), which means that ad = bc. Then ϕ([(a, b)]) = ab−1 = cd−1 = ϕ((c, d)). So ϕ is well-defined as a function. Furthermore, the function is injective since (a, b) ∼ (c, d) if and only if ad = bc if and only if ab−1 = cd−1 . Also, ϕ is surjective since ϕ([(a, e)]) = a for all a ∈ A. e Finally, for all [(a, b)], [(c, d)] ∈ A, ϕ([(a, b)][(c, d)]) = ϕ([(ac, bd)]) = ac(bd)−1 = ab−1 cd−1 = ϕ([(a, b)])ϕ([(c, d)]). Thus, ϕ is a homomorphism as well. Hence, it is an isomorphism.

4 | Quotient Groups 4.1 – Cosets and Lagrange’s Theorem Exercise: 1 Section 4.1 Question: List all the distinct left cosets of H = h3̄i in Z/21Z and list all the elements in each coset. ¯ 15, ¯ 18}, ¯ 1̄ + h3̄i = {1̄, 4̄, 7̄, 10, ¯ 13, ¯ 16, ¯ 19}, ¯ and Solution: All of the distinct left cosets are 0̄ + h3̄i = {0̄, 3̄, 6̄, 9̄, 12, ¯ ¯ ¯ ¯ 2̄ + h3̄i = {2̄, 5̄, 8̄, 11, 14, 17, 20}. Exercise: 2 Section 4.1 Question: Suppose that Z20 is generated by the element z. List all the distinct left cosets of hz 5 i and list all the elements in each coset. Solution: Let H = hz 5 i = {1, z 5 , z 10 , z 15 }, then the remaining left cosets are zH = {z, z 6 , z 11 , z 16 }, z 2 H = {z 2 , z 7 , z 12 , z 17 }, z 3 H = {z 3 , z 8 , z 13 , z 18 }, and z 4 H = {z 4 , z 9 , z 14 , z 19 }. Exercise: Section 4.1 Question: In the group Q8 , list the distinct left cosets of the following subgroups and in each case list all the elements in each coset: (a) hii; (b) h−1i. Solution: a) Let H = hii = {i, −1, −i, 1} and the only other coset is jH = {−k, −j, k, j}. b) Let H = h−1i = {−1, 1} then the left cosets are iH = {−i, i}, jH = {−j, j}, and kH = {−k, k}. Exercise: 4 Section 4.1 Question: List all the distinct left cosets of H = h4̄i in U (35) and list all elements in each coset. Solution: Let us first list the elements in U (35). There will be φ(35) = (5 − 1)(7 − 1) = 24 elements. For simplicity, we do not write the bars over top of the integers but understand them as elements in U (35). U (35) = {1, 2, 3, 4, 6, 8, 9, 11, 12, 13, 16, 17, 18, 19, 22, 23, 24, 26, 27, 29, 31, 32, 33, 34}. The subgroup H has for elements H = {1, 4, 16, 29, 11, 9}. Then the other cosets are 2H = {2, 8, 32, 23, 22, 18} 3H = {3, 12, 13, 17, 33, 27} 6H = {6, 24, 26, 34, 31, 19}.

Exercise: 5 Section 4.1 Question: List all the distinct left cosets of H = h7̄, 47i in U (48) and list all elements in each coset. Solution: For convenience we drop the bars and understand that we are working in mod 48. First we list out the elements of U (48) = {1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47}. Let H = h7̄, 47i = {1, 7, 41, 47}. The remaining left cosets are 5H = {5, 13, 35, 43}, 11H = {11, 19, 29, 37}, and 17H = {17, 23, 25, 31}. Exercise: 06 Section 4.1 Question: List all the distinct left cosets of H = h(1 2 3)i in A4 and list all elements in each coset. Following the interpretation given in Excercise 3.5.43 of A4 as the group of rigid motions of a tetrahedron, describe each left coset by what all the elements of the coset do to the tetrahedron. Solution: Let H = h(1 2 3)i = {1, (1 2 3), (1 3 2)} which corresponds to rotating by 120◦ and 240◦ around the axis which goes through the 4th vertex and the midpoint of the opposite face and the identity tranformation. For cosets we have, (1 2)(3 4)H = {(1 2)(3 4), (1 4 2), (1 2 4)} (all the 3-cycles represent rotating by 120◦ and 240◦ around the axis that goes through the 3rd vertex and the midpoint of the opposite face, the other element corresponds to rotating by 180◦ around the axis which connects the midpoints of 12 and 34), (1 3)(2 4)H = 175

176

CHAPTER 4. QUOTIENT GROUPS

{(1 3)(2 4), (1 3 4), (1 4 3)} (all the 3-cycles represent rotating by 120◦ and 240◦ around the axis that goes through the 2nd vertex and the midpoint of the opposite face, the other element corresponds to rotating by 180◦ around the axis which connects the midpoints of 13 and 24), and (1 4)(2 3)H = {(1 4)(2 3), (2 4 3), (2 3 4)} (all the 3-cycles represent rotating by 120◦ and 240◦ around the axis that goes through the 1st vertex and the midpoint of the opposite face, the other element corresponds to rotating by 180◦ around the axis which connects the midpoints of 14 and 23). We have covered all the elements of A4 . Exercise: 7 Section 4.1 Question: Consider the group G = Z2 ⊕ Z2 ⊕ Z2 with Z2 = hz | z 2 = 1i. There is a bijection between the elements of G and the (vertices of the) unit cube C via f (z a , z b , z c ) = (a, b, c), where a, b, and c are 0 or 1. (See Figure 4.2.) For each of the following subgroups, list all the distinct left cosets, describe them as subsets of the unit cube (via the mapping f ), and label the cosets on a sketch of the unit cube. a) H = h(1, z, 1)i. b) K = h(z, z, z)i. c) L = h(z, 1, 1), (1, z, 1)i. Solution: The cosets of H, K, L will be subsets of the cube. a) For H, the cosets are (1, 1, 1)H, (z, 1, 1)H, (1, 1, z)H,

(z, 1, z)H.

As vertices of the cube the cosets respectively contain the points {(0, 0, 0), (0, 1, 0)},

{(1, 0, 0), (1, 1, 0)},

{(0, 0, 1), (0, 1, 1)},

{(1, 0, 1), (1, 1, 1)}.

These subsets correspond to the edges of the cube that are parallel to the y-axis. b) For K, the cosets are (1, 1, 1)K, (z, 1, 1)K, (1, z, 1)K, (1, 1, z)K. As vertices of the cube the cosets respectively contain the points {(0, 0, 0), (1, 1, 1)},

{(0, 1, 1), (1, 0, 0)},

{(1, 0, 1), (0, 1, 0)},

{(1, 1, 0), (0, 0, 1)}.

These subsets correspond to the four longest diagonals through the center of the cube. c) For L, the cosets are (1, 1, 1)L, (1, 1, z)L. As vertices of the cube the cosets respectively contain the points {(0, 0, 0), (0, 1, 0), (1, 1, 0), (1, 0, 0)},

{(0, 0, 1), (0, 1, 1), (1, 1, 1), (1, 0, 1)}.

These subsets correspond to the faces of the cube that are parallel to the xy-plane. Exercise: 8 Section 4.1 Question: Consider the group C× with multiplication and define the subgroup H = {z ∈ C× |z| = 1}. Describe geometrically (as a subset of the plane) each coset of H. √ Solution: We interpret a complex number z = a+bi as the point (a, b) in the plane and recall that |z| = a2 + b2 which corresponds to the distance that the point (a, b) is from the origin. Therefore, the subgroup H is the unit circle. Now, let r = c + di ∈ C× and we will consider the left coset rH. For any element a + bi ∈ H, we know that (c + di)(a + bi) = (ac − bd) + (ad + bc)i ∈ rH, which corresponds to the point (ac − bd, ad + bc). Now, if we calculate p |(ac − bd) + (ad + bc)i| = (ac − bd)2 + (ad + bc)2 p = a2 c2 − 2abcd + b2 d2 + a2 d2 + 2abcd + b2 c2 p = a2 (c2 + d2 ) + b2 (c2 + d2 ) p = (a2 + b2 )(c2 + d2 ) p = (1)(c2 + d2 ) = |r|.

4.1. COSETS AND LAGRANGE’S THEOREM

177

So any element in rH is on the circle of radius |r| centered about the origin. So for every left coset rH corresponds to the circle of radius |r| about the origin. Exercise: 09 Section 4.1 Question: Consider the group C× with multiplication and consider the subgroup H = R× . Describe geometrically (as a subset of the plane) each coset of H. Solution: Let p ∈ C× where p = a + bi. Then consider the coset pH which geometrically contains all real number multiples of the point (a, b). This corresponds to the line which goes through the origin and the point (a, b). Exercise: 10 Section 4.1 Question: Prove the last claim in Example 4.1.8. With the group G = S5 and the same subgroup H, prove also that the right cosets H(1 2 3 4 5)i correspond to H(1 2 3 4 5)i = {τ ∈ S5 | τ (5 − i) = 5}. Solution: For various i, consider the coset (1 2 3 4 5)i H, where H = {σ ∈ S5 | σ(5) = 5}. Any permutation in τ ∈ (1 2 3 4 5)i H, is of the form (1 2 3 4 5)i σ, where σ leaves 5 fixed. Hence τ (5) = (1 2 3 4 5)i (5) = i. Furthermore, suppose that τ is any permutation that maps 5 to i. Then (1 2 3 4 5)5−i τ maps 5 to 5. Thus there exists σ ∈ H with (1 2 3 4 5)5−i τ = σ and hence τ = (1 2 3 4 5)i σ. Now for various i, consider the right coset H(1 2 3 4 5)i , where H = {σ ∈ S5 | σ(5) = 5}. Any permutation in τ ∈ H(1 2 3 4 5)i , is of the form σ(1 2 3 4 5)i , where σ leaves 5 fixed. Hence τ (5−i) = σ(1 2 3 4 5)i (5−i) = σ(5) = 5. Furthermore, suppose that τ is any permutation that maps 5 − i to 5. Then τ (1 2 3 4 5)5−i maps 5 (to 5 − i and then) to 5. Thus there exists σ ∈ H with τ (1 2 3 4 5)5−i = σ and hence τ = σ(1 2 3 4 5)i . Exercise: 11 Section 4.1 Question: Consider the group (Q, +). a) Show that |Q : Z| is infinite. b) Show that there is no proper subgroup of Q of finite index. Solution: a) Let H = Z we will provide a way to construct an infinite number of distinct left cosets. For any prime n ∈ Z}. We will now prove that for any two distinct primes p ∈ Z consider the coset p1 + H = { np+1 p 1 1 p, q, the cosets p + H and q + H are disjoint. Assume for the sake of contradiction that the cosets are not disjoint. Then for some n, m ∈ Z we have np+1 = mq+1 ⇒ (np + 1)q = (mq + 1)p. So the left side p q is a multiple of p, since p and q are distinct primes this must imply that p|(np + 1) for some n ∈ Z. This implies that p|1, which is a contradiction, so the cosets must be disjoint. From Euclid, we know that there are an infinite amount of primes, so there also must be an infinite amount of disjoint left cosets of Z which implies that |Q : Z| is infinite. b) Let H be some proper subgroup of Q. Assume that |Q : H| = n. For any r ∈ Q, consider the n + 1 left cosets {0r + H, 1r + H, . . . nr + H}. Since |Q : H| = n, two of these cosets must be the same. We will denote them as kr + H and mr + H where k > m. Then by Proposition 4.1.4, we know that (k − m)r ∈ H where 1 ≤ (k − m) ≤ n. Since for every rational r ∈ Q we know that there is some l between 1 and n where lr ∈ H, we have n!Q ⊆ H. By the density property of the rationals, we know that n!Q = Q. So Q ⊆ H ⊆ Q ⇒ Q = H. Which contradicts the fact that H is a proper subgroup of Q. So there is no proper subgroup of Q of finite index. Exercise: 12 Section 4.1 Question: Consider the cosets of H = hsi in D4 as described in Example 4.1.2. We write AB = {ab | a ∈ A and b ∈ B} for any subsets A, B ⊆ G. Prove that (rH)(r2 H) is a left coset of H, but that (rH)(r3 H) is not. Solution: rH = {r, rs} and r2 H = {r2 , r2 s}. By examination, (rH)(r2 H) = {r3 , r3 s} = r3 H but (rH)(r3 H) = {1, s, r2 s, r2 } which is not the size of H and so cannot be a left coset of H. Exercise: 13 Section 4.1 Question: Let H be a subgroup of G. Suppose one attempted to define the function F from the set of left

178

CHAPTER 4. QUOTIENT GROUPS

cosets of H to the set of right cosets of H via F (gH) = Hg. Explain why F is not a function. [Hint: we also say that F is not well-defined.] Solution: Since we are actually mapping equivalence classes, we need to make sure that regardless of which representative of our equivalence class we choose, we will map to the same right coset. We seek a counter example by working in one of the noncommutative groups that we know of, D4 . Consider H = {1, s}. The elements r, rs ∈ D4 form the same left cosets of H, rH = rsH = {r, rs}. Now, under F we should have F (rH) = F (rsH) since r ∼1 rs. However, F (rH) = Hr = {r, r3 s} and F (rsH) = Hrs = {rs, r3 }. So, even though r ∼1 rs, we get F (rH) 6= F (rsH) showing that F is not well defined and not a function. Exercise: 14 Section 4.1 Question: Show that a complete set of distinct representative of ∼1 is not necessarily a complete set of distinct representatives of ∼2 . Solution: Since ∼1 corresponds to representatives of equivalent left cosets and ∼2 corresponds to representatives of equivalent right cosets, we will look for a complete set of representatives of left cosets that do not represent all right cosets in D4 since it is noncommutative. Let H = {1, s} and {1, rs, r2 , r3 } be a complete set of representatives of all the left cosets of H. Now we observe that H1 = H, Hrs = {rs, r3 }, Hr2 = {r2 , r2 s}, and Hr3 = {r3 , rs}. We notice that Hrs = Hr3 and that there is no representative for the right coset Hr = {r, r3 s}. So {1, rs, r2 , r3 } is not a complete set of representatives for all the right cosets of H and this provides a counter example. Exercise: 15 Section 4.1 Question: Let G be a group and do not assume it is finite. Prove that if H ≤ K ≤ G, then |G : H| = |G : K| · |K : H|.

Solution: Proof 1: Note that the cosets of H inside the subgroup K are cosets of H in G. Hence in |K : H| is infinite, then |G : H| is also infinite. Also, for every coset gK of K in G, there is at least one coset of H in G, namely the coset gH. Hence, if |G : K| is infinite, then so is |G : H|. So the formula |G : H| = |G : K| · |K : H| holds if any of the indices on the right hand side are infinite. Now suppose that both indices are finite. Let us call m = |K : H| and n = |G : K|. Let g1 K, g2 K, . . . , gn K be the n distinct cosets of K in G. Also, let k1 H, k2 H, . . . , km H be the m distinct left cosets of H in K. Every set of the form gi kj H, for i = 1, 2, . . . , n and j = 1, 2, . . . , m is a left coset of H in G. Now since the left cosets of a subgroup partition the whole group, then m [

kj H = K.

j=1

Thus n [ m [ i=1 j=1

gi kj H =

n [ i=1

 gi 

m [

 kj H  =

j=1

n [

gi K = G.

i=1

Thus the union of all the cosets of the form gi kj H is G. Hence, every coset of H in G is of the form gi kj H for some i and j. However, we can also show that for distinct i and j, these cosets are also distinct. Suppose that gi0 kj 0 H = gi kj H. Since kj 0 H and kj H are both subsets of K, then gi0 kj 0 H and gi kj H are subsets of the same left coset of K in G so gi0 K = gi K. Thus i0 = i. Since i0 = i, we also deduce that kj 0 H = kj H as cosets of H in K. Hence, kj 0 = kj . In other words, all the left cosets of H in G given by gi kj H with i = 1, 2, . . . , n and j = 1, 2, . . . , m are distinct. Hence, the cosets of H in G are indexed uniquely by (i, j) ∈ {1, 2, . . . , n} × {1, 2, . . . , m}, so |G : H| = mn.

4.1. COSETS AND LAGRANGE’S THEOREM

179

Proof 2: Let gi K with i in some indexing set I give all the distinct cosets of K in G. So the set {gi ∈ G | i ∈ I} is a complete set of distinct representatives for the partition created by the coset of K. Similarly, let kj H with j in some indexing set J and kj ∈ K give all the distinct cosets of H in K. Note that I and J may or may not be finite (or even countable). If there are n cosets of K in G, then we can use the set I = {1, 2, . . . , n} and similarly with J . We will show that gi kj H with (i, j) ∈ I × J gives all the distinct cosets of H in G. Obviously, gi ki H is a coset of H in G for every pair (i, j) ∈ I × J . We note that   [ [ [ [ gi kj H = gi  kj H  = gi K = G. (i,j)∈I×J

i∈I

j∈J

i∈I

So the union of all cosets of H of the form gi kj H is G. Furthermore, we prove they are all distinct. Suppose that gi kj H = gs kt H for two pairs (i, j) and (s, t) in I × J . Then gi kj H = gs kt H ⇐⇒ gs−1 gi kj H = kt H ⇐⇒ gs−1 gi ∈ K ⇐⇒ gi K = gs K ⇐⇒ i = s. But then also kj H = kt H so j = t, since the kj give distinct cosets of H in K for different j in J . Hence, all different pairs (i, j) ∈ I × J correspond to unique cosets of H as gj kj H. Hence, as cardinalities of sets |G : H| = |I × J | = |I| |J | = |G : K| |K : H|. These identities on cardinalities hold, whether or not I and J are finite. Exercise: 10 Section 4.1 Question: Let G be a group and let H and K be subgroups with |G : H| = m and |G : K| = n finite. Prove that a) |G : H ∩ K| ≤ mn; b) lcm(m, n) ≤ |G : H ∩ K|. Solution: We remark first that H ∩ K is a subgroup of G. Also a difficulty of this problem comes from the fact that since G could be infinite, we cannot simply use Lagrange’s Theorem. However, we need to count the possible number of cosets. By Exercise 4.1.15, |G : H ∩ K| = |G : H| |H : H ∩ K| and similarly with K in the middle. a) We try to put an upper bound on |H : H ∩ K| which means we determine how many different cosets of H ∩ K we can have in H. If h1 , h2 ∈ H then for cosets in H we have h1 (H ∩ K) = h2 (H ∩ K) if and only −1 if h−1 2 h1 ∈ H ∩ K. Thus h1 (H ∩ K) and h2 (H ∩ K) are not equal if an only if h2 h1 ∈ H̄ ∪ K̄ where the −1 Ā means complement. This implies that h2 h1 ∈ / K which means that h1 K and h2 K are distinct cosets of K in G. This shows that |H : H ∩ K| can be at most |G : K|. This proves that |G : H ∩ K| = |G : H||H : H ∩ K| ≤ |G : H||G : K| = mn. b) Since |G : H ∩ K| = |G : H| |H : H ∩ K| and |G : H ∩ K| = |G : K| |K : H ∩ K|, we deduce that |G : H ∩ K| is a multiple of both m and n. Hence |G : H ∩ K| is a multiple of lcm(m, n). In particular, |G : H ∩ K| ≥ lcm(m, n). Exercise: 17 Section 4.1 Question: Let ϕ : G → H be a group homomorphism. a) Prove that the left cosets of Ker ϕ are the fibers of ϕ. b) Deduce that for all g ∈ G, the left coset g(Ker ϕ) is equal to the right coset (Ker ϕ)g. Solution: a) Suppose that g1 , g2 ∈ G where ϕ(g1 ) = ϕ(g2 ) = h so that g1 and g2 are in the same fiber of h. Then ϕ(g1 )−1 ϕ(g2 ) = 1 ⇒ϕ(g1−1 g2 ) = 1 ⇒g1−1 g2 ∈ Ker ϕ. By Proposition 4.1.4, this implies that g1 and g2 represent the same left coset of Ker ϕ. Since the reasoning works in both directions, this shows that the fibers of ϕ are the left cosets of Ker ϕ.

180

CHAPTER 4. QUOTIENT GROUPS

b) For any ag ∈ (Ker ϕ)g we have ϕ(ag) = ϕ(a)ϕ(g) = 1ϕ(g) = ϕ(g). So then every element of (Ker ϕ)g is in the fiber of ϕ(g). From part a, we can deduce that every element of (Ker ϕ)g is also an element of g(Ker ϕ). Since we know that the sets have the same cardinality, we know they must be equivalent. Exercise: 18 Section 4.1 Question: Consider the Cayley graph for S4 given in Figure 3.12. a) Prove that the triangles (whose edges are blue) correspond to right cosets of h(1 2 3)i. b) Prove that the squares with all red edges correspond to the right cosets of h(1 2 3 4)i. c) Prove that the squares with mixed colors for edges are not the left or right cosets of any subgroup. Solution: We revisit the Cayley graph for S4 . a) By definition of the blue arrows, one passes from a vertex g to h along a blue arrow if h = (1 2 3)g. Consequently, a give triangle consists of {g, (1 2 3)g, (1 3 2)g} = h(1 2 3)ig, i.e., a right coset of h(1 2 3)i. b) By definition of the red arrows, one passes from a vertex g to h along a blue arrow if h = (1 2 3 4)g. Consequently, a give square consists of {g, (1 2 3 4)g, (1 3)(2 4)g, (1 4 3 2)g} = h(1 2 3 4)ig, i.e., a right coset of h(1 2 3 4)i. c) Consider the mixed square in the rhombicuboctahedron with vertices S = {1, (132), (14), (1432)}. Assume that the mixed square for the left cosets of some subgroup. Then this subgroup would need to be the set S since it contains the identity 1. However, (14)(1432) = (243) ∈ / S so S is not a subgroup. Contradiction. Exercise: 19 Section 4.1 Question: Suppose that a group G has order |G| = 105. List all possible orders of subgroups of G. Solution: By Lagrange’s Theorem, all possible orders of subgroups of G are just the divisors of |G|. So all possible orders of subgroups of a group which has order 105 are {1, 3, 5, 7, 15, 21, 35, 105}. Exercise: 20 Section 4.1 Question: Suppose that a group G has order |G| = 48. List all possible orders of subgroups of G. Solution: By Lagrange’s Theorem, all possible orders of subgroups of G are just the divisors of |G|. So all possible orders of subgroups of a group which has order 48 are {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}. Exercise: 21 Section 4.1 Question: Prove that n − 1 ∈ U (n) for all integers n ≥ 3. Apply Lagrange’s Theorem to hn − 1i to deduce that Euler’s totient function φ(n) is even for all n ≥ 3. Solution: We know that if gcd(n − 1, n) = 1 for all integers n ≥ 3 then n − 1 ∈ U (n) as well. Suppose that n − 1 and n are not relatively prime. Then there is some d > 1 where d|(n − 1) and d|n. Then d must also divide (n) − (n − 1) = n − n + 1 = 1. This implies that d must be one and contradicts the assumption that n and n − 1 are not relatively prime. This shows that for n ≥ 3, gcd(n, n − 1) = 1 ⇒ n − 1 ∈ U (n). Since (n − 1)2 = n2 − 2n + 1 = 1 we see that |hn − 1i| = 2. By Lagrange’s Theorem, |U (n)| = φ(n) must be even for all n ≥ 3. Exercise: 22 Section 4.1 Question: Prove or disprove that Z6 ⊕ Z10 has: a) a subgroup of order 4; b) a subgroup isomorphic to Z4 . Solution: We will let Z6 ⊕ Z10 = {(xi , y j ) x6 = y 1 0 = 1}.

4.1. COSETS AND LAGRANGE’S THEOREM

181

a) The subgroup H = h(x3 , 1), (1, y 5 )i = {1, (x3 , 1), (1, y 5 ), (x3 , y 5 )} of Z6 ⊕ Z10 has order 4. b) Consider the subgroup generated by any element (xi , y j ) ∈ Z6 ⊕ Z10 . The subgroup will have order lcm(|xi |, |y j |). Now, if we were to find a cyclic group of order 4, this would imply that at least one of xi or y j would have order 4. However, examining the elements Z6 and Z10 , we see that neither contains an element of order 4, so no cyclic subgroup can exist of order 4. Exercise: 23 Section 4.1 Question: Suppose that G is a group with |G| = pq, where p and q are primes, not necessarily distinct. Prove that every proper subgroup of G is cyclic. Solution: By Lagrange’s Theorem, every proper subgroup of G has order p or q. Since both p and q are primes, by Proposition 4.1.14 any subgroup would be cyclic. Exercise: 24 Section 4.1 Question: Let G = GL2 (F5 ). 1. Use Lagrange’s Theorem to determine all the possible sizes of subgroups of G. 2. Show that the orders of the following elements is respectively 3, 4, and 5, 0 2 3 0 1 A= , and B = , and C = 2 4 0 3 0

1 . 1

3. Determine the order of AB and BC without performing any matrix calculations. Solution: a) First we determine the size of G. G is the set of all 2 × 2 invertible matrices with elements in F5 . The first column can be anything except both zeroes. So the first column has 52 − 1 options. The second column can not be any multiple of the first column, so we have 52 − 5 options. In total, we have (52 − 1)(52 − 5) = (24)(20) = 480 options. So |G| = 480. By Lagrange’s theorem, all possible sizes of subgroups are the divisors of the order of G, so 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 80, 96, 120, 160, 240, 480. b)

0 2

3 0

1 0

A= B= C=

3 2 16 40 1 0 = = 4 40 96 0 1 4 0 81 0 1 0 = = 3 0 81 0 1 5 1 1 5 1 0 = = . 1 0 1 0 1

c) Since B is a multiple of the identity matrix, it is in the center of G. Then, |AB| = lcm(|A|, |B|) = 12 and |BC| = lcm(|B|, |C|) = 20. Exercise: 25 Section 4.1 Question: Let G = GL2 (F5 ) and let H be the subgroup of upper triangular matrices. [See Exercise 3.5.22] Prove that |G : H| = 6 and find 6 different matrices g1 , g2 , . . . , g6 such that the cosets gi H for i = 1, 2, . . . , 6 are all the left cosets of H. Solution: We start by finding the size of H, For any A ∈ H, det(A) = ad − bc. Since A is upper triangular, c = 0. So det(A) = ad. So the determinant is nonzero as long as both a and d are nonzero. So we have 5 options for b and 4 nonzero options for a and d. So |A| = 5 · 4 · 4 = 80. From problem 24 we know that |G| = 480. So |G : H| = |G|/|H| = 480/80 = 6. For the six different matrices we will choose the identity matrix and all the matrices of the form k 1 1 0

182

CHAPTER 4. QUOTIENT GROUPS

where k ∈ F5 . Now, we will show the cosets are disjoint by examining the form of the matrices in each coset. The first coset is just the upper triangular matrices. For the second through sixth cosets, we will examine what multiplying on the left by these types of matrices do to upper triangular matrices. k 1 a b ka kb + d = 1 0 0 d a b Since k is different for each coset, we know that they must be disjoint since a property for every matrix in each coset is that the top left corner is k times the bottom left corner. These five cosets are obviously disjoint from the identity coset since a 6= 0 so that none of these cosets contain upper triangular matrices. Since |G : H| = 6 and we have found 6 disjoint cosets, these cosets must be all the left cosets of H. Exercise: 26 Section 4.1 Question: Let p be a prime number. Prove that the subgroups of Zp ⊕ Zp consist of {1}, Zp ⊕ Zp and p + 1 subgroups that are cyclic and of order p. Solution: We know that {1} and Zp ⊕ Zp are both subgroups. Since |Zp ⊕ Zp | = p2 , any other subgroup will have order p. This implies that the subgroups will be cyclic. Now, consider an arbitrary subgroup H = h(xi , y j )i. |H| = |(xi , y j )| = lcm(|xi |, |y j |). Now we have two options. If the element is the identity, |H| = 1. Otherwise, one or both of the elements will have order p and then |H| = p. So for every non-identity element, we can create a cyclic subgroup H of order p by using it as a generator. Since prime cyclic subgroups intersect trivially (in only the identity), each subgroup will hold p − 1 different elements of Zp ⊕ Zp . Then we can make p + 1 different subgroups until we run out of non-identity elements to generate subgroups. Exercise: 27 Section 4.1 Question: Let G be a group of order 21. a) Prove that G must have an element of order 3. b) By the strategy of Example 4.1.15, can we determine whether G must have an element of order 7? Solution: a) If G is cyclic, then it has a group of order 3, H = hx7 i. So assume G is not cyclic. Then G is made up of subgroups of size 7. Each subgroup can intersect with all the other subgroups trivially in the identity. Therefore, if we have n distinct subgroups, we have n · 6 + 1 different elements. However, n · 6 + 1 = 21 implies that 6|20. Which is not true. So G cannot be made up of subgroups of only size 7 and so must have another subgroup of size 3. Since this subgroup has a prime order, it must be cyclic and generated by some element, and therefore there is an element that has order 3. b) The only way G would not have an element of order 7 is if G were made up entirely of subgroups of order 3. Since each subgroup intersects with every other subgroup only in the identity, if we have n subgroups, we have n(3 − 1) different non-identity elements. So with the identity we would have n(3 − 1) + 1 = 2n + 1 elements. This value can equal 21 if we let n = 10. So we cannot determine whether or not G must have an element of order 7 using this strategy. Exercise: 28 Section 4.1 Question: Let G be a group of order 3p, where p is a prime number. Prove that G has an element of order 3. Solution: If G is cyclic and generated by x, then |xp | = lcm(3p, p)/ gcd(3p, p) = 3p/p = 3. If G has any subgroups of order 3 then it has an element of order 3. Assume that G is not cyclic and does not have any subgroups of order 3. Then it must be made up of subgroups of order p. Each subgroup intersects with every subgroup trivially and contains p − 1 non-identity elements. So if we have n different subgroups we have n(p − 1) + 1 different elements. In order for this to work we will need n(p − 1) + 1 = 3p. Rearranging we get, n−1 . So, if n = 1, 2, 3 we get either p = 0, -1 and undefined. If np − n + 1 = 3p ⇒ (n − 3)p = n − 1 ⇒ p = n−3 n = 4, we get p = 3 and we will have an element of order 3. If n = 5 we get p = 2 and then |G| = 6. However, from Exercise 3.3.31 part d and e, we know that the only groups of order 6 are Z6 , which is cyclic and has an element of order 3 (x2 ) and D3 which also has an element of order 3 (r). If n ≥ 6, we see that 1 < p < 2 from graphing the function and observing the limit as n approaches infinity. So we have exhausted all options for n which have either resulted in a contradiction or finding an element of order 3. We have checked all possibilities for how G could be formed and have found that each results in an element of order 3. Exercise: 29 Section 4.1 Question: Let G be a group of order pq, where p and q are distinct odd primes such that (p − 1) - (q − 1) and (q − 1) - (p − 1). Prove that G contains an element of order p and an element of order q.

4.1. COSETS AND LAGRANGE’S THEOREM

183

Solution: We will prove that G has an element of order p. The proof that G has an element of order q will have the exact same format, except p and q will be swapped. If G has at least one subgroup of order p, then any non-identity element of that subgroup has order p. So we will assume that G is made up entirely of subgroups of order q. Since each distinct subgroup has q − 1 distinct elements, if we have n subgroups we have n(q − 1) + 1 elements in our group. Since |G| = pq, we know that n(q − 1) + 1 = pq or equivalently n(q − 1) + 1 = p(q − 1) + p. Now if we take both sides mod q − 1, 1̄ = p̄ ⇒ 0̄ = p − 1. However, this implies that q − 1|p − 1, which is a contradiction. So G cannot be made up entirely of subgroups of order q and must have a subgroup of order p and therefore an element of order p. We have checked all possible cases and have found that G must have an element of order p in all of them. Since the proof that G has an element of order q will be identical to the one above except swapping p and q, we have proved that G must contain an element of order p and an element of order q. Exercise: 30 Section 4.1 Question: Let G be a group and let H, K ≤ G. Prove that if gcd(|H|, |K|) = 1, then H ∩ K = {1}. Solution: We will prove the statement’s contrapositive which is, If H ∩ K contains a non-identity element, then gcd(|H|, |K|) > 1. Let g ∈ H ∩ K and |g| = n > 1. By Lagrange’s Theorem, since g ∈ H this implies that n||H| and since g ∈ K this implies that n||K|. This implies that n| gcd(|H|, |K|). So gcd(|H|, |K|) > 1 and we have proved the contrapositive and the original statement follows. Exercise: 31 Section 4.1 Question: Let G be a group and let H be a subgroup with |G : H| = p, a prime number. Prove that if K is a subgroup of G that strictly contains H, then K = G. |G| |G| |K| = = p. Since the index of subgroups within groups are always integers, we Solution: Consider |K| |H| |H| |K| |G| have limited options. Since K strictly contains H, we must have that = p which implies that = 1. This |H| |K| shows that K = G. Exercise: 32 Section 4.1 Question: Let G be a group of order pqr, where p, q, r are distinct primes. Let A be a subgroup of order pq and let B be a subgroup of order qr. Prove that AB = G and that |A ∩ B| = q. Solution: First, recall that if A and B are subgroups of a group G, then A ∩ B is also a subgroup of G. Furthermore, A ∩ B is a subgroup both of A and of B. By Lagrange’s Theorem, |A ∩ B| divides both |A| and |B| Thus |A ∩ B| divides gcd(|A|, |B|) = gcd(pq, qr). Since p, q, r are distinct primes, then |A ∩ B| is equal to 1 or q. Now consider the subset AB that is the product set of the two subgroups. By Proposition 4.1.16, |AB| =

pq 2 r |A| |B| = . |A ∩ B| |A ∩ B|

Assume that |A ∩ B| = 1. Then |AB| = pq 2 r. However, this is greater than the cardinality of G so we have a contradiction. Consequently, we deduce that |A ∩ B| = q and |AB| = pqr. Since |G| = pqr, then |AB| is a subset of G with the same finite cardinality as G. Hence G = AB. Exercise: 33 Section 4.1 Question: Use Lagrange’s Theorem applied to U (n) to prove Euler’s Theorem (a generalization of Fermat’s Little Theorem), which states that if gcd(a, n) = 1 then aϕ(n) ≡ 1

(mod n).

Solution: We know that if gcd(a, n) = 1 then ā ∈ U (n). Now |U (n)| = ϕ(n) and by Corollary 4.1.12 of Lagrange’s Theorem, āϕ(n) = 1̄. This is equivalent to saying that aϕ(n) ≡ 1 (mod n). Exercise: 34 Section 4.1 Question: Show that there exists a subgroup of order d for each divisor d of |S4 | = 24 and give an example of a subgroup for each divisor.

184

CHAPTER 4. QUOTIENT GROUPS

Solution: The divisors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. For 1 and 24 we have the identity subgroup and the whole group trivially. For 2, we have {id, (1 2)}. For 3, we have {id, (1 2 3), (1 3 2)}. For 4, we have {id, (1 2), (3 4), (1 2)(3 4)}. For 6, we have {id, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)} ∼ = S3 . For 8, we have {id, (1 2 3 4), (1 3)(2 4), (1 4 3 2), (1 3), (2 4), (1 4)(2 3), (1 2)(3 4)} ∼ = D8 . For 12, we have {id, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)} ∼ = A4 .

Exercise: 35 Section 4.1 Question: Let p be a prime number. This exercise guides the proof that a group of order 2p is isomorphic to Z2p or Dp . Let G be an arbitrary group with |G| = 2p. a) Without using Cauchy’s Theorem, prove that G contains an element a of order 2 and an element b of order p. b) Prove that if ab = ba, then G ∼ = Z2p . c) Prove that if a and b do not commute, then aba = b−1 . Deduce in this case that G ∼ = Dp . Solution: Let p be a prime number and let G be an arbitrary group with |G| = 2p. We point out here that if p = 2, then there are only two groups of order 2p = 4, namely Z4 or Z2 ⊕ Z2 ∼ = D2 . Hence, the result of the exercise holds. From now on, we assume that p is odd. a) Assume that the nonidentity elements of G are only of order p. Let a ∈ G − {1} and let b be another element not in hbi. Then it is easy to see that am = bn if and only if m and n are both multiples of p. Then hai ∪ hbi contains 2p − 1 elements, since they have a nontrivial intersection of {1}. Let c be the one element in C − (hai ∪ hbi). By the assumption it must have order p. Again, we can show that none of the elements c, c2 , . . . cp−1 are in hai ∪ hbi, which contradicts the assumption that |G| = 2p. Assume now that the nonidentity elements of G are only of order 2. Exercise 3.2.28 shows that G must be abelian. By induction on the minimum number of generators a group H whose nonidentity elements are only of order 2, we prove that such a group must have order 2n . Suppose that H is generated by a single element. This element has order 2 and thus H ∼ = Z2 . Now suppose that for some positive integer n, all groups with the above property and with a minimum number of generators of n has order 2n . Consider now a group K with that property but with n + 1 minimum number of generators. Let {g1 , g2 , . . . , gn+1 } be a set of generators of K. Then hg1 , g2 , . . . , gn i has a n minimum number of generators so by the induction hypothesis, has 2n elements. Since K is abelian, every element in K can be written uniquely as h ∈ hg1 , g2 , . . . , gn i or h0 gn+1 for some h ∈ hg1 , g2 , . . . , gn i. Hence, K contains 2n elements. In particular, if p is odd, then G cannot contain exclusive elements or order dividing 2. We conclude that G must contain at lease one element of order p and one element of order 2. b) Suppose that ab = ba. By order considerations, am = bn if and only if m is a multiple of 2 and n is a multiple of p. Suppose that (ab)k = 1. Then since a and b commute, ak bk = 1. Thus, ak = b−k = bk . Consequently, k is a multiple of 2 and a multiple of the odd prime p. Therefore, k is a multiple of 2p. Thus, since |G| = 2p, we have |ab| = 2p so G is cyclic with G ∼ = Z2p . c) Now suppose that a and b do not commute. Since |G : hbi| = 2, then the coset ahbi = G − hbi = hbia. Thus, ahbia−1 = hbi and, in particular, aba−1 ∈ hbi. Thus aba−1 = bk for some integer k with 2 ≤ k ≤ p − 1. Since a has order 2, k times

k times

z }| { z }| { 2 b = a2 ba−2 = a(aba−1 )a−1 = abk a−1 = (aba−1 ) · · · (aba−1 ) = bk · · · bk = bk . Hence, k must satisfy k 2 ≡ 1 (mod p). The solutions so this congruence equation are k ≡ 1 and k ≡ −1. However, the case k ≡ 1 corresponds to a and b commuting so we do not consider it. Thus, we have aba−1 = b−1 . In this case, all the elements of G are of the form bn or abn . We conclude that G = ha, b | a2 = bp = 1, aba−1 = b−1 i ∼ = Dp . Exercise: 36 Section 4.1 Question: Let G be a group and let H, K ≤ G. Prove that that HK ≤ G if and only if HK = KH as sets.

4.2. CONJUGACY AND NORMAL SUBGROUPS

185

Solution: Let G be a group and let H and K be subgroups of G. Suppose that HK ≤ G. Let k ∈ K and h ∈ H be arbitrary. Then kh = (h−1 k −1 )−1 . But h−1 k −1 ∈ HK and since HK is closed under taking inverses we conclude that kh ∈ HK. Thus KH ⊆ HK. Stronger, since HK ≤ G, then hk = (h0 k 0 )−1 for some h0 ∈ H and k 0 ∈ K. Then hk = (k 0 )−1 (h0 )−1 so hk ∈ KH. Thus HK ⊆ KH. We conclude that HK = KH. Conversely, suppose that HK = KH. We know that HK is nonempty since both H and K are nonempty. Let h1 k1 and h2 k2 be in HK. Then (h1 k1 )(h2 k2 )−1 = h1 (k1 k2−1 )h2 . By the assumption, we conclude that (k1 k2−1 )h2 = h3 k3 for some h3 ∈ H and k3 ∈ K. Thus, (h1 k1 )(h2 k2 )−1 = (h1 h3 )k3 ∈ HK. By the One-Step Subgroup Criterion, HK is a subgroup.

4.2 – Conjugacy and Normal Subgroups Exercise: 1 Section 4.2 Question: Prove that An is a normal subgroup of Sn . Solution: Consider any element σ ∈ An , then we can write σ = σ1 σ2 · · · σ2k where each σi is a transposition. For any element τ ∈ Sn , we know that we can decompose τ into a composition of transpositions, so let τ = τ1 τ2 · · · τn where each τi is a transposition. Then τ −1 = (τ1 τ2 · · · τn )−1 = τn τn−1 . . . τ1 . For any σ ∈ An and τ ∈ Sn , consider the conjugacy, τ στ −1 = (τ1 τ2 · · · τn )(σ1 σ2 · · · σ2k )(τn τn−1 . . . τ1 ) We have written the composition as a product of 2k + n + n = 2k + 2n = 2(k + n) transpositions so we must have that τ στ −1 ∈ An for any σ ∈ An and τ ∈ Sn . Which by Theorem 4.2.4 implies that An is a normal subgroup of Sn . Exercise: 2 Section 4.2 Question: Determine whether hr2 i is normal in D8 . Solution: Let H = hr2 i. By Theorem 4.2.8, we just have to check the generators of G (r and s) with the generators of our subgroups (r2 ). So sr2 s = ssr6 = r6 ∈ H and rr2 r7 = r2 ∈ H. By Theorem 4.2.8, we know that H E G. Exercise: 3 Section 4.2 Question: Determine whether h(1 2)(3 4), (1 3)(2 4)i is normal in A4 . Solution: Let H = h(1 2)(3 4), (1 3)(2 4)i = {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3), 1}. From Example 4.2.13 we know that the conjugacy classes of Sn are exactly all the cycles of a certain cycle type. In H, we have all the cycles of type (a b)(c d) of S4 , which implies that it is a conjugacy class of S4 . By Proposition 4.2.14, since H is a union of one conjugacy class, we know that H E S4 . Since A4 ≤ S4 , we know that H E A4 as well. Exercise: 4 Section 4.2 Question: Find all normal subgroups of D6 . Solution: First, we know that {1} and D6 are normal subgroups. Also, all the subgroups of index 2 are subgroups: hri, hr2 , si, hr2 , sri. Otherwise the subgroups hr2 i and hr3 i are also normal, with hr3 i = Z(D6 ). We can confirm that we have all the normal subgroups as follows. If a normal subgroup contains sra , then it also contains, rsra r−1 = sra−2 . Hence it contains sra−2 sra = ssr2−a ra = r2 . Thus, we have already found all the normal subgroups that only involve powers of r but the other two subgroups hr2 , si and hr2 , sri are the only possible normal subgroups that contain a reflection. Exercise: 5 Section 4.2 Question: Prove that H E G for all subgroups H ≤ Z(G).

186

CHAPTER 4. QUOTIENT GROUPS

Solution: Let H ≤ Z(G) and let g ∈ G and h ∈ H be any elements. Consider ghg −1 . Since h ∈ H we know that h ∈ Z(G) which implies that h commutes with g and g −1 . So we get ghg −1 = gg −1 h = 1h = h ∈ H. Since this holds for any arbitrary h ∈ H and g ∈ G, by Theorem 4.2.4 we have H E G for all H ≤ Z(G). Exercise: 6 Section 4.2 Question: Prove that every subgroup of Q8 is a normal subgroup. [This shows that the converse to Proposition 4.2.5 is false.] Solution: The only subgroups of Q8 are hii, hji, hki, and h−1i. Since −1 and 1 are both in Z(Q8 ), from Exercise 4.2.5 we know that h−1i E Q8 . Let H be any of the remaining three subgroups. Then |H| = 4 which implies that |Q8 : H| = 2. So by Proposition 4.2.2, each of the remaining three subgroups is a normal subgroup as well. This shows that every subgroup of Q8 is a normal subgroup. Exercise: 7 Section 4.2 Question: Let F = Q, R, C, or Fp where p is prime. Prove that for all positive integers n, SLn (F ) E GLn (F ). Solution: From Example 3.7.7, we know that the determinant function, or det: GLn (F ) → F ∗ , is a homomorphism for Q, R, C, and Fp (where p is prime). The kernal of the determinant homomorphism is precisely those matrices who have a determinant of one, which is how SLn (F ) is defined. By Proposition 4.2.7, since SLn (F ) is the kernel of a homomorphism, SLn (F ) E GLn (F ). Exercise: 8 Section 4.2 Question: Let F = Q, R, C, or Fp where p is prime. Denote by Tn (F ) ≤ GLn (F ) the upper triangular matrices in GLn (F ). a) Prove that the function ϕ : Tn (F ) → (F × )n defined by ϕ(A) = (a11 , a22 , . . . , ann ) is a homomorphism, where by (F × )n we mean the nth direct sum of the multiplicative group F × = F −{0}. b) Conclude that the subgroup of Tn (F ) consisting of matrices with 1s down the diagonal is a normal subgroup. Solution: Let F = Q, R, C, or Fp where p is prime. Denote by Tn (F ) ≤ GLn (F ) the upper triangular matrices in GLn (F ). a) Let A, B ∈ Tn (F ). Since A and B are upper triangular, then A = (aij ) and B = (bij ) with aij = 0 and bij = 0 whenever j < i. If we call C the product, then C = AB with cij =

n X

aik bkj .

(4.1)

k=1

The diagonal elements of C are cii . In the summation of (4.1), if k < i, then aik = 0 and if k > i, then bki = 0. Hence cii = aii bii . Consequently, the function ϕ satisfies ϕ(AB) = (a11 b11 , a22 b22 , . . . , ann bnn ) = (a11 , a22 , . . . , ann )(b11 , b22 , . . . , bnn ) = ϕ(A)ϕ(B). Hence ϕ is a homomorphism. b) The subgroup of Tn (F ) with 1s down the diagonal is the kernel of ϕ. Hence, it is a normal subgroup of T( F ). Exercise: 9 Section 4.2 Question: Let n be a positive integer and let G = GLn (R). a) Prove that H = GLn (Q) is a subgroup that is not a normal subgroup. b) Define the subset K as K = {A ∈ GLn (R) | det(A) ∈ Q}. Prove that K is a normal subgroup that contains H.

4.2. CONJUGACY AND NORMAL SUBGROUPS

187

Solution: a) We just need to find one counter-example where we conjugate an element of H and end up not in H. Consider the conjugation 1 π 1 0 1 −π 1 π = ∈ / GLn (Q). 0 1 0 2 0 1 0 2 This shows that GLn (Q) is not a normal subgroup of GLn (R). b) Since for any M ∈ GLn (Q), det(M ) ∈ Q, we have H ⊆ K. Now, consider any D ∈ G and M ∈ K. Let det(D) = n and det(M ) = q ∈ Q. Then consider DM D−1 . Now det(DM D−1 ) = det(D) det(M ) det(D−1 ) 1 = det(D) det(M ) det(D) 1 n = nq = q = q ∈ Q. n n This shows that DM D−1 ∈ K for all D ∈ G and M ∈ K. By Theorem 4.2.4 this shows that K is a normal subgroup of G. Exercise: 10 Section 4.2 Question: Prove Proposition 4.2.12. Solution: We will prove that the property of being conjugate, or ∼c , is an equivalence relation. ∼c is reflexive since x = 1x1 for any x. If x ∼c y then y = gxg −1 ⇒ g −1 yg = x and by letting h = g −1 we get hyh−1 = x which shows that y ∼c x so that ∼c is symmetric. If y = gxg −1 and z = hyh−1 then z = hyh−1 = h(gxg −1 )h−1 = (hg)x(hg)−1 . This shows that if x ∼c y and y ∼c z then x ∼c z. So ∼c is transitive. We have verified that all the properties hold, so ∼c is an equivalence relation. Exercise: 11 Section 4.2 Question: Prove Corollary 4.2.10 Solution: Let G be a group, K ≤ G, and H ≤ Ng (K). Recall that the set HK = {hk h ∈ H and k ∈ K}. We will prove that HK ≤ G using the one step subgroup criteria. HK is nonempty since 1 · 1 = 1 ∈ HK. Consider two elements h1 k1 , h2 k2 ∈ HK. We will show that h1 k1 (h2 k2 )−1 ∈ HK. Now, h1 k1 (h2 k2 )−1 = h1 k1 k2−1 h−1 2 . −1 Now, k1 k2−1 = k3 ∈ K. So h1 k1 k2−1 h2 = h1 k3 h2 = h1 h2 (h−1 k h ). Since H ≤ N (K), we know that h k h 3 2 g 3 2 = 2 2 k4 ∈ K. So then h1 h2 (h−1 k h ) = h h k = (h h )k ∈ HK. So by the one step subgroup criteria, HK ≤G 3 2 1 2 4 1 2 4 2 for all H ≤ Ng (K). This completes the proof of Corollary 4.2.10. Exercise: 12 Section 4.2 Question: Let G be a group. Prove that if H ≤ G is the unique subgroup of a given order n, then H E G. Solution: In Exercise 37 of Section 3.7, we show that gHg −1 is another subgroup of H and it is isomorphic to H. In particular |gHg −1 | = |H|. However, if H is the unique subgroup of a given size, then gHg −1 = H for all g ∈ G. Hence H is a normal subgroup. Exercise: 13 Section 4.2 Question: Let ϕ : G → H be a group homomorphism and let N E H. Prove that ϕ−1 (N ) E G. [Note that this generalizes the fact that kernels of homomorphisms are normal subgroups of the domain group.] Solution: Let H = {h ∈ G ϕ(h) ∈ N }. We first prove that H is a subgroup. Consider h, k ∈ H. We will use the one step subgroup criteria. Consider ϕ(hk −1 ) = ϕ(h)ϕ(k)−1 ∈ N since ϕ(h), ϕ(k) ∈ N and N is a subgroup. This implies that hk −1 ∈ H. So H is a subgroup. Now we will prove that H is a normal subgroup of G. Let g ∈ G and h ∈ H. Consider ghg −1 . We will show that this element is in the fiber of N . Consider ϕ(ghg −1 ) = ϕ(g)ϕ(h)ϕ(g −1 ) = ϕ(g)ϕ(h)ϕ(g)−1 . Now, since ϕ(h) ∈ N and N E G, we have ϕ(g)ϕ(h)ϕ(g)−1 ∈ N . This implies that ghg −1 ∈ H since H is the fiber of N . By Theorem 4.2.4, H E G and equivalently ϕ−1 (N ) E G. Exercise: 14 Section 4.2 Question: Let G be a group, H ≤ G and N E G. Prove that H ∩ N E H.

188

CHAPTER 4. QUOTIENT GROUPS

Solution: Let h ∈ H and k ∈ H ∩ N be any elements and consider hkh−1 . Since k ∈ N , h ∈ G and N is a normal subgroup, we must have hkh−1 ∈ N . Since k, h ∈ H, we must have hkh−1 ∈ H. By the definition of intersection, this shows that hkh−1 ∈ H ∩ N . By Theorem 4.2.4, H ∩ N E H. Exercise: 15 Section 4.2 Question: Prove that the intersection of two normal subgroups N1 , N2 of a group G is again a normal subgroup N1 ∩ N2 E G. Solution: We use Theorem 4.2.4. Let g ∈ G and let n ∈ N1 ∩ N2 . Since N1 E G and n ∈ N1 , then gng −1 ∈ N1 . Also, since N2 E G and n ∈ N2 , then gng −1 ∈ N2 . Hence gng −1 ∈ N1 ∩ N2 . We conclude that N1 ∩ N2 E G. Exercise: 16 Section 4.2 Question: Let N1 and N2 be normal subgroups is G. Prove that N1 N2 is the join of N1 and N2 and that it is a normal subgroup in G. Solution: We first prove that N1 N2 is the join of N1 and N2 by showing that it is a subgroup using the one step −1 −1 subgroup criteria. Let g1 h1 , g2 h2 ∈ N1 N2 and consider g1 h1 (g2 h2 )−1 = g1 h1 h−1 2 g2 . Now, h1 h2 = h3 ∈ N2 . So −1 −1 −1 −1 −1 −1 g1 h1 h2 g2 = g1 h3 g2 = g1 g2 g2 h3 g2 . Since h3 ∈ N2 and N2 is a normal subgroup, g2 h3 g2 = h4 ∈ N2 . So g1 g2−1 g2 h3 g2−1 = g1 g2−1 h4 = (g1 g2−1 )h4 ∈ N1 N2 . This proves that N1 N2 is a subgroup and so is the join of N1 and N2 . Now we will show that it is normal. Let g ∈ G and xy ∈ N1 N2 . Consider g(xy)g −1 = gxg −1 gyg −1 . Since x ∈ N1 and N1 is a normal subgroup, gxg −1 = x0 ∈ N1 . Likewise, gyg −1 = y 0 ∈ N2 . So that gxg −1 gyg −1 = x0 y 0 ∈ N1 N2 . By Theorem 4.2.4, N1 N2 is a normal subgroup in G. Exercise: 17 Section 4.2 Question: Let {Ni }i∈I be a collection of normal subsets of G. Prove that the intersection \ Ni i∈I

is a normal subgroup. Do not assume that I is finite. Solution: T −1 Let n ∈ i∈I . Since n exists in every Ni , gng −1 ∈ Ni for all i. This implies T Ni and g ∈ G. Consider gng T −1 that gng ∈ i∈I Ni . By Theorem 4.2.4, i∈I Ni is a normal subgroup. Exercise: 18 Section 4.2 Question: Suppose that a subgroup H ≤ G is such that if h ∈ H with |h| = n, then H contains all the elements in G of order n. Prove that H is a normal subgroup. Solution: Consider any elements h ∈ H and g ∈ G and the conjugation ghg −1 . From Exercise 3.3.25, we know that |ghg −1 | = |h| ⇒ ghg −1 ∈ H. By Theorem 4.2.4, H is a normal subgroup. Exercise: 19 Section 4.2 Question: Let A be any subset of a group G. Prove that CG (A) E NG (A). Solution: Let g ∈ NG (A) and let h ∈ CG (A). We need to determine whether ghg −1 is in CG (A). For all a ∈ A (ghg −1 )a(ghg −1 )−1 = ghg −1 agh−1 g −1 = g(h(g −1 ag)h−1 )g −1 . Since NG (A) is a subgroup of G, then g −1 ∈ NG (A). Hence g −1 ag = a0 ∈ A. Since h ∈ CG (A), then hah−1 = a0 . Furthermore, since g −1 ag = a0 then ag = ga0 so a = ga0 g −1 . Thus, (ghg −1 )a(ghg −1 )−1 = ga0 g −1 = a0 . Hence ghg −1 ∈ CG (A). We have shown that CG (A) E NG (A). Exercise: 20 Section 4.2 Question: We say that a subgroup H of a group G is a characteristic subgroup if ψ(H) = H for all automorphisms ψ ∈ Aut(G). Prove that every characteristic subgroup is a normal subgroup. Solution: Let G be any group and H be any characteristic subgroup of G. Let g ∈ G be any element. By Proposition 4.2.15, the function ψg : G → G defined by ψg (x) = gxg −1 is an automorphism of G. Since H is

4.2. CONJUGACY AND NORMAL SUBGROUPS

189

a characteristic subgroup of G, this implies that ψg (H) = gHg −1 = H for all g ∈ G. By Theorem 4.2.4, this implies that H is a normal subgroup. Exercise: 21 Section 4.2 Question: Prove that if g ∈ Z(G), then the conjugacy class of g is the singleton set {g}. Solution: Since 1g1 = g, g exists in the conjugacy class of g. Suppose that some other h exists in the conjugacy class of g. Then for some x ∈ G, h = xgx−1 , but since g ∈ Z(G) we have h = xgx−1 = xx−1 g = 1g = g. So no other element can exist in the conjugacy class and the conjugacy class of g is the singleton set {g}. Exercise: 22 Section 4.2 Question: Prove that in an abelian group, all the conjugacy classes consist of a single element. Solution: Let G be an abelian group and consider the conjugacy class of any element g ∈ G. Since Z(G) = G, we know that g ∈ Z(G). By Exercise 4.2.21, we know that the conjugacy class of g is the singleton set {g}. This shows that every conjugacy class consists of a single element. Exercise: 23 Section 4.2 Question: Prove that the conjugacy classes of Dn are: a) if n is even: {1}, {rn/2 }, {ra , r−a } for 1 ≤ a ≤ n2 − 1, {s, sr2 , . . . , srn−2 }, and {sr, sr3 , . . . , srn−1 }; and 2 n−1 b) if n is odd: {1}, {ra , r−a } for 1 ≤ a ≤ n−1 }. 2 , and {s, sr, sr , . . . , sr Solution: a) We know that the conjugacy class of 1 is {1}. Consider any ra ∈ Dn . ra commutes with all powers of r, so the only elements we have to check are of the form srb . So (srb )(ra )(srb ) = srb sr−a rb = ssr−b r−a rb = r−a . Since conjugating r−a with any other element will either give us r−a or r−(−a) = ra we conclude that the conjugacy classes of powers of r are {ra , r−a } and {rn/2 } (since n/2 = −n/2 mod n). Now we will find the conjugacy class of sr. Consider conjugation by sra , we get sra srsra = ssr−a sr−1 ra = 1sra r−1 ra = sr2a−1 . So we can get any odd number for the power of r by choosing different values of a, so the conjugacy class becomes {sr, sr3 , . . . srn−1 }. Now we conjugate by ra to see if any other elements belong in the conjugacy class. ra srr−a = sr−a rr−a = sr−2a+1 and we get another element already belonging to the conjugacy class. Now we will find the conjugacy class of s. Consider ra sr−a = sr−a r−a = sr−2a . So we get all elements of the form srk where k is any even number. So our conjugacy class is the remaining elements {s, sr2 , . . . srn−2 }. b) We know that the conjugacy class of 1 is {1}. Consider any ra ∈ Dn . ra commutes with all powers of r, so the only elements we have to check are of the form srb . So (srb )(ra )(srb ) = srb sr−a rb = ssr−b r−a rb = r−a . Since conjugating r−a with any other element will either give us r−a or r−(−a) = ra we conclude that the conjugacy classes of powers of r are {ra , r−a }. Now we will find the conjugacy class of sr. Consider conjugation by sra , we get sra srsra = ssr−a sr−1 ra = 1sra r−1 ra = sr2a−1 . So we can get any odd number for the power of r by choosing different values of a. If we make a big enough so that 2a − 1 > n, then we get the elements of the form sr2a−1−n where the power of r is now even since n and 2a − 1 is odd. So we can get all elements of the form srk and the conjugacy class is {s, sr, sr2 , . . . srn−1 }. Exercise: 24 Section 4.2 Question: List all the conjugacy classes in A4 . Solution: From Example 4.2.13, we know that any cycle, when conjugated, will remain the same cycle type. The first conjugacy class is trivial, {1}. Now we will find the conjugacy class of (1 2)(3 4). With a little experimentation, we observe that (1 2 3)(1 2)(3 4)(1 3 2) = (1 4)(2 3) and (1 2 3)(1 4)(2 3)(1 3 2) = (1 3)(2 4) and since we have hit all the cycles of type (a b)(c d) we know that we must have found the entire conjugacy class which is {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. Before we find the conjugacy classes of the 3-cycles, it is important to note that the only σ ∈ S4 where σ(a b c)σ −1 = (a c b) is (b c) which is an odd permutation and therefore not an

190

CHAPTER 4. QUOTIENT GROUPS

element of A4 . This implies that any 3-cycles that are inverses cannot be in the same conjugacy class. Now we will find the conjugacy class of (1 2 3). After some experimenting we see that (1 2)(3 4)(1 2 3)(1 2)(3 4) = (1 4 2) and (1 3)(2 4)(1 4 2)(1 3)(2 4) = (2 4 3) and also (2 3 4)(1 2 3)(2 4 3) = (1 3 4). Since the conjugacy class contains half of the 3-cycles, we cannot get anymore from conjugation because then we would include an inverse of one of the elements. So this conjugacy class becomes {(1 2 3), (1 4 2), (1 3 4), (2 4 3)}. By taking the inverse of each equation above we see that the inverses of those 3-cycles are conjugate as well. So the last conjugacy class gets all of the remaining 3-cycles, {(1 3 2), (1 2 4), (1 4 3), (2 3 4)}. Exercise: 25 Section 4.2 Question: List all the conjugacy classes in the group G2 of Example 3.8.6. Solution: Let G2 = ha, b | a3 = b7 = 1, ab = b2 ai. First, {1} is a conjugacy class by itself. For the conjugacy class of x we need 1a1−1 = a bab−1 = bab6 = b13 a = b6 a b2 ab−2 = b2 ab5 = b12 a = b5 a b3 ab−3 = b3 ab4 = b11 a = b4 a b4 ab−4 = b4 ab3 = b10 a = b3 a b5 ab−5 = b5 ab2 = b9 a = b2 a b6 ab−6 = b6 ab = b8 a = ba (bm xn )a(bm xn )−1 = bm an aa−n b−m = bm ab−m Hence we have shown that [a] = {a, ba, b2 a, b3 a, b4 a, b5 a, b6 a}. Similarly, we can show with the same reasoning that the conjugacy class of a2 is [a2 ] = {a2 , ba2 , b2 a2 , b3 a2 , b4 a2 , b5 a2 , b6 a2 }. So far we have taken into account 2 × 7 + 1 = 15 elements. The conjugacy class of b involves aba−1 = b2 ,

a2 ba−2 = b4 ,

(bm x)b(bm x)−1 = bm b2 b−m = b2 ,

(bm a2 )b(bm a2 )−1 = b4 .

So [b] = {b, b2 , b4 }. Similarly, [b3 ] = {b3 , b5 , b6 }. We have found 5 conjugacy classes. Exercise: 26 Section 4.2 Question: Let G be a group. Consider the group of automorphisms Aut(G). Prove that the group Inn(G) of inner automorphisms (see Exercise 3.7.38) is a normal subgroup of Aut(G). Solution: Let ϕ ∈ Aut(G) be any automorphism and ψg ∈ Inn(G) be an inner-automorphism defined by ψg (x) = gxg −1 . Let x ∈ G and consider the conjugation (ϕ ◦ ψg ◦ ϕ−1 )(x) = ϕ(ψg (ϕ−1 (x))) = ϕ(g(ϕ−1 (x))g −1 ) = ϕ(g)ϕ(ϕ−1 (x))ϕ(g −1 ) = ϕ(g)(x)ϕ(g)−1 = hxh−1 = ψh (x) This implies that (ϕ ◦ ψg ◦ ϕ−1 ) ∈ Inn(G) for any automorphism ϕ and any inner automorphism ψg . By Theorem 4.2.4, Inn(G) is a normal subgroup of Aut(G). Exercise: 27 Section 4.2 Question: In this exercise, we guide the reader to prove that A5 is a simple group. a) Prove that, as a partition of A5 , the set of conjugacy classes in A5 is a strict refinement of the partition of A5 into cycle types. b) Prove that A5 has 5 conjugacy classes, 2 consisting of equal parts of the subset of 5-cycles, all of the 3-cycles, the whole subset of permutations with cycle type (2, 2), and the identity. c) After determining the sizes of all of the conjugacy classes of A5 , use Proposition 4.2.14 and Lagrange’s Theorem to conclude that A5 has no normal subgroups besides {1} and the whole group.

4.3. QUOTIENT GROUPS

191

Solution: We consider the group A5 . a) Recall that for all cycles (a1 a2 · · · an ) and any permutation σ ∈ Sn , we have σ(a1 a2 · · · an )σ −1 = (σ(a1 ) σ(a2 ) · · · σ(an )). This observation generalizes not just to cycles, but to products of cycles. Consequently, the equivalence class by cycle types in A5 corresponds to conjugacy classes of A5 but conjugated by elements in S5 . When considering conjugacy classes entirely inside A5 , there are fewer permutations σ when considering the conjugacy class of some element in A5 . Consequently, each conjugacy class in A5 of a given element will have the same cycle type of that element but in A5 the conjugacy class of τ might no be the whole set of permutations of same cycle type of τ . For example, consider the conjugacy class of (1 2 3 4 5). If σ is an odd permutation that fixes 1, then σ(1 2 3 4 5)σ −1 = (1 σ(2) σ(3) σ(4) σ(5)). Furthermore, if τ is any permutation such that τ (1 2 3 4 5)τ −1 = (1 σ(2) σ(3) σ(4) σ(5)), then τ = ρσ, where ρ is an appropriate 5-cycle. In any case, ρσ is an odd permutation. In particular, in A5 , there are two distinct conjugacy classes of 5-cycles in A5 each of 12 = 24/2 elements consisting of either the odd or the even permutations σ that fix 1. In particular, the conjugacy classes form a partition of A5 that is a strict refinement of the partition of cycle types. b) We already determined two of the conjugacy classes, each containing 5-cycles and containing 12 elements apiece. There is also the conjugacy class consisting of just the identity. For the 3-cycles: Suppose that we know the values σ(1), σ(2) and σ(3) of a permutation σ ∈ S5 . There are 2 remaining options for σ(4) and then 1 option for σ(5). Let σ1 and σ2 be these two remaining options. Then σ1−1 σ2 fixes 1, 2, and 3 but transposes 4 and 5. Thus, σ2 = σ1 (4 5) and hence one of the permutations is even and the other is odd. In particular, there exists an even permutation σ such that σ(1 2 3)σ −1 = (a b c) for any 3-cycle (a b c). Hence, the 3-cycles form a conjugacy class in A5 . Finally, consider the permutations consisting of two 2-cycles. Let a, b, c, d be distinct elements in {1, 2, 3, 4, 5}. We prove that (a b)(c d) is in the conjugacy class of (1 2)(3 4) in A5 , thereby showing that the set of (2, 2)cycles is a conjugacy class. Let σ be a permutation such that σ(1) = a, σ(2) = b, σ(3) = c, and σ(4) = d. Then σ(1 2)(3 4)σ −1 = (a b)(c d). If σ is even then we are done. If σ is odd, then σ(1 2) is an even permutation and, (σ(1 2))(1 2)(3 4)(σ(1 2))−1 = σ(3 4)(1 2)σ −1 = (c d)(a b) = (a b)(c d). Hence, every (2, 2)-cycle is in the conjugacy class of (1 2)(3 4). c) The conjugacy classes of A5 have orders 1, 12, 12, 20, 15. Note that the sum is indeed |A5 | = 60. By Proposition 4.2.14, every normal subgroup is the union of conjugacy classes. Also every normal subgroup must contain the identity. Let’s consider options for nontrivial normal subgroups of A5 based on the number of nonidentity conjugacy classes it contains. If a normal subgroup N contains one nonidentity conjugacy classes, then the options for its order are, 13, 21, and 16, none of which divide 60. By Lagrange’s Theorem, A5 does not have a subgroup of any of these orders. If a normal subgroup N contains two nonidentity conjugacy classes, then the options for its order are, 25, 33, 28, and 36, none of which divide 60. By Lagrange’s Theorem, A5 does not have a subgroup of any of these orders. If a normal subgroup N contains three nonidentity conjugacy classes, then all the options for orders are greater than 30 and less than 60 so do not divide 60. Finally, if we take all four of the nonidentity conjugacy classes, we obtain all of A5 . Hence, the only normal subgroups of A5 are {1} and itself. Thus A5 is a simple group.

4.3 – Quotient Groups Exercise: 1 Section 4.3 Question: Prove that Sn /An ∼ = Z2 .

192

CHAPTER 4. QUOTIENT GROUPS

Solution: We know that |An | = |Sn |/2 since An is the kernel of the homomorphism of the sign of a permutation which can only be either 1 or −1. Since |Sn : An | = 2, we have that |Sn /An | = 2. Since this group has a prime order, we know that it is cyclic. Therefore, by Proposition 3.7.25, Sn /An ∼ = Z2 .

Exercise: 2 Section 4.3 Question: Prove that hr2 i E D4 , list the elements in D4 /hr2 i, and show that D4 /hr2 i ∼ = Z2 ⊕ Z2 . Solution: Let H = hr2 i = {1, r2 , r4 , r6 }. Since n is even, we will use the results of Exercise 4.2.23 and show that H is the union of the conjugacy classes {1}, {r2 , r6 }, and {r4 }. By Proposition 4.2.14, H is a normal subgroup of D4 . The elements in D4 /H are {1̄, r̄, s̄, sr}. ¯ We will set up the isomorphism ϕ : D4 /H → Z2 ⊕ Z2 by ϕ(r̄) = (x, 1) and ϕ(s̄) = (1, y). The order of each element is preserved and we show it is a homomorphism by testing the generators of our quotient group (r̄ and s̄), ϕ(r̄)ϕ(s̄) = (x, 1)(1, y) = (x, y) = ϕ(rs). ¯ By observation, it is clearly injective and surjective and so is an isomorphism.

Exercise: 3 Section 4.3 Question: Consider the multiplicative group R× and the subgroup {1, −1}. Prove that R× /{1, −1} ∼ = R>0 , where the latter set is given the group structure of multiplication. Solution: Let H = {1, −1} be a subgroup which is certainly normal since R× is abelian. Consider the quotient group K = R× /H. We will define the isomorphism as ϕ : K → R>0 by ϕ(n̄) = |n|. This is certainly surjective and is also injective since when ϕ(n̄) = ϕ(k̄) ⇒ n = k or n = −k but a property of our cosets is that −nH = nH. Now ϕ(nm) = |nm| = |n||m| = ϕ(n̄)ϕ(m̄) for all n, m ∈ R× so that ϕ is an isomorphism and we have shown that R× /{1, −1} ∼ = R>0 .

Exercise: 4 Section 4.3 Question: Consider the group G = U (33) and consider the subgroup H = h4i. Since G is abelian, H E G. List the elements of G/H and determine its isomorphism type (i.e., find to which group in the table of Section A.2.1. it is isomorphic.) Solution: The group G = U (33) has order 20 H = {1, 4, 16, 31, 25}. The quotient group will have four elements. They are G/H = {H, 2H, 5H, 7H}. We look at the operation in G/H. We have 2H · 2H = H

5H · 5H = 25H

7H · 7H = 16H = H.

Thus all the non-identity elements in G/H have order 2. Hence G/H ∼ = Z2 ⊕ Z2 .

Exercise: 5 Section 4.3 Question: Explicitly list the elements in the quotient group U (33)/h10i. Show that this quotient group is cyclic and find an explicit generator. Solution: Let H = h10i = {1, 10}. Now |U (33)| = (3 − 1)(11 − 1) = 20, so we should have 10 different cosets. ¯ 17, ¯ 25, ¯ 29, ¯ 31, ¯ 32} ¯ = h2̄i. Recognizing that ā now refers to the left coset aH we have U (33)/h10i = {1̄, 2̄, 4̄, 8̄, 16,

Exercise: 6 Section 4.3 Question: Consider the quotient group construction described in Example 4.3.7. Fill out the Cayley table of Q8 and shade boxes of this Cayley table with 4 different colors to mimic the visual illustration of the quotient group process as shown in Example 4.3.6. Solution: The desired Cayley table and its reduction in a quotient group is:

4.3. QUOTIENT GROUPS

193 Q8

1

−1

i

−i

j

−j

k

−k

1

−1

i

−i

j

−j

k

−k

−1 −1

1

−i

i

−j

j

−k

k

i

−i −1

1

k

−k −j

j

−i −i

i

−1 −k

k

j

j

−j −k

k

−1

1

−j −j

j

k

−k

1

k

k

−k

j

−j −i

−k −k

k

−j

j

1

i

j

1

i

Q8 /h−1i 1̄

ī

j̄

k̄

1̄

1̄

ī

j̄

k̄

−j

ī

ī

1̄

k̄

j̄

i

−i

j̄

j̄

k̄

1̄

ī

−1 −i

i

i

−1

1

k̄

k̄

j̄

ī

1̄

−i

1

−1

Exercise: 7 Section 4.3 Question: Consider the function π : R2 → R given by π(x, y) = 2x + 3y. a) Show that π is a homomorphism from (R2 , +) to (R, +) and describe the fibers geometrically. b) Determine Ker π. c) Interpret geometrically the addition operation in R2 / Ker π. Solution: We consider the function π : R2 → R given by π(x, y) = 2x + 3y. a) Let (x1 , y1 ) and (x2 , y2 ) be two pairs in R2 . Then ϕ((x1 , y1 ) + (x2 , y2 )) = ϕ(x1 + x2 , y1 + y2 ) = 2(x1 + x2 ) + 3(y1 + y2 ) = 2x1 + 2x2 + 3y1 + 3y2 ϕ(x1 , y1 ) + ϕ(x2 , y2 ) = 2x1 + 3y1 + 2x2 + 3y2 . Since these two are equal, then ϕ is a homomorphism. b) The kernel ker ϕ is the set of all pairs in R2 such that 2x + 3y = 0. This is a line in R2 . c) The cosets (x0 , y0 ) Ker ϕ correspond to the fibers of ϕ. The fibers of ϕ are ϕ−1 (c) for all c ∈ R. Points in ϕ−1 (c) are pairs on the lines 2x + 3y = c. Addition in R2 / Ker ϕ corresponds to adding lines, as if adding along the axis with origin 0 and direction (2, 3). Exercise: 8 Section 4.3 Question: Let N be a normal subgroup of G. Prove that for all g ∈ G and all k ∈ Z, (gN )k = (g k )N in G/N . Solution: Since we are in a quotient group, this is just an application of Corollary 4.3.3. If k is positive or zero then, (gN )k = gN gN . . . gN (k times) = (g k )N . If k is negative than, (gN )k = ((gN )−1 )|k| = (g −1 N )|k| = (g −1 )|k| N = g −|k| N = g k N . Exercise: 9 Section 4.3 Question: Let N be a normal subgroup of G. Prove that for all g ∈ G and all k ∈ Z, (gN )k = (g k )N in G/N . Solution: Since we are in a quotient group, this is just an application of Corollary 4.3.3. If k is positive or zero then, (gN )k = gN gN . . . gN (k times) = (g k )N . If k is negative than, (gN )k = ((gN )−1 )|k| = (g −1 N )|k| = (g −1 )|k| N = g −|k| N = g k N . Exercise: 10 Section 4.3 Question: Consider the group (R, +) and the quotient group R/Z. Recall that the torsion subgroup of an abelian group is the subgroup of elements of finite order. (See Exercise 3.5.18) a) Prove that Tor(R/Z) = Q/Z.

194

CHAPTER 4. QUOTIENT GROUPS

b) Prove that R/Z is isomorphic to the circle group, defined as cos θ − sin θ ∈ GL2 (R) θ ∈ R . sin θ cos θ Solution: a) Let r̄ ∈ R/Z. If r̄ has a finite order, this implies that, for some positive n ∈ Z, r̄ + r̄ · · · + r̄(n times) = (r + r . . . + r(n times)) = nr = 0̄, which can only happen when nr ∈ Z. So for some k ∈ Z we have nr = k ⇒ r = nk ⇒ r ∈ Q and r̄ ∈ Q/Z. Which shows that Tor(R/Z) ⊆ Q/Z. For any q/r ∈ Q/Z, we know that q/r + q/r + . . . + q/r(r times) = qr/r = q̄ and, since q ∈ Z, q̄ = 0̄ so that q/r has a finite order. This shows that Q/Z ⊆ Tor(R/Z) which implies that Q/Z = Tor(R/Z). b) Let C represent the circle group. We define the homomorphism ϕ : R/Z → C by ϕ(r̄) =

cos 2rπ sin 2rπ

− sin 2rπ . cos 2rπ

First we show that ϕ is well defined in terms of the equivalence relation on our quotient group. Suppose that for some r, s ∈ R we have r̄ = s̄. This implies that for some integer k ∈ Z, r + k = s. Consider cos 2(r + k)π − sin 2(r + k)π ϕ(s̄) = ϕ(r + k) = sin 2(r + k)π cos 2(r + k)π cos 2rπ + 2kπ − sin 2rπ + 2kπ = sin 2rπ + 2kπ cos 2rπ + 2kπ Since cos and sin are periodic by any integer multiple of 2π this is equivalent to cos 2rπ + 2kπ − sin 2rπ + 2kπ cos 2rπ − sin 2rπ = sin 2rπ + 2kπ cos 2rπ + 2kπ sin 2rπ cos 2rπ = ϕ(r̄). So ϕ is well defined with respect to the equivalence relation on our quotient group. To show that this is a homomorphism consider any r̄, q̄ ∈ R/Z. Then cos 2rπ − sin 2rπ cos 2qπ − sin 2qπ ϕ(r̄)ϕ(q̄) = sin 2rπ cos 2rπ sin 2qπ cos 2qπ cos 2rπ cos 2qπ − sin 2rπ sin 2qπ − cos 2rπ sin 2qπ − cos 2qπ sin 2rπ = cos 2qπ sin 2rπ + cos 2rπ sin 2qπ cos 2rπ cos 2qπ − sin 2rπ sin 2qπ By double-angle formulas this is cos 2rπ + 2qπ − sin 2rπ + 2qπ cos 2(r + q)π = sin 2rπ + 2qπ cos 2rπ + 2qπ sin 2(r + q)π

− sin 2(r + q)π cos 2(r + q)π

= ϕ(r + q).

Now we just need to check that order is preserved by ϕ. Let q̄ ∈ R/Z be an element of finite order. From part a we know that for some n ∈ Z, nq ∈ Z. So n cos 2qπ − sin 2qπ ϕ(q̄)n = sin 2qπ cos 2qπ cos 2nqπ − sin 2nqπ = sin 2nqπ cos 2nqπ 1 0 = 0 1 = ϕ(0̄). This shows that ϕ is a homomorphism. Now we will show that ϕ is a bijection by using Proposition 3.7.15. First we will show that the kernel of ϕ is the identity coset, or 0 + Z. Suppose that 1 0 ϕ(r̄) = , 0 1

4.3. QUOTIENT GROUPS

195

Then we must have that cos(2rπ) = 1 which implies that r ∈ Z and therefore r̄ = 0̄. So that ϕ is injective. For surjectivity, consider any θ ∈ R we will show that there is an r̄ ∈ R/Z where ϕ(r̄) =

cos θ sin θ

−sinθ . cos θ

θ This comes down to finding a r̄ where 2rπ = θ. Let r = 2π , then 2rπ = θ. So ϕ is surjective and injective. This shows that ϕ is an isomorphism.

Exercise: 11 Section 4.3 Question: Prove that if G is generated by a subset {g1 , g2 , . . . , gn }, then the quotient group G/N is generated by {ḡ1 , ḡ2 , . . . , ḡn }. Solution: Let k̄ ∈ G/N be any element. Now, since k ∈ G we know that k = g1α1 g2α2 · · · gnαn for some product of powers of generators. Using the canonical projection (which is a homomorphism) we see that k̄ = π(k) = π(g1α1 g2α2 · · · gnαn ) = π(g1α1 )π(g2α2 ) · · · π(gnαn ) = π(g1 )α1 π(g2 )α2 · · · π(gn )αn = ḡ1α1 ḡ2α2 · · · ḡnαn . This shows that the quotient group G/N is generated by {ḡ1 , ḡ2 , . . . , ḡn }. Exercise: 12 Section 4.3 Question: Consider the dihedral group Dn and let d be a divisor of n. a) Prove that hrd i E Dn . b) Show that Dn /hrd i ∼ = Dd . c) Give a geometric interpretation of this last result. (What information is conflated when taking the quotient group?) Solution: Let N = hrd i. a) We will check the generators of Dn with the generators of N . r(rd )r−1 = rd ∈ N and s(rd )s = ssr−d = r−d ∈ N . So N E Dn . b) To avoid confusion, we will let Dd = {x, y xd = y 2 = 1 and xy = yx−1 }. From Exercise 4.3.11, we know that {r̄, s̄} will be a generating set for Dn /N . So |r̄| = d, |s̄| = 2, and rs = r̄s̄ = s̄r̄−1 = s̄r̄d−1 = srd−1 . The orders of the elements and the relations are equivalent so that Dn /N ∼ = Dd from the isomorphism defined by ϕ : Dn /N → Dd where ϕ(r̄) = x and ϕ(s̄) = y. c) It is important to note the difference between the geometric interpretation of the quotient Dn /N and the group Dd . Although they are isomorphic, they have different meanings. The geometric interpretation of the group Dd is the standard collection of rotations and symmetries of a regular d-gon in the plane. The geometric interpretation of the quotient group Dn /N results from the kernel of the quotient, N. When we quotient by hrd i, we are creating a quotient group that does not distinguish between any two elements that differ by operating by an element of N . This is equivalent to viewing every dth point as indistinguishable from one another. So if we rotate the n-gon d times, we view it as returning to it’s original orientation. If we reflect it around one axis, and then reflect it around another axis that is 2dπ/n radians away from that axis, we view it as returning to it’s original orientation.

Exercise: 13 Section 4.3 Question: Let G be a group and let ∼ be an equivalence relation on G that behaves well with respect to the operation. Prove that (G/ ∼, ·) is a group. Solution: We start by showing that · is a binary operation. Let [x], [y] ∈ G/ ∼, then [x] · [y] = [xy], since xy ∈ G, we know that [xy] ∈ G/ ∼ since any equivalence relation must partition G so that xy must be in some equivalence class. So · is a binary operation.

196

CHAPTER 4. QUOTIENT GROUPS

Associativity Consider [x] · ([y] · [z]) = [x] · [yz] = [x(yz)] = [(xy)z] = [xy] · [z] = ([x] · [y]) · [z]. This shows that · is associative on G/ ∼. Identity Since 1 ∈ G acts as an identity for all group elements, and ∼ behaves well with respect to the operation, we know that the equivalence class [1] will act as the identity since [x] · [1] = [x1] = [x] = [1x] = [1] · [x]. Inverses For any [x] ∈ G/ ∼, we know that [x−1 ] ∈ G/ ∼ since x−1 ∈ G, so that [x] · [x−1 ] = [xx−1 ] = [1] = [x−1 x] = [x−1 ] · [x]. So G/ ∼ has inverses. So the pair (G/ ∼, ·) is a group. Exercise: 14 Section 4.3 Question: Let N be a normal subgroup of a group G and write g for the coset gN in the quotient group G/N . a) Show that for all g ∈ G, if the order of g is finite, then |g| is the least positive integer k such that g k ∈ N . b) Deduce that the element order |g| is equal to |hgiN |/|N |. Solution: Let N be a normal subgroup of a group G. a) If the order of g is finite in G/N , then the order of g = gN is the first positive integer such that (gN )k = g k N = N . Hence the order is the first positive integer such that g k ∈ N . b) If k is the first positive integer such that g k ∈ N , then hgi ∩ N = hg k i. Set |g| = n. We have k|n. From the product set hgiN we have |hgiN | =

|hgi| |N | |hgiN | |g| n ⇐⇒ = = = gcd(n, k) = k. |hgi ∩ N | |N | |hgi ∩ N | n/ gcd(n, k)

Exercise: 15 Section 4.3 Question: Consider the group called G which is given in generators and relations as G = hx, y | x4 = y 3 = 1, x−1 yx = y −1 i. a) Prove that G is a nonabelian group of order 12. b) Prove that hyi is a normal subgroup and that G/hyi ∼ = Z4 . c) Prove that G is not isomorphic to A4 or D6 . Solution: Let G = hx, y | x4 = y 3 = 1, x−1 yx = y −1 i. a) From the identity x−1 yx = y 2 , we have yx = xy 2 . Consequently, in any word involving the generators, we can move the x element to the left, possibly changing the powers on y as needed. Hence, every element in G can be written as xa y b with 0 ≤ a ≤ 3 and 0 ≤ b ≤ 2. Now suppose that xa y b = xc y d . Then xa−c = y d−b . The powers of x have orders 1, 2, or 4 and the powers of y have order 1 or 3. Hence, the only time xn = y m is when n is a multiple of 4 and m is a multiple of 3. Hence xa y b = xc y d if and only a ≡ c (mod 4) and b ≡ d (mod 3). Hence xa y b with 0 ≤ a ≤ 3 and 0 ≤ b ≤ 2 are all distinct and thus G has 12 elements. b) Obviously yyy −1 = y ∈ hyi. Also the identity x−1 yx = y 2 leads to x−2 yx2 = x−1 y 2 x = x−1 yxx−1 yx = y 4 = y. Also xyx−1 = x−3 yx3 = x−1 yx = x−1 yx = y 2 ∈ hyi. Thus, since both of the generators of G normalize hyi, we have hyi E G. Furthermore, the quotient group G/hyi is generated by xhyi so G/hyi ∼ = Z4 .

4.3. QUOTIENT GROUPS

197

c) A4 does not have a normal subgroup of order 3 so A4 is not isomorphic to G. The group D6 has not elements of order 4, whereas G does not. Hence G is not isomorphic to D6 . Exercise: 16 Section 4.3 Question: Consider the group G given in generators and relations as G = hx, y | x4 = y 5 = 1, x−1 yx = y 2 i. a) Prove that G is a nonabelian group of order 20. b) Prove that hyi is a normal subgroup and that G/hyi ∼ = Z4 . c) Prove that G is not isomorphic to D10 . [Note: This group is called the Frobenius group of order 20 and is denoted by F20 .] Solution: a) First we will prove that G is nonabelian. Assume that xy = yx. By operating on the left by x−1 we get x−1 xy = x−1 yx ⇒ y = x−1 yx. By the relation provided in the group presentation we know that x−1 yx = y 2 . However this implies that y = y 2 ⇒ y = 1. This contradicts the fact that the order of y in G is 5. So G must be nonabelian. Now we will prove that |G| = 20. The relation in our presentation x−1 yx = y 2 can be rewritten as yx = xy 2 . So we can move any powers of y to the right of x. This implies that every element can be written as xa y b where 0 ≤ a ≤ 3 and 0 ≤ b ≤ 4. So there are no more than 4 · 5 = 20 distinct elements of G. Assume that there were less than 20 distinct elements. Then for some 0 ≤ a, c ≤ 3 and 0 ≤ b, d ≤ 4 we have xa y b = xc y d where either a and c or b and d are distinct. Then we can rewrite this as xa−c = y d−b . Since all powers of y and x are distinct except when they are the identity, we must have a − c = 0̄ mod 4 and d − b = 0̄ mod 5. Given the ranges of a, c and b, d, this implies that a = c and b = d. Which contradicts the fact that at least one of the pairs of powers must be distinct. So we must have exactly 20 elements in G. b) Let N = hyi. We will show N is a normal subgroup by conjugating by the generators of our group. Consider y(y)y −1 = y ∈ N and x(y)x−1 = x−3 yx3 = x−1 (x−1 (x−1 yx)x)x = x−1 (x−1 (y 2 )x)x = x−1 (x−1 yx)(x−1 yx)x = x−1 (y 2 y 2 )x = x−1 y −1 x which is the inverse of the left side of the relation presented, and so it must equal the inverse of the right side, (y 2 )−1 = y −2 = y 3 ∈ N . So N is a normal subgroup. Consider the quotient group G/N . By Exercise 4.3.11, we know that the set {x̄, ȳ} must generate G/N . However ȳ = 1̄ and x̄4 = 1̄ so that we only need the generator x̄. So G/N is cyclic. Furthermore |G/N | = |G|/|N | = 20/5 = 4. Since same sized cyclic groups are isomorphic, we have G/N ∼ = Z4 . c) In G, we only have an element of order 4, namely x. In D10 , we only have elements of order 2, 5, and 10. So they cannot be isomorphic since order must be preserved under isomorphism. Exercise: 17 Section 4.3 Question: Let G be the group defined by G = hx, y | x2 = y 8 = 1, yx = xy 5 i. a) Show that hy 4 i E G. b) Show that G/hy 4 i is a group of order 8. c) Find the isomorphism type of G/hy 4 i (i.e., find to which group in the table of Section A.2.1 it is isomorphic.) Explain. [Note: This group is called the modular group of order 16.] Solution:

198

CHAPTER 4. QUOTIENT GROUPS

a) Let N = hy 4 i. We will show this subgroup is normal by conjugating by the generators of G. So yy 4 y −1 = y 4 ∈ N and xy 4 x = (xyx)(xyx)(xyx)(xyx) = y 5 y 5 y 5 y 5 = y 20 = y 4 ∈ N . So N is a normal subgroup. b) Consider the quotient group G/N . We will find the order of G first. Because of the relation presented, we know that any element can be written as xi y j so that |G| ≤ 16. Let i be 0 or 1, and 0 ≤ j ≤ 7. Suppose that xi y j = xa y b ⇒ xi−a = y b−j , which only occurs when i − a ≡ 0 mod 2 and b − j ≡ 0mod8 which only happens within the ranges provided when i = a and b = j. So for any two distinct pairs of numbers in the ranges provided, we get a different element. In total we get 8 · 2 = 16 different elements. Now |N | = 2 which implies that |G/N | = |G : N | = 16/2 = 8. c) From our original relation we get that ȳx̄ = yx = xy 5 = x̄ȳ 5 = x̄ȳ and so our group is abelian. Since it has an element of order 4 (ȳ) and is not cyclic, it must be Z4 × Z2 . Exercise: 18 Section 4.3 Question: Consider the group SL2 (F3 ), i.e. the special linear group of 2 × 2 matrices with determinant 1, with entries in F3 , modular arithmetic modulo 3. a) Show that this group is non abelian with 24 elements. b) Prove that the center of SL2 (F3 ) is the subgroup of two elements 1 0 2 0 Z(SL2 (F3 )) = , . 0 1 0 2 c) The projective special linear group PSL2 (F3 ) = SL2 (F3 )/Z(SL2 (F3 )) has order 12. Determine, with proof, to which group in the table of Section A.2.1 it is isomorphic. Solution: a) We will first find | GL2 (F3 )|. The only restriction on the first column is that it cannot be all zeroes, the next column cannot be a multiple of the first, so we get | GL2 (F3 )| = (32 − 1)(32 − 3) = 48. Now SL2 (F3 ) is the kernel of the homomorphism det : GL2 (F3 ) → (F3 , ×). Since |(F3 , ×)| = 2, we must have | SL2 (F3 )| = | GL2 (F3 )|/2 = 48/2 = 24. An examination of the product of two matrices of SL2 (F3 ) in different orders shows that 1 2 2 1 1 0 = 0 1 1 1 1 1 but 2 1

1 1

1 0

2 1

=

2 1

2 . 0

So SL2 ((F )3 ) is a nonabelian group of order 24. b) Let A, C ∈ SL2 (F3 ), we will let A be an arbitrary matrix and solve for C in the equation CA − AC = 0. a b e f e f a b ae + bg af + bh ae + f c be + f d − = − c d g h g h c d ce + dg cf + dh ga + ch bg + dh bg − f c f (a − d) + b(h − e) = c(e − h) + g(d − a) cf − bg 0 0 = 0 0 Now, we have the equation cf − bg = 0 ⇒ cf = bg. Since this must hold for all possible f and g, we must have c = b = 0. Plugging these values back into our equation we get, 0 f (a − d) + 0(h − e) 0 f (a − d) = 0(e − h) + g(d − a) 0 g(d − a) 0 0 0 = 0 0

4.3. QUOTIENT GROUPS

199

Now f (a − d) = g(d − a) = 0 for all possible values of f and g, so we have that a = d. So Z(SL2 (F3 )) contains only matrices of the form x 0 0 x where x ∈ F3 and x2 ≡ 1 (mod 3). So we get Z(SL2 (F3 )) =

1 0

0 2 , 1 0

0 . 2

c) We note that in PSL2 (F3 ), matrices are considered equal if they differ by a multiplicative constant. First we show that PSL2 (F3 ) is nonabelian. Consider the two products: 1 1 0 1 2 2 1 1 = = 0 1 2 1 2 1 1 2 0 1 1 1 0 1 = . 2 1 0 1 2 0 Since they do not equal one another and both elements are members of PSL2 (F3 ), we find that the group is nonabelian. According to the table of groups of small order, the only nonabelian groups of order 12 are D6 , A4 , and another group denoted by Z3 o Z4 , which has a presentation of hx, y | x3 = y 4 = 1, yx = x2 yi. The elements of PSL2 (F3 ) are 1 0 2 1 1 1 0 1 1 1 2 0 , , , , , 0 1 1 1 1 2 2 0 2 0 1 1 1 1 1 0 1 2 1 0 0 1 1 2 , , , , , 0 1 1 1 0 1 2 1 2 1 1 0 These elements have orders 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, and 3. The group D6 has an element of order 3 whereas PSL2 (F3 ) does not. The group Z3 o Z4 has an element of order 4, whereas PSL2 (F3 ) does not. Hence, we conclude that PSL2 (F3 ) ∼ = A4 . Exercise: 20 Section 4.3 Question: Find an example of a group G in which the center of G/Z(G) is not trivial. Solution: Let G = D8 . Then Z(G) = {1, r4 } = hr4 i and from Exercise 4.3.12, G/Z(G) ∼ = D4 ⇒ Z(G/Z(G)) ∼ = 2 Z(D4 ), and Z(D4 ) = {1, r }. So the center of G/Z(G) is not trivial. Exercise: 21 Section 4.3 Question: Prove that if G/Z(G) is cyclic, then G is abelian. Give an example to show that G is not necessarily abelian if we only assume that G/Z(G) is abelian. Solution: Let G/Z(G) = hḡi. Then every element in G can be written in the form of g a zi where zi ∈ Z(G). Consider the product of two arbitrary elements, g a zi g b zj . Since zi and zj are in the center of G. We have (g a zi )(g b zj ) = zj g a g b zi = zj g a+b zi = zj g b g a zi = (g b zj )(g a zi ). Which shows that G is abelian. For a counter example to only assuming that G/Z(G) is abelain, consider D4 /Z(D4 ) = D4 /hr2 i ∼ = D2 . We know that D2 is isomorphic to Z2 ⊕ Z2 which is abelian. However D4 is non abelian. Exercise: 22 Section 4.3 Question: Prove that if |G| = pq, where p and q are two primes, not necessarily distinct, then G is either abelian or Z(G) = {1}. [Hint: Use Exercise 4.3.21]

200

CHAPTER 4. QUOTIENT GROUPS

Solution: Since Z(G) is a subgroup of G, by Lagrange’s theorem it must be of order pq, p, q, or 1. If |Z(G)| = pq, then G is abelian. If |Z(G)| = p, then |G/Z(G)| = |G|/|Z(G)| = pq/p = q. Since G/Z(G) has a prime order, it must be cyclic. By Exercise 4.3.21, this implies that G is abelian. By the exact same reasoning, if |Z(G)| = q, then G is abelian. The only remaining order for Z(G) is 1. So either G is abelian or Z(G) = {1}. Exercise: 23 Section 4.3 Question: Let N be a normal subgroup of a finite group G and let g ∈ G. Prove that if gcd(|g|, |G/N |) = 1, then g ∈ N . Solution: Consider the canonical projection of g into G/N . Since π is a homomorphism, we have that |π(g)| = |ḡ| which must divide |g|. Which implies that gcd(|ḡ|, |G/N |) must divide gcd(|g|, |G/N |) ⇒ gcd(|ḡ|, |G/N |) = 1. Since |ḡ| must divide |G/N |, we have |ḡ| = 1. This shows that ḡ = 1̄ ⇒ g ∈ N . Exercise: 24 Section 4.3 Question: Let G be a group. The commutator subgroup, denoted G0 , is defined as the subgroup generated by all products x−1 y −1 xy, for any x, y ∈ G. In other words, G0 = hx−1 y −1 xy | x, y ∈ Gi. a) Prove that G0 E G. b) Prove (without using part a) that G0 is a characteristic subgroup of G. (See Exercise 4.2.20) c) Prove that G/G0 is abelian. Solution: a) Let g ∈ G and x−1 y −1 xy ∈ G0 be any one of the generators of the subgroup and consider the conjugation g(x−1 y −1 xy)g −1 = (gx−1 y −1 )(xyg −1 ) = (gx−1 y −1 g)(g −1 xyg −1 ). Letting h = g −1 x and k = yg −1 we see that this is equal to h−1 k −1 hk ∈ G0 . By Theorem 4.2.8, we have G0 E G. b) Recall that a characteristic subgroup is a subgroup H that is fixed by any automorphism ψ (not just the inner automorphisms which are conjugation). Let ψ be any automorphism and x−1 y −1 xy ∈ G0 be any generator of G0 . Consider ψ(x−1 y −1 xy) = ψ(x−1 )ψ(y −1 )ψ(x)ψ(y) = ψ(x)−1 ψ(y)−1 ψ(x)ψ(y) = g −1 h−1 gh ∈ G0 . Since any generator will end up back in G0 and ψ is a homomorphism, this shows that ψ(G0 ) ⊆ G0 . Now, since ψ is an isomorphism, we know that in the finite (order of G0 is finite) case, ψ(G0 ) = G0 . In the case when |G0 | is infinite, we know that ψ(G0 ) ⊆ G0 . Now operating by ψ −1 gets us, (G0 ) ⊆ ψ −1 (G0 ). c) Consider the quotient G/G0 . For any two element x, y ∈ G, the product x−1 y −1 xy ∈ G0 ⇒ x−1 y −1 xy = 1̄. Splitting up the elements we get x̄−1 ȳ −1 x̄ȳ = 1̄. By moving the two inverses to the other side of the equation we get x̄ȳ = ȳx̄, which shows that G/G0 is abelian. Exercise: 25 Section 4.3 Question: Let A and B be two groups and let G = A ⊕ B. The subgroup A × {1} ≤ G is isomorphic to A. Prove that A × {1} E G and that G/(A × {1}) ∼ = B. Solution: Let (a, b) ∈ G and let (x, 1) ∈ A × {1}. Then (a, b)(x, 1)(a, b)−1 = (axa−1 , bb−1 ) = (axa−1 , 1) ∈ A × {1}. Hence A × 1 E G. Consider the function ψ : G/A × {1} → B defined by ψ (a, b)(A × {1}) = b. This is a well-defined function since (a, b)(A × {1}) = (a0 , b0 )(A × {1}) if and only if b = b0 . The only if condition implies that the function is also injective. Furthermore, it is surjective and also a homomorphism. Hence ψ gives an isomorhism between G/A × {1} and B. Exercise: 26 Section 4.3 Question: In Example 4.3.13, prove that the subgroups h6i, h8i, and h16i satisfy the conditions of the Direct Sum Decomposition Theorem. Solution: Let G = U (35) where |G| = 24. First we find the elements of each subgroup. N1 = h6i = {6̄, 1̄}, ¯ 22, ¯ 1̄}, and N3 = h16i = {16, ¯ 11, ¯ 1̄}. Now, we see that N1 ∩ N2 = {1}, so we will find the N2 = h8i = {8̄, 29, ¯ ¯ ¯ ¯ ¯ subgroup N1 N2 = {1̄, 8̄, 22, 29, 13, 34, 27, 6̄}. Now we see that N1 N2 ∩ N3 = {1}. So consider the subgroup

4.4. ISOMORPHISM THEOREMS

201

N1 N2 N3 , by Proposition 4.1.16, |N1 N2 N3 | = |N1 N2 ||N3 | = 8 · 3 = 24. This implies that N1 N2 N3 = G. So both conditions of Theorem 4.3.12 are satisfied and we know that the G ∼ = N1 ⊕ N2 ⊕ N3 . Exercise: 27 Section 4.3 Question: Use the direct sum decomposition to show that U (100) ∼ = Z20 ⊕ Z2 ∼ = Z5 ⊕ Z4 ⊕ Z2 . 2 2 Solution: Let G = U (100). Then |G| = (5 − 5)(2 − 2) = 40. Consider the subgroups N1 = h7̄i = ¯ 43, ¯ 1̄}, N2 = h41i ¯ = {41, ¯ 81, ¯ 21, ¯ 61, ¯ 1̄} and N3 = {1̄, 99}. ¯ {7̄, 49, It’s clear that N1 ∩ N2 = {1̄}, so we calculate ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ N1 N2 = {1̄, 41, 81, 21, 61, 7̄, 47, 87, 27, 67, 29, 9̄, 89, 69, 3̄, 63, 23, 83}. Now, we can see that N1 N2 ∩ N3 = {1}. By Proposition 4.1.16, |N1 N2 N3 | = 5 · 4 · 2 = 40 so N1 N2 N3 = G. So both conditions of the Direct Sum Decomposition Theorem are satisfied and we see that G ∼ = N1 ⊕ N2 ⊕ N3 . Now, since cyclic groups of a given order are isomorphic, we also know that G ∼ = Z4 ⊕ Z5 ⊕ Z2 . Looking back at our subgroup N1 N2 we see that it is also cyclic and generated by 87. So let H = h87i. Then we know that H ∩ N3 = {1̄} and |HN3 | = 20 ∗ 2 = 40 which implies that HN3 = G. So by the Direct Sum Decomposition Theorem, G ∼ = H ⊕ N3 . Since all cyclic groups of a given order are isomorphic, we have G ∼ = Z20 ⊕ Z2 . Exercise: 28 Section 4.3 Question: Let p be a prime number and let k be a positive integer. Prove that Zpk is not isomorphic to the direct product of any other groups. Solution: Let G = U (35) where |G| = 24. First we find the elements of each subgroup. N1 = h6i = {6̄, 1̄}, ¯ 22, ¯ 1̄}, and N3 = h16i = {16, ¯ 11, ¯ 1̄}. Now, we see that N1 ∩ N2 = {1}, so we will find the N2 = h8i = {8̄, 29, ¯ 29, ¯ 13, ¯ 34, ¯ 27, ¯ 6̄}. Now we see that N1 N2 ∩ N3 = {1}. So consider the subgroup subgroup N1 N2 = {1̄, 8̄, 22, N1 N2 N3 , by Proposition 4.1.16, |N1 N2 N3 | = |N1 N2 ||N3 | = 8 · 3 = 24. This implies that N1 N2 N3 = G. So both conditions of Theorem 4.3.12 are satisfied and we know that the G ∼ = N1 ⊕ N2 ⊕ N3 . Exercise: 29 Section 4.3 Question: Prove that Q8 is not isomorphic to the direct product of any other groups. Solution: A group G is isomorphic to a direct product of two groups if and only if it has a nontrivial strict normal subgroup A such that G ∼ = A ⊕ G/A. The group Q8 has only one nontrivial strict normal subgroup, namely A = h−1i. Furthermore, Q8 /h−1i ∼ = Z2 ⊕ Z2 . Hence, h−1i ⊕ Q8 /h−1i ∼ = Z2 ⊕ Z2 ⊕ Z2 but in this group the largest order of any element is 2. On the other hand, the largest order of an element in Q8 is 4. Thus, there is no nontrivial strict normal subgroup A such that G ∼ = A ⊕ G/A. Thus, Q8 is not the direct product of other groups.

4.4 – Isomorphism Theorems Exercise: 01 Section 4.4 Question: Show that the only homomorphism from D7 to Z3 is trivial. Solution: Since |D7 | = 14 and |Z3 | = 3, then gcd(|D7 |, |Z3 |) = 1. By Corollary 4.4.2, the only homomorphism from D7 to Z3 is trivial. Exercise: 02 Section 4.4 Question: Show that the only homomorphism from S4 to Z3 is trivial. Solution: We know from Example 4.2.13 that all the conjugacy classes of S4 are the cycle types (4-cycles, 3-cycles, 2-cycles, 2 2-cycles, and the identity). From Proposition 4.2.14, a subgroup is normal if and only if it is the union of conjugacy classes. Now, the only subgroups of Z3 are {1} and Z3 . So in order for there to be a non-trivial homomorphism, it will have to be surjective. This implies that the order of the normal subgroup we’re looking for is |S4 |/|Z3 | = 24/3 = 8. Now, the orders of all of the conjugacy classes are: {(a b c d) − 6, (a b c) − 8, (a b) − 6, (a b)(c d) − 3, (a) − 1}. The only possible union of conjugacy classes in which the orders sum to 8 is just the 3-cycles. However, this won’t be a group since it does not include the identity. So we cannot find a normal subgroup of size 8 which implies that the only homomorphism from S4 to Z3 is trivial.

202

CHAPTER 4. QUOTIENT GROUPS

Exercise: 3 Section 4.4 Question: Let ϕ : G → H. Prove that |ϕ(G)| divides |G|. Solution: By the First Isomorphism Theorem, Ker ϕ is a normal subgroup of G and G/ Ker ϕ ∼ = Im ϕ. By Lagrange’s Theorem, |G : Ker ϕ| = |G|/| Ker ϕ|, but by the definition of a quotient group |G : Ker ϕ| = |G/ Ker ϕ| = | Im ϕ|. Further, Im ϕ = ϕ(G), so |ϕ(G)| = |G|/| Ker ϕ| or |G| = |ϕ(G)|| Ker ϕ|. Hence |ϕ(G)| divides |G| with quotient | Ker ϕ|. Exercise: 04 Section 4.4 Question: Let F be Q, R, C, or Fp . Prove that GLn (F )/ SLn (F ) ∼ = F ×. × Solution: Consider the homomorphism det : GLn (F ) → F . This homomorphism is surjective since for any k ∈ F × we can construct a matrix with 1’s down the entire diagonal except one k in the upper left hand corner and that matrix will have determinant k. Furthermore, SLn (F ) is defined as the kernel of this homomorphism. By the first isomorphism theorem, this implies that GLn (F )/ SLn (F ) ∼ = F ×. Exercise: 05 Section 4.4 Question: Let p be an odd prime and let H be any group of odd order. Prove that the only homomorphism from Dp to H is trivial. Solution: Let H be a group of any odd order. Suppose that gcd(|Dp |, |H|) = 1, then by Corollary 4.4.2, the only homomorphism is the trivial homomorphism. The only other option when |H| is odd is gcd(|Dp |, |H|) = p. Then we could have a homomorphism ϕ : Dp → H where | Im ϕ| = p. This would imply that | Ker ϕ| = 2. However the only subgroups of order 2 in Dp are of the form hsri i for some i none of which are normal by examining the conjugacy classes in Exercise 4.2.23 part b. So no such homomorphism could exist. So the only possibility is the trivial homomorphism. Exercise: 6 Section 4.4 Question: Let G be a group and consider homomorphisms ϕ : G → Z2 . a) Prove that if ϕ is surjective, then exactly half of the elements of G map to the identity in Z2 . b) Show that there does not exist a surjective homomorphism from A5 to Z2 . Solution: Let G be a group and consider homomorphisms ϕ : G → Z2 . a) Suppose that ϕ is surjective. Then ϕ(G) = Z2 so in particular |ϕ(G)| = 2. By the first isomorphism theorem, G/ ker ϕ ∼ = Z2 so [G : ker ϕ] = 2. Hence, exactly half of the elements in G are mapped to the identity. b) Assume there exists a surjective homomorphism ϕ : A5 → Z2 . Then by part (a), exactly half of the elements in A5 must map to the generator of Z2 , which has order 2. Now if |ϕ(g)| = 2, then 2 divides |g|. Recall that |A5 | = 60. The alternating group A5 contains all 5-cycles of S5 . This accounts for 24 elements. Since the 5-cycles have order 5, which is not divisible by 2, all of these must map to the identity in Z2 . Also, A3 has 53 × 2 = 10 3-cycles. These have order 3 so they must map to the identity as well. This accounts for 34 elements. Hence, more that 30 elements must map to the identity which, by part (a), contradicts the assumption that ϕ is surjective. Exercise: 7 Section 4.4 Question: Let G1 and G2 be groups and let N1 and N2 be normal subgroups respectively in G1 and G2 . Prove that N1 ⊕ N2 E G1 ⊕ G2 and that (G1 ⊕ G2 )/(N1 ⊕ N2 ) ∼ = (G1 /N1 ) ⊕ (G2 /N2 ). Solution: Consider the function f : G1 ⊕ G2 → (G1 /N1 ) ⊕ (G2 /N2 ) defined by f (g1 , g2 ) = (g1 N1 , g2 N2 ). Since the projection G → G/N is a homomorphism, so is the function f . However, the kernel of f is ker f = {(g1 , g2 ) ∈ G1 ⊕ G2 | (g1 N1 , g2 N2 ) = (N1 , N2 )} = N1 ⊕ N2 . By the first isomorphism theorem, we deduce that (G1 ⊕ G2 )/(N1 ⊕ N2 ) ∼ = (G1 /N1 ) ⊕ (G2 /N2 ).

4.4. ISOMORPHISM THEOREMS

203

Exercise: 08 Section 4.4 Question: Let N1 and N2 be normal subgroups of G. Prove that N1 N2 /(N1 ∩ N2 ) ∼ = (N1 N2 /N1 ) ⊕ (N1 N2 /N2 ). Solution: Since N1 and N2 are both normal subgroups of G, by Exercise 4.2.15, we know that their intersection N1 ∩ N2 is a normal subgroup as well. By Exercise 4.2.14, we know that N1 ∩ N2 E N1 and N1 ∩ N2 E N2 . Now, consider H1 = N1 /(N1 ∩ N2 ) and H2 = N2 /N1 ∩ N2 which are both subgroups of N1 N2 /(N1 ∩ N2 ). Assume that for some non-identity element g, g ∈ H1 and g ∈ H2 , this implies that g ∈ H1 ∩ H2 which implies that g ∈ N1 ∩ N2 . However the only element that is in N1 ∩ N2 and H1 ∩ H2 is the identity element which contradicts the fact that g is a non-identity element, so g must not exist in both H1 and H2 . This shows that H1 ∩ H2 = {1}. By Proposition 4.1.16, |H1 H2 | = |N1 /(N1 ∩ N2 )||N2 /(N1 ∩ N2 )| = |N1 |/|N1 ∩ N2 | · |N2 |/|N1 ∩ N2 | = |N1 ||N2 |/2(|N1 ∩ N2 |) = (|N1 ||N2 |/|N1 ∩ N2 |)/(|N1 ∩ N2 |) = |N1 N2 |/(|N1 ∩ N2 |) = |N1 N2 /(N1 ∩ N2 )| This implies that H1 H2 = N1 N2 /(N1 ∩N2 ). We have satisfied all the conditions of the Direct Sum Decomposition Theorem and we conclude that N1 N2 /(N1 ∩ N2 ) ∼ = H1 ⊕ H2 ∼ = N1 /(N1 ∩ N2 ) ⊕ N2 /(N1 ∩ N2 ). By the Second Isomorphism Theorem, N1 N2 /(N1 ∩ N2 ) ∼ = N1 N2 /N1 ⊕ N1 N2 /N2 .

Exercise: 9 Section 4.4 Question: Let F = Q, R, C, or Fp where p is prime. Let T2 (F ) be the set of 2 × 2 upper triangular matrices with nonzero determinant. We consider T2 (F ) as a group with the operation of multiplication. n 1 b o a) Prove that U2 (F ) = | b ∈ F is a normal subgroup that is isomorphic to (F, +). 0 1 b) Prove that the corresponding quotient group is T2 (F )/U2 (F ) ∼ = F × ⊕ F ×. Solution: a) Since it is not obvious, we first check that U2 (F ) is a subgroup using the one step subgroup criteria.

1 0

1 a 1 b , ∈ U2 (F ) 0 1 0 1 a 1 1 −b 1 −b + a = ∈ U2 (F ) 1 1 0 1 0 1

By the one step subgroup criteria, U2 (F ) is a subgroup. Now we will see if it is normal conjugating an arbitrary element of U2 (F ) with an arbitrary element in T2 (F ) and show that it ends up in U2 (F ).

a 0

b 1 c 0

e c (ac)−1 1 0

−b a

= (ac)−1

a 0

ae + b c

c 0

−b a

ca −ba + a2 e + ab = (ac)−1 0 ca 1 a2 e(ac)−1 = ∈ U2 (F ) 0 1 By Theorem 4.2.4 part 4, U2 (F ) is a normal subgroup. Now we will show that U2 (F ) is isomorphic to Fp . Consider the function ϕ : U2 (F ) → (F, +) by, ϕ

1 0

a = a. 1

204

CHAPTER 4. QUOTIENT GROUPS We first show that this is a homomorphism, 1 a 1 b 1 ϕ =ϕ 0 1 0 1 0

a+b 1

= (a + b) = a + b 1 a 1 =ϕ ϕ 0 1 0

b 1

,

which shows that ϕ is a homomorphism. Since Ker ϕ is the identity matrix, we know that ϕ is injective by Proposition 3.7.15. Since the groups have the same order, ϕ is also surjective making ϕ an isomorphism. So U2 ∼ = (F, +). b) Let π : T2 (F ) → T2 (F )/U2 (F ) be the canonical projection. Consider the set of matrices n o a 0 A= a, b ∈ F × 0 b We will show that any two matrices from A under the canonical projection are distinct. To show this, assume that a 0 c 0 π =π 0 b 0 d Well this implies

a 0

0 c b 0

−1 0 ∈ U2 (F ). d

However

a 0

0 c b 0

−1 0 a = d 0

0 −1 d 0 cd b 0 c ad 0 = cd−1 0 bc −1 cd ad 0 = 6∈ U2 (F ). 0 cd−1 bc

Which shows that all matrices of A are distinct under the canonical projection. Now we will show that for any matrix M ∈ T2 (F ) there is some N ∈ A and T ∈ U2 (F ) where M = N T . Consider the arbitrary matrix a b a 0 1 a−1 b = . 0 c 0 c 0 1 Finally, we will show that for any matrix M ∈ T2 (F ), π(M ) = π(N ) for some N ∈ A. We will utilize the fact that we can rewrite M as N · T for N ∈ A and T ∈ U2 (F ). Then, π(M ) = π(N T ) = π(N )π(T ) = π(N ) · 1 = π(N ). So we have shown that, under our homomorphism π, Im π ∼ = A or T2 (F )/U2 (F ) ∼ = A. Additionally, we have now shown that A is a subgroup. Now consider the function ϕ : A → F × ⊕ F × by a 0 ϕ = (a, b). 0 b For any two elements in A we have a 0 c ϕ 0 b 0

0 ac 0 =ϕ d 0 bd = (ac, bd) = (a, b) · (c, d) a 0 c 0 =ϕ ϕ . 0 b 0 d

4.4. ISOMORPHISM THEOREMS

205

Which shows that ϕ is a homomorphism. The kernel is clearly the identity matrix which implies that ϕ is injective and for any (c, d) ∈ F × ⊕ F × we have c 0 ϕ = (c, d) 0 d so that ϕ is also surjective. This shows ϕ to be a isomorphism and we have A ∼ = F × ⊕ F × . By transitivity × × ∼ we have T2 (F )/U2 (F ) = F ⊕ F . Exercise: 10 Section 4.4 Question: Let G be a group and let N E G such that |G| and | Aut(N )| are relatively prime. Then N ≤ Z(G). [Hint: Use Proposition 4.2.15] Solution: For any g ∈ G, consider the functions of the type ψg : N → N where ψg (n) = gng −1 . By Proposition 4.2.15, these are automorphisms of N . Additionally, the association Ψ : NG (N ) → Aut(N ) is a homomorphism where Ψ(g) = ψg . Since NG (N ) = G, the association becomes Ψ : G → Aut(N ). Now since, gcd(|G|, | Aut(N )|) = 1, we know from Corollary 4.4.2 that this homomorphism must be trivial. This implies that for all g ∈ G, Ψ(g) = ψg = ψ1 . This implies that for any g ∈ G and n ∈ N , ψg (n) = ψ1 (n) ⇒ gng −1 = 1n1 ⇒ gng −1 = n ⇒ gn = ng so that n ∈ Z(G) which implies that N ≤ Z(G). Exercise: 11 Section 4.4 Question: Suppose that H and K are distinct subgroups of G, each of index 2. Prove that H ∩ K is a normal subgroup of G and that G/(H ∩ K) ∼ = Z2 ⊕ Z2 . Solution: Since H and K both have index 2 in G, by Proposition 4.2.2 both subgroups are normal. Since both are normal, by Exercise 4.2.15 we know that H ∩ K is also a normal subgroup of G. Now we will show that the product HK is G. From Corollary 4.2.10, we know that HK is a subgroup. Since each is index 2, we deduce that |H| = |K| = |G|/2. Since they are distinct and H ∩ K is a subgroup of both, we know that |H ∩ K| = |G|/2d for some d ≥ 2. By Proposition 4.1.16, |HK| = |H||K|/|H ∩ K| = (|G|2 /4)/(|G|/2d) |G|2 2d 4 |G| |G|d . = 2 =

Now for sure d ≥ 2 which implies that |HK| ≥ |G|. Since it must be a subgroup, we must have equality. Which implies that HK = G. Now we can consider G/H ∩ K which we will rewrite as HK/(H ∩ K). By Exercise 4.4.8, we know that HK/(H ∩ K) ∼ = HK/H ⊕ HK/K. Now HK/H is equivalent to G/H. Since |G : H| = 2, this implies that |G/H| = 2. So HK/H = G/H ∼ = Z2 . Likewise, HK/K = G/K ∼ = Z2 . So we have ∼ HK/(H ∩ K) ∼ Z ⊕ Z or G/(H ∩ K) Z ⊕ Z . = 2 = 2 2 2 Exercise: 12 Section 4.4 Question: Prove all parts of the Fourth Isomorphism Theorem. Solution: Let G be a group with a normal subgroup N . We first prove that there is a bijection between the subgroups A of G which contain N and the set of subgroups of G/N . Call ϕ : G → G/N . Let A be a subgroup of G that contains N . Obviously, ϕ(A) = A/N is not empty since it contains 1̄. Not let xN and yN be in ϕ(A). Then (xN )(yN )−1 = (xy −1 )N ∈ ϕ(A) since xy −1 ∈ A. Thus ϕ(A) is a subgroup of G/N . Let H be a subgroup of G/N . It is easy to show that ϕ−1 (H) is a subgroup of G (by virtue of ϕ being a homomorphism) and that it contains N . By properties of functions, ϕ(ϕ−1 (H)) = H. Hence, the map defined by ϕ from subgroups of G containing N to subgroups of G/N is surjective. Now suppose that A and B are

206

CHAPTER 4. QUOTIENT GROUPS

subgroups of B that contain N and suppose that ϕ(A) = ϕ(B). In particular, for all a ∈ A, there exists b ∈ B such that aN = bN . But then b−1 a ∈ N . Since N ⊆ A, we deduce that b−1 a = a0 for some a0 ∈ A, so b = a(a0 )−1 and thus B ⊆ A. By an identical reasoning, A ⊆ B. Hence, ϕ(A) = ϕ(B) implies A = B. This establishes that there is a bijection between the subgroups of G/N and the subgroups of G that contain N . Part (2): In order to prove part (2), we need to set up a homomorphism between the set of left cosets of A in B and the set of left cosets of A/N in B/N . The natural function to consider is f (bA) = b̄(A/N ), where b̄ = bN . We first prove that f is well-defined as a function. Recall b1 A = b2 A if and only if b−1 2 b1 ∈ A. −1 So if b1 A = b2 A, then (b−1 (b1 N ) ∈ A/N so b1 (A/N ) = b2 (A/N ). 2 b1 )N ∈ A/N , which means that (b2 N ) Therefore, f is a well-defined function. Now, among all cosets bA, we can choose b arbitrarily in B, then f is surjective. Now suppose that f (b1 A) = f (b2 A). Then b1 (A/N ) = b2 (A/N ), which means that (b2 N )−1 (b1 N ) ∈ A/N and thus (b−1 2 b1 )N ∈ A/N . −1 −1 −1 −1 0 In particular, (b−1 b )N = aN for some a ∈ A. Thus, a (b b ) ∈ N ⊆ A. Since a (b b 1 1 1 ) = a for some 2 2 2 −1 0 a0 ∈ A, then b−1 b = aa so b b ∈ A. Therefore, b A = b A and we conclude that f is injective. This 1 1 1 2 2 2 establishes the desired bijection, which implies that, assuming the indices are finite, |A : B| = |A/N : B/N |. Part (3): Let gN ∈ hA/N, B/N i. Then gN = (g1 N )(g2 N ) · · · (gk N ) = (g1 g2 · · · gk )N, for some cosets gi N ∈ A/N ∪ B/N . (We do not need to take into account inverses, since the generating sets involved are subgroups themselves.) However, the element g1 g2 · · · gk ∈ hA, Bi. Thus, hA/N, B/N i ⊆ hA, BiN . The reverse inclusion is nearly identical. Together, they establish the desired equality. Part (4): Let g/N ∈ A/N ∩ B/N . Then gN ∈ A/N and gN ∈ B/N . Then g ∈ A and g ∈ B. Thus g/N ∈ (A ∩ B)/N . Thus A/N ∩ B/N ⊆ (A ∩ B)/N . Conversely, let gN ∈ (A ∩ B)/N . Then g ∈ A ∩ B so g ∈ A and g ∈ B. Hence, gN ∈ A/N ∩ B/N . Therefore, A/N ∩ B/N ⊆ (A ∩ B)/N and we conclude that A/N ∩ B/N = (A ∩ B)/N . Part (5): Suppose that A E G. Then gag −1 = a0 ∈ A for all a ∈ A and g ∈ G. Thus (gag −1 )N = (gN )(aN )(gN )−1 = a0 N ∈ A/N. Thus, A/N E G/N . Conversely, if A/N E G/N , then for all g ∈ G and all a ∈ A, there exists a0 ∈ A with (gN )(aN )(gN )−1 = a0 N. Thus, a0 N = (gag −1 )N . Therefore (a0 )−1 (gag −1 ) ∈ N and since N ≤ A, we have gag −1 ∈ A. This proves part (5).

Exercise: 13 Section 4.4 Question: Let G be the group of isometries of Rn . Prove that the subgroup D of direct isometries is a normal subgroup and that G/D ∼ = Z2 . Solution: Let f be a direct isometry in D and let g be any isometry in G. By Theorem 3.9.5, f (~x) = A~x + ~b, ~ where A is an n × n matrix with A> A = I with det A = 1 and where ~b is a constant vector. Also g(~x) = C~x + d, > −1 ~ where C is also an orthogonal matrix (with C C = I) and d is a constant vector. Then gf g satisfies ~ + ~b + d~ gf g −1 (~x) = C A C > (~x − d) = C AC > ~x + −AC > d~ + ~b = CAC > ~x + (I − CAC > )d~ + C~b. Obviously, this is an isometry. Furthermore, det(CAC > ) = det(C) det(A) det(C)−1 = det(A) = 1. Hence, gf g −1 is a direct isometry. By the conjugate condition, we have shown that D E G. Consider the function ϕ : G → {1, −1}, where we view {1, −1} as a group (subgroup of R) with multiplication, defined by ϕ(f ) = det(A), where f (~x) = A~x +~b It is easy to check that ϕ is a homomorphism and that ker ϕ = D. This gives another proof that D E G but the first isomorphism theorem tells us that G/D ∼ = Z2 .

4.5. FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS

207

4.5 – Fundamental Theorem of Finitely Generated Abelian Groups Exercise: 1 Section 4.5 Question: List all the possible invariant factors list for the following integers: a) 45; b) 480; c) 900. Solution: We give all the groups of the state order: a) n = 45: Z15 ⊕ Z3

Z45 b) n = 480 = 25 · 3 · 5 Z480

Z240 ⊕ Z2

Z60 ⊕ Z4 ⊕ Z2

Z120 ⊕ Z4

Z60 ⊕ Z2 ⊕ Z2 ⊕ Z2

Z120 ⊕ Z2 ⊕ Z2 Z30 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2

c) n = 900 = 22 · 32 · 52 Z900

Z450 ⊕ Z2

Z180 ⊕ Z5

Z300 ⊕ Z3

Z90 ⊕ Z10

Z150 ⊕ Z6

Z60 ⊕ Z15

Z30 ⊕ Z30

Exercise: 2 Section 4.5 Question: List all the possible elementary divisors for the following integers: a) 45; b) 480; c) 900. Solution: a) Note that 45 = 32 5. So that there only exist two possible elementary divisors, 45 = 32 · 5 and 45 = 3 · 3 · 5. b) 480 = 25 35. So all possible elementary divisors are 480 = 25 · 3 · 5, 480 = 24 · 2 · 3 · 5, 480 = 23 · 2 · 2 · 3 · 5, 480 = 23 · 22 · 3 · 5, 480 = 22 · 2 · 2 · 2 · 3 · 5, 480 = 22 · 22 · 2 · 3 · 5, and 480 = 2 · 2 · 2 · 2 · 2 · 3 · 5. c) 900 = 22 32 52 . So all possible elementary divisors are 900 = 22 · 32 · 52 , 900 = 22 · 32 · 5 · 5, 900 = 22 · 3 · 3 · 52 , 900 = 22 · 3 · 3 · 5 · 5, 900 = 2 · 2 · 32 · 52 , 900 = 2 · 2 · 32 · 5 · 5, 900 = 2 · 2 · 3 · 3 · 52 , and 900 = 2 · 2 · 3 · 3 · 5 · 5.

Exercise: 3 Section 4.5 Question: List all the non-isomorphic abelian groups of order 945 both in invariant factors form and elementary divisors form. Show which decompositions correspond to which in the different forms. Solution: First we note that 945 = 33 57. We notice that the only possible partitioning will occur with the power of three. All possible non-isomorphic abelian groups of order 945 are (listing the elementary divisors form on the left and the invariant factors form on the right): Z27 ⊕ Z5 ⊕ Z7 ∼ = Z945 Z9 ⊕ Z3 ⊕ Z5 ⊕ Z7 ∼ = Z315 ⊕ Z3 Z3 ⊕ Z3 ⊕ Z3 ⊕ Z5 ⊕ Z7 ∼ = Z105 ⊕ Z3 ⊕ Z3 .

Exercise: 4 Section 4.5 Question: List all the non-isomorphic abelian groups of order 864 both in invariant factors form and elementary divisors form. Show which decompositions correspond to which in the different forms. Solution:

We first note that 864 = 25 33 . All non-isomorphic abelian groups of order 864 are (elementary

208

CHAPTER 4. QUOTIENT GROUPS

divisor forms on left and invariant factors form on right): Z32 ⊕ Z27 ∼ = Z864 Z32 ⊕ Z9 ⊕ Z3 ∼ = Z288 ⊕ Z3 Z32 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z96 ⊕ Z3 ⊕ Z3 ∼ Z16 ⊕ Z2 ⊕ Z27 = Z432 ⊕ Z2 Z16 ⊕ Z2 ⊕ Z9 ⊕ Z3 ∼ = Z144 ⊕ Z6 Z16 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z48 ⊕ Z6 ⊕ Z3 Z8 ⊕ Z2 ⊕ Z2 ⊕ Z27 ∼ = Z216 ⊕ Z2 ⊕ Z2 ∼ Z8 ⊕ Z2 ⊕ Z2 ⊕ Z9 ⊕ Z3 = Z72 ⊕ Z6 ⊕ Z2 Z8 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z24 ⊕ Z6 ⊕ Z6 ∼ Z216 ⊕ Z4 Z8 ⊕ Z4 ⊕ Z27 = Z8 ⊕ Z4 ⊕ Z9 ⊕ Z3 ∼ = Z72 ⊕ Z12 Z8 ⊕ Z4 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z24 ⊕ Z12 ⊕ Z3 Z4 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z27 ∼ = Z108 ⊕ Z2 ⊕ Z2 ⊕ Z2 ∼ Z36 ⊕ Z6 ⊕ Z2 ⊕ Z2 Z4 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z9 ⊕ Z3 = Z4 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z12 ⊕ Z6 ⊕ Z6 ⊕ Z2 ∼ Z4 ⊕ Z4 ⊕ Z2 ⊕ Z27 = Z108 ⊕ Z4 ⊕ Z2 Z4 ⊕ Z4 ⊕ Z2 ⊕ Z9 ⊕ Z3 ∼ = Z36 ⊕ Z12 ⊕ Z2 ∼ Z12 ⊕ Z12 ⊕ Z6 Z4 ⊕ Z4 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z3 = Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z27 ∼ = Z54 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ∼ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z9 ⊕ Z3 = Z18 ⊕ Z6 ⊕ Z2 ⊕ Z2 ⊕ Z2 Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3 ⊕ Z3 ⊕ Z3 ∼ = Z6 ⊕ Z6 ⊕ Z6 ⊕ Z2 ⊕ Z2

Exercise: 5 Section 4.5 Question: What is the smallest value of n such that there exist 5 abelian groups of order n? List out the specific groups. Solution: Using Corollary 4.5.23, we need n = q1β1 q2β2 · · · qtβt with p(β1 )p(β2 ) · · · p(βt ) = 5. The smallest integer n that satisfies this condition is n = 16. The five abelian groups are Z16

Z8 ⊕ Z2

Z4 ⊕ Z4

Z4 ⊕ Z2 ⊕ Z2

Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 .

Exercise: 6 Section 4.5 Question: What is the smallest value of n such that there exist 4 abelian groups of order n? List out the specific groups. αn 1 α2 Solution: If n = pα 1 p2 . . . pn , then by Corollary 4.5.23 there are p(α1 )p(α2 ) . . . p(αn ) abelian groups of order n. Now we observe that p(1) = 1, p(2) = 2, p(3) = 3, and p(4) = 4. Now, we know that 16 = 24 has 4 different abelian groups. A quick look at all numbers lower than 16 (most of them are prime or multiples of 2 primes each taken to the first power with the exception of 4, 8, 9, and 12) convinces us that this is the lowest n possible. Then all abelian groups of order 16 are: Z16 , Z8 ⊕ Z2 , Z4 ⊕ Z4 , Z4 ⊕ Z2 ⊕ Z2 , and Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 .

Exercise: 7 Section 4.5 Question: What is the smallest value of n such that there exist 13 abelian groups of order n? List out the specific groups.

4.5. FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS

209

Solution: We first create a list of numbers and there corresponding partitions: 1 : p(1) = 1 2 : p(2) = 2 3 : p(3) = 3 4 : p(4) = 5 5 : p(5) = 7 6 : p(6) = 11 7 : p(7) = 15. Now we see that the number 27 will certainly have at least 13 abelian groups of its order. We will now show that there could not be any smaller number. If we have some powers α1 α2 · · · αn so that p(α1 )p(α2 ) . . . p(α3 ) ≥ 13 but where α1 + α2 + · · · + αn ≥ 7 then we can conclude that for any combination of distinct primes, since each αn α1 +α2 +...αn 1 α2 must be greater than or equal to 2, pα ≥ 27 . So now we just need to show that any 1 p2 . . . pn ≥ 2 collection of powers which add to any number less than 7 have partitions that multiply to a number less than 13. For any collection of powers α1 , α2 , · · · , αn which satisfy α1 + α2 + · · · + αn < 7, there exists a collection of powers β1 , β2 , · · · , βm where βi = αi when i ≤ n and βi = 1 when i > n so that β1 + β2 + · · · + βm = 6 and p(α1 )p(α2 ) . . . p(αn ) = p(β1 )p(β2 ) . . . p(βm ). This implies that we only need to check partitions of 6. So the listed result is: 6 →p(6) = 11 5 + 1 →p(5)p(1) = 7 4 + 2 →p(4)p(2) = 10 4 + 1 + 1 →p(4)p(1)p(1) = 5 3 + 2 + 1 →p(3)p(2)p(1) = 6 3 + 1 + 1 + 1 →p(3)p(1)p(1)p(1) = 3 2 + 2 + 2 →p(2)p(2)p(2) = 8 2 + 2 + 1 + 1 →p(2)p(2)p(1)p(1) = 4 2 + 1 + 1 + 1 + 1 →p(2)p(1)p(1)p(1)p(1) = 2 1 + 1 + 1 + 1 + 1 + 1 →p(1)p(1)p(1)p(1)p(1)p(1) = 1 and we see that none of the results are greater than thirteen so that n = 128 must be the smallest order. We now list the abelian groups of order 128: Z128 , Z64 ⊕ Z2 , Z32 ⊕ Z4 , Z32 ⊕ Z2 ⊕ Z2 Z16 ⊕ Z8 , Z16 ⊕ Z4 ⊕ Z2 , Z16 ⊕ Z2 ⊕ Z2 ⊕ Z2 Z8 ⊕ Z8 ⊕ Z2 , Z8 ⊕ Z4 ⊕ Z4 , Z8 ⊕ Z4 ⊕ Z2 ⊕ Z2 , Z8 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 Z4 ⊕ Z4 ⊕ Z4 ⊕ Z2 , Z 4 ⊕ Z4 ⊕ Z2 ⊕ Z2 ⊕ Z2 , Z 4 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2

Exercise: 8 Section 4.5 Question: Suppose that an abelian group has order 100 and has at least 2 elements of order 2. What are the possible groups that satisfy these conditions? Solution: Since 100 = 22 52 , we see that all possible abelian group of 100 are Z100 , Z50 ⊕ Z2 , Z20 ⊕ Z5 , and Z10 ⊕ Z10 . After examination, we see that the only groups that satisfy the condition of having two elements of order 2 are Z10 ⊕ Z10 and Z50 ⊕ Z2 . Exercise: 9 Section 4.5 Question: List all the abelian groups of order 72 that contain an element of order 6.

210

CHAPTER 4. QUOTIENT GROUPS

Solution: We note that 72 = 23 32 . So we have 6 different possible abelian groups, Z72 , Z36 ⊕ Z2 , Z18 ⊕ Z2 ⊕ Z2 , Z24 ⊕ Z3 , Z12 ⊕ Z6 , and Z6 ⊕ Z6 ⊕ Z2 . Since each direct sum has as a component, a group whose order is divisible by 6, we see that every abelian group of order 72 contains an element of order 6. Exercise: 10 Section 4.5 Question: Find the isomorphism type (by invariant factors decomposition) of U (21). ¯ 11, ¯ 13, ¯ 16, ¯ 17, ¯ 19, ¯ 20}. ¯ ¯ = 2. Solution: U (21) = {1̄, 2̄, 4̄, 5̄, 8̄, 10, We note that |U (21)| = 12 and that |8̄| = |20| Now all possible groups of order 12 are Z12 and Z6 ⊕Z2 . Since Z12 only has one element of order 2, by elimination we know U (21) ∼ = Z6 ⊕ Z2 . Exercise: 11 Section 4.5 Question: Find the isomorphism type (by invariant factors decomposition) of U (27). Solution: We know that |U (27)| = φ(27) = 27 − 9 = 18. According to FTFGAG, this can be isomorphic to only two groups: Z18 or Z6 ⊕ Z3 . Consider powers of the element 2 ∈ U (27). 1

2

2 =2

2 =4

3

2 =8

4

5

2 = 16 2 = 5

6

2 = 10

and we can already stop. In Z6 ⊕ Z3 every elements has an order that divides 6. Since 2 has an order that is greater than 6, we can conclude that U (27) is not isomorphic to Z6 ⊕ Z3 . Hence U (27) ∼ = Z18 . Exercise: 12 Section 4.5 Question: Suppose that G is an abelian group of order 176 such that the subgroup H = {g 2 | g ∈ G} has order 22. What are the possible groups G with this property? Solution: Suppose that A1 ⊕ A2 ⊕ . . . ⊕ An is an abelian group written in invariant factors form. Suppose that x1 , x2 , x3 , . . . xn are generators where xi generates the ith cyclic group in the direct sum. Then the order of the abelian group as a whole is |x1 | · |x2 | . . . · |xn |. Now, the subgroup H defined on this group would be generated by x21 , x22 . . . x2n and the order would be |H| = |x21 ||x22 | . . . |x2n | =

|x1 | |x2 | |xn | ... gcd(|x1 |, 2) gcd(|x2 |, 2) gcd(|xn |, 2)

Now since the order of the generators is equivalent to the order of the corresponding cyclic group we have |x2 | |xn | |x1 | ... gcd(|x1 |, 2) gcd(|x2 |, 2) gcd(|xn |, 2) |A1 | |A2 | |An | = ... . gcd(|A1 |, 2) gcd(|A2 |, 2) gcd(|An |, 2)

|H| =

Now we are ready to examine G. We note that 176 = 24 11 so that all possible abelian groups of order 176 are Z176 , Z88 ⊕ Z2 , Z44 ⊕ Z4 , Z44 ⊕ Z2 ⊕ Z2 , and Z22 ⊕ Z2 ⊕ Z2 ⊕ Z2 . We use our formula to calculate the order of H in each of these groups respectively: Z176 :|H| = 176/2 = 88 Z88 ⊕ Z2 :|H| = 88/2 · 2/2 = 44 Z44 ⊕ Z4 :|H| = 44/2 · 4/2 = 22 · 2 = 44 Z44 ⊕ Z2 ⊕ Z2 :|H| = 44/2 · 2/2 · 2/2 = 22 Z22 ⊕ Z2 ⊕ Z2 ⊕ Z2 : |H| = 22/2 · 2/2 · 2/2 · 2/2 = 11. So we see that the only abelian group of order 176 with the property is Z44 ⊕ Z2 ⊕ Z2 . Exercise: 13 Section 4.5 Question: Prove Lemma 4.5.16. Solution: Let q be a prime number and G be an abelian group of order q α . By Theorem 4.5.11, we know that G can be written uniquely as G ∼ = Zm1 ⊕ Zm2 ⊕ . . . ⊕ Zmk . We also know that q α = m1 m2 . . . mk which implies

4.5. FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS

211

that mi = q αi for every i since they must divide q α . Now since q α = q α1 q α2 . . . q αk = q α1 +α2 +...αk this implies that α = α1 + α2 + . . . + αk . Additionally we know that mi+1 |mi which in our case means that q αi+1 |q αi which is equivalent to αi ≥ αi+1 . Written in light of all the αi ’s this means that α1 ≥ α2 ≥ . . . ≥ αk ≥ 1. Finally, we reiterate that an abelian group of order q α can be uniquely written as G ∼ = Zα1 ⊕ Zα2 ⊕ . . . ⊕ Zαk where the αi ’s retain all the properties previously mentioned. This completes the proof of Lemma 4.5.16. Exercise: 14 Section 4.5 Question: Let p be a prime number. For all abelian groups of order p3 , list how many elements there are of each order. Solution: From Lemma 4.5.16 we can list all abelian groups of order p3 quickly: Zp3 , Zp2 ⊕Zp and Zp ⊕Zp ⊕Zp . We recall that if x is a generator of some cyclic group of order n then |xd | = |x|/ gcd(d, n) and that if x is an element of A1 and y is an element of A2 then the element (x, y) ∈ A1 ⊕ A2 has order lcm(|x|, |y|). First we examine Zp3 . Let x ∈ Zp3 be a generator. We will examine the equation |xd | = |x|/ gcd(|x|, d) = p3 / gcd(p3 , d). If we want the order to be p, then we need d to be a multiple of p2 . We have p − 1 different multiples of p2 . If we want the order to be p2 , then we need d to be a multiple of p but not of p2 . There are p2 − 1 multiples of p, but p − 1 of those are also multiples of p2 , so we have p2 − 1 − (p − 1) = p2 − p elements of order p2 . We also have 1 element of order 1 and the remaining (p3 ) − (p2 − p + p − 1 + 1) = p3 − p2 elements have order p3 . Now we examine the group Zp2 ⊕ Zp . In Zp2 we have p − 1 multiples of p which have order p, one element of order 1, and the remaining p2 − p elements have order p2 . In Zp , we have one element of order 1 and p − 1 elements of order p. If we have an element (x, y), then |(x, y)| = lcm(|x|, |y|) = lcm(pi , pj ) for 0 ≤ i, j. Then |(x, y)| = max(|x|, |y|). So then we have (p2 − p)(p) = p3 − p2 elements of order p2 , (p − 1)(p) + 1(p − 1) = p2 − p + p − 1 = p2 − 1 elements of order p, and 1 element of order 1. Now in Zp ⊕ Zp ⊕ Zp all of our element will have order p except 1 so we have p3 − 1 element of order p and 1 element of order 1. Exercise: 16 Section 4.5 Question: Find all integers n such that there exists a unique abelian group of order n. Solution: Let n = q1α1 q2α2 . . . qkαk and p represent the partition function. Then we know that the number of abelian groups of n is equal to p(α1 )p(α2 ) . . . p(αk ). Then we need p(αi ) = 1 for every i. The only number that has one partition is one, so none of our primes can have a power greater than one. So the integers that have a unique abelian group associated with them are the square-free integers. Exercise: 17 Section 4.5 Question: Let λ be a partition of an integer and denote by Gλ the group Gλ = Z2λ1 ⊕ Z2λ2 ⊕ · · · ⊕ Z2λr . 0

a) Prove that Gλ contains 2`(λ) − 1 = 2λ1 − 1 elements of order 2. b) Prove that Gλ contains 2|λ| − 2|λ|−m elements of order 2λ1 , where λ1 appears m times in the partition λ. Solution: Let λ be a partition of an integer and denote by Gλ the group Gλ = Z2λ1 ⊕ Z2λ2 ⊕ · · · ⊕ Z2λr . a) Every group Z2λi in the direct sum has exactly one element of order 2 and one element of order 1 (the only divisor of 2). Every element of order 2 or 1 in Gλ is of the form (a1 , a2 , . . . , ar ) where |ai | divides 2. There are two such options for each ai for all i with 1 ≤ i ≤ r. Hence, there are 2r = 2`(λ) elements of order 2 or 0 1. The identity is the only element of order 1, so there are exactly 2`(λ) = 2λ1 elements of order 2. b) In the cyclic group Z2λ1 , there are 2λ1 elements of order 2λ1 . If λ2 < λ1 , then an element in Gλ is of the form (a1 , a2 , . . . , ar ) where a1 has order 2λ1 and all other entry in the r-tuple can be anything. There are 2|λ|−λ1 elements in the (r − 1)-tuple (a2 , a3 , . . . , ar ). Hence, if λ2 < λ1 , there are 2λ1 −1 2|λ|−λ1 = 2|λ|−1 elements of order 2λ1 in Gλ . However, it is possible that some of the parts of the partition are equal. Suppose that in the partition λ, we have λ1 = λ2 = · · · = λm and λm+1 < λm . We use an inclusion-exclusion argument to obtain the result.

212

CHAPTER 4. QUOTIENT GROUPS For 1 ≤ i ≤ m, let Ai be the set of elements (a1 , a2 , . . . , ar ) ∈ Gλ such that ai is an element of order 2λ1 . Note that the intersection of any k of these sets has exactly k |Ai1 ∩ Ai2 ∩ · · · ∩ Aik | = 2λ1 −1 2|λ|−kλ1 = 2|λ|−k . The set of elements of order 2λ1 in Gλ is A1 ∪ A2 ∪ · · · ∪ Am . By the Inclusion-Exclusion Principle, |A1 ∪ A2 ∪ · · · ∪ Am | =

m X

|As | −

s=1

There are exactly

m k

X

|As ∩ As | + · · · − |A1 ∩ A2 ∩ · · · ∩ Am |.

1≤s<t≤m

terms in the k summation and each term is 2|λ|−k . Hence,

|A1 ∪ A2 ∪ · · · ∪ Am | =

m X

m X m |λ|−k m m−k 2 = 2|λ|−m (−1)k−1 2 k k k=1 ! X m m k m−k 2 − (−1) 2 = 2|λ|−m (2m − (−1 + 2)m ) k

(−1)k−1

k=1

= 2|λ|−m

k=0

|λ|−m

=2

(2 − 1) = 2|λ| − 2|λ|−m . m

Exercise: 18 Section 4.5 Question: Let G = hx, y | x12 = y 18 = 1, xy = yxi. Consider the subgroup H = hx4 y −6 i. Find the isomorphism type of G/H. Solution: Note that G = Z12 ⊕ Z18 ∼ = Z36 ⊕ Z6 and that |G| = 216. Furthermore, |H| = 3 with H = {1, x4 y −6 , x8 y 6 }. Thus G/H is an abelian group of order 72 = 23 · 32 . Now there is no element in G of order 8, so there cannot possibly an element of order 8 in G/H. Just using FTFGAG, this still leaves us with 4 possibilities. Note that |y 2 | = 9 and consider the order of y 2 H in G/H. We know that in any quotient group G/N , the order |gN | divides |g|. Since (y 2 H)3 = y 6 H 6= H, since y 6 ∈ / H, then y 2 H has order greater than 3 so it must have order 9. By FTFGAG, this leaves us with only two possibilities: Z36 ⊕ Z2 or Z18 ⊕ Z2 ⊕ Z2 . Consider the element x3 H. We note that (x3 H)2 = x6 H 6= H since x6 ∈ / H. Thus, x3 H must have an order 3 3 that divides |x | = 4 but its order is greater than 2. Thus, x H has order 4. Hence, G/H contains an element of order 4. By a process of elimination, and using FTFGAG, we conclude that G/H ∼ = Z36 ⊕ Z2 . Exercise: 19 Section 4.5 Question: Let G = (Z/12Z)2 and let H = {(a, b) | 2a + 3b = 0}. Determine the isomorphism type of G/H. Solution: The group G has |G| = 144. We need to calculate the order of the subgroup H. Leaving off the bars on the congruence classes, we have H = {(0, 0), (0, 4), (0, 8), (3, 2), (3, 6), (3, 10), (6, 0), (6, 4), (6, 8), (9, 2), (9, 6), (9, 10)} Since H has 12 elements, then G/H also has 12 elements. According to FTFGAG, G/H can be isomorphism to Z12 or Z6 ⊕ Z2 . /H We consider the element (1, 1) ∈ G/H. The 6th “power” of that element is (6, 6) in (Z/12Z)2 . But (6, 6) ∈ so the order of (1, 1) does not divide 6. Hence G/H is not isomorphic to Z6 ⊕ Z2 . Therefore G/H ∼ = Z12 . Exercise: 20 Section 4.5 Question: Let G = (Z/12Z)2 and let H = {(a, b) | a + 5b = 0}. Determine the isomorphism type of G/H. Solution: We note that |G| = 144. Also, in the relation in the group H, we note that a + 5b = 0 implies that a = 7b. Hence, H = {(7b, b) | b ∈ Z/12Z}. Thus, |H| = 12 and in particular |G/H| = 12. By FTFGAG and just this information, we conclude that there can be 2 different groups, namely Z12 or Z6 ⊕ Z2 . Consider the element (3, 0) + H, which is not equal to H since (3, 0) ∈ / H. It is easy to check that this has order 4. Since G/H contains an element of order 4, then by we have conclude already, we deduce that G/H ∼ = Z/12Z. Exercise: 21 Section 4.5 Question: Prove Cauchy’s Theorem for abelian groups. In other words, let G be an abelian group and prove that if p is a prime number dividing |G| then G contains an element of order p.

4.5. FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS

213

Solution: Let G be a group and p a prime number where p | |G|, so G must be a finite abelian group. We will construct an element of order p. Consider the invariant factors decomposition of G, G ∼ = Zm1 ⊕ Zm2 ⊕ . . . Zmk . Since p divides the order of the group, by FTFGAG it must divide m1 . So suppose that m1 = pi d for some other i−1 divisor d not divisible by p. Let x be a generator of Zm1 and consider the order of the element xp d , |xp

i−1

d

|x| gcd(|x|, pi−1 d) pi d = i gcd(p d, pi−1 d) pi d = i−1 p d = p.

|=

i−1

Now, the element (xp d , 1, 1 . . . , 1) ∈ G will have the same order. So we have shown how to find an element of order p whenever p divides the order of an abelian group G. Exercise: 22 Section 4.5 Question: Prove Corollary 4.5.23. Solution: Corollary 4.5.23 depends on Theorem 4.5.18. By Lemma 4.5.16, there are p(q βi ) possibilities for the the component Ai in the summand of Theorem 4.5.18. Here, p is the partition function as described in the section. Since these choices for the groups Ai are independent of each other, the number of possibilities for a group of order q1β1 q2β2 · · · qtβt is p(q1β1 )p(q2β2 ) · · · p(qtβt ).

Exercise: 23 Section 4.5 Question: Let p be a prime number. Prove that Aut(Zp ⊕ Zp ) ∼ = GL2 (Fp ). Solution: Let z be a generator for Zp . An automorphism ψ ∈ Aut(Zp ⊕Zp ) must map (z, 1) to another element of order p, which means any element (z a , z b ) with (a, b) 6= (0, 0), where we only consider a and b modulo p. This is also true for (1, z), i.e., ψ((1, z)) = (z c , z d ). Hence, for any element (z m , z n ) in Zp ⊕ Zp , we have ψ((z m , z n )) = ψ((z, 1))m ψ((1, z))n = (z a , z b )m (z c , z d )n = (z am+cn , z bm+dn ). So if ψ((z m , z n )) = (z r , z s ), then r a = s b

c m , d n

where all entries are taken in Z/pZ = Fp . In order for this homomorphism to be invertible, we need the matrix a c b d to be invertible. This defines a function from Aut(Zp ⊕ Zp ) and it is not hard to see that this function is bijective. The above matrix multiplication identity shows that composition of such automorphisms corresponds to multiplication of the matrices. Hence, the association described above defines an isomorphism from Aut(Zp ⊕Zp ) to GLp (Fp ). Exercise: 24 Section 4.5 Question: Let G be an abelian group. Prove that Aut(G) is abelian if and only if G is cyclic. Solution: Let G be cyclic with G ∼ = Zn . We saw in Proposition 3.7.28 that Aut(Zn ) = U (n), which is abelian. The challenging part of this exercise is to prove the converse. So we suppose that G is abelian but not cyclic and prove that Aut(G) is not abelian. Then by FTFGAG, G ∼ = A⊕Fm ⊕Fn , where A is an abelian group, possibly trivial, and n | n. Note that an automorphism ψ 0 on Fm ⊕ Fn leads to an automorphism ψ(a, g) = (a, ψ 0 (g)) on A ⊕ (Fm ⊕ Fn ). In particular, the automorphisms on Fm ⊕ Fn form a subset of Aut(G). Consequently, we reduce to the case when G = Zm ⊕ Zn , where n | m.

214

CHAPTER 4. QUOTIENT GROUPS

Suppose that x generates Zm and the y generates Zn . Let d = m/n. Consider the automorphisms ψ1 and ψ2 that satisfy ψ1 ((x, 1)) = (x, y)

ψ1 ((1, y)) = (1, y)

ψ2 ((x, 1)) = (x, 1)

ψ2 ((1, y)) = (xd , y).

It is an easy check that functions defined this way on the generators of Zm and Zn are homomorphisms and that they are surjective, and hence bijections. But then but ψ2 (ψ1 ((x, 1))) = ψ2 ((x, y)) = (xd+1 , y).

ψ1 (ψ2 ((x, 1)) = ψ1 ((x, 1)) = (x, y)

Since d ≤ m/2, we know that d + 1 < m and so xd+1 6= x. Since ψ1 and ψ2 do not commute, Aut(G) is not abelian. Exercise: 25 Section 4.5 Question: What is the first integer n such that there exist two partitions λ and µ of n such that λ0 = λ and µ0 = µ? Solution: We note that for a partition to equal it’s conjugate implies that the Young diagram is symmetric by reflecting through the line that slashes along the diagonal. For any odd number n, an easy way to form a partition that is equal to it’s conjugate is by letting λ = (bn/2c + 1, 1, 1, . . . , 1). Observing the Young diagram is a quick way to see that this will hold. For any even number greater than 2, we can form the partition λ = (n/2, 2, 1, 1, . . . , 1). With these tools, we can quickly check and find that the first number to have two partitions where λ = λ0 and µ = µ0 is 8 where λ = (4, 2, 1, 1) and µ = (3, 3, 2). Exercise: 26 Section 4.5 Question: Let p(n) be the partition function of n. Prove that as power series ∞ X

n

p(n)x =

n=0

∞ Y k=1

1 1 − xk

.

Solution: Consider the nth term in the product ∞ Y k=1

1 1 − xk

=

∞ Y k=1

  X  xkj  . j=0

In the product, all coefficients are 1 but will lead to nonunit coefficients once we gather like terms. Furthermore, in order to obtain a term of degree n, we only need to consider k ≤ n is the infinite product and j ≤ n in each of the infinite series. Consequently, determining the coefficient of xn requires only a finite calculation. We obtain a terms xn by picking a term from each infinite series so they arrive as the form. xn = xj1 +2j2 +···+kjk +···+njn with the jk ≥ 0. Each solution (j1 , j2 , . . . , jn ) to the equation n = j1 + 2j2 + · · · + kjk + · · · + njn corresponds to the partition of n of the form j1 times

j2 times

jn times

z }| { z }| { z }| { n = 1 + 1 + · · · + 1 + 2 + 2 + · · · + 2 + · · · + n + n + · · · + n. This covers uniquely all possible partitions of n. Hence, the number of solutions to n = j1 +2j2 +· · ·+kjk +· · ·+njn with ji ≥ 0 is precisely the number p(n) of partitions of n. The result follows. Exercise: 27 Section 4.5 Question: Find the error in the proof of the following erroneous proposition. (See text for proof)

4.5. FUNDAMENTAL THEOREM OF FINITELY GENERATED ABELIAN GROUPS

215

Solution: The proof fails due to the fallacy of affirming the conclusion. Observe the following excerpt of the proof: (g1 , a1 ) Ker h = (g2 , a2 ) Ker h ⇒ (g2−1 g1 , a−1 2 a1 ) ∈ Ker h ⇒ g2−1 g1 a−1 2 a1 = 1 ⇒ g2−1 g1 ∈ A ⇒ g2 A = g1 A. This shows that the association ϕ : G/A⊕A → (G ⊕ A)/ Ker h defined by ϕ(gA, a) = (g, a) Ker h is a function. Now, the proof-writer only proves the implications in one direction and then assumes that the association is a function. However consider two representative g1 , g2 where g1 A = g2 A. Then under this function ϕ we have, ϕ(g1 A, a) = ϕ(g2 A, a) ⇒ (g1 , a) Ker h = (g2 , a) Ker h Now, since both g1 and g2 represent the same coset, this implies that g2 = g1 x for some x ∈ A. And since we are saying that the cosets (g1 , a) Ker h and (g2 , a) Ker h are equivalent, we must have for some (d, d−1 ) ∈ Ker h: (g1 , a)(d, d−1 ) = (g2 , a) ⇒ (g1 d, ad−1 ) = (g1 x, a) ⇒ ad−1 = a ⇒ d−1 = d = 1 ⇒ (g1 , a) = (g1 x, a) ⇒ g1 = g1 x ⇒ x = 1. Now, since this holds for any elements that are representatives of the coset of A, we must have that A is the trivial subgroup, or {1}. So the proposition only holds for the trivial subgroup and not any subgroup. Exercise: 28 Section 4.5 Question: Prove that the set of functions from N to Z, denoted Fun(N, Z), equipped with function addition + is not a free abelian group. Solution: [Errata: This question is too challenging for this stage of the book and requires content from module theory. At this stage, the reader should be able to prove that the set ei with i ∈ N where ( 1 if i = j ei (j) = 0 otherwise is not a basis of Fun(N, Z). This is because linear combinations are always finite linear combinations. Hence, any linear combination of elements in the set {ei }i∈N give sequences of integers that are 0 on all by finitely many entries. Thus, this set of elements does not span Fun(N, Z). The reader can also prove that Fun(N, Z) cannot have a countable basis. Consider the function ψ : Fun(N, Z) → R defined by j a k X n 2−n . ψ((an )) = a0 + an − 2 2 n=1 an The expression an − 2 2 gives 0 if an is even and 1 if an is odd. The infinite series converges to a real number in [0, 1], expressing the number in binary expansion. this shows that ψ is surjective. Since R is uncountable, then Fun(N, Z) is uncountable. Assume now that Fun(N, Z) has a countable basis S = {f1 , f2 , . . .}. Consider the set of finite linear combinations of S in Fun(N, Z). Let Sn = {f1 , f2 , . . . , fn }. Since Sn is linearly independent, then Span(Sn ) is isomorphic to Zn , which is countable by Exercise 1.2.7. Let fn : N∗ → Zn a bijection, which gives a bijection gn : N∗ → Span(Sn ). Let h : N∗ → N∗ × N∗ be a bijection (see Exercise 1.2.7), then the function F : N∗ → gh(n)1 (h(n)2 ) defines a surjection of N∗ onto Span(S), first by selecting the number m in which the linear combination involving only the first m elements f1 , f2 , . . . , fm and then by selecting which specific linear

216

CHAPTER 4. QUOTIENT GROUPS

combination in Span(Sm ). Thus Span(S) is countable. However, Fun(N, Z) is not countable. This contradicts that S is a basis since S does not span Fun(N, Z). ] Exercise: 29 Section 4.5 Question: Consider the subgroup H = {(x1 , x2 , x3 ) ∈ Z3 | 6 divides 2x1 + 3x2 + 4x3 } of the free abelian group Z3 . Find a basis of H. Solution: Note that by considering the parity of 2x1 + 3x2 + 4x3 , in order for this to be divisible by 6, the variable x2 must be divisible by 2. Setting x2 = 2x02 , the condition 6 divides 2x1 + 3x2 + 4x3 is equivalent to 3 divides x1 + 3x02 + 2x3 . The linearly independent elements (−3, 1, 0) and (−2, 0, 1) solve this condition. Furthermore, by taking linear combinations of these we see that s(−3, 1, 0) + t(−2, 0, 1) = (−3s − 2t, s, t) allow us to get all possible options for x02 and x3 . However, these solutions give x1 + 3x02 + 2x3 = 0. In order to get divisible by 3, we find that r(3, 0, 0) give all possibilities for these multiples. Putting this discussion back into (x1 , x2 , x3 ), we claim that {(3, 0, 0), (−6, 2, 0), (−2, 0, 1)} is a basis of H.

5 | Rings 5.1 – Introduction to Rings Exercise: 01 Section 5.1 Question: Let R = R>0 , with the addition of x y = xy, and the multiplication x ⊗ y = xy . Solution: It is clear that for any a, b ∈ R we have a b = ab ∈ R and a ⊗ b = ab ∈ R. So that both operations are binary operations on R. Now we check that (R, ) is an abelian group. Associativity a (b c) = a (bc) = a(bc) = (ab)c = (a b)c = (a b) c So is associative. Identity Since 1 ∈ R, we know that for any a ∈ R, we have a 1 = a1 = a = 1a = 1 a. So that (R, ) has an identity. Inverses For any a ∈ R, 1/a ∈ R and 1/a a = (1/a)(a) = 1 and a 1/a = a(1/a) = 1 so that (R, ) has inverses. Commutativity For any a, b ∈ R, a b = ab = ba = b a. So is commutative. This shows that the pair (R, ) is an abelian group. Now we check that ⊗ is associative. Let a, b, c ∈ R then a ⊗ (b ⊗ c) = a ⊗ (bc ) c

= ab However

(a ⊗ b) ⊗ c = (ab ) ⊗ c = (ab )c = abc 2

Setting a = 2, b = 1 and c = 2 shows us that the first expression is equal to 21 = 21 = 2 while the second expression is 21·2 = 22 = 4. So ⊗ is not associative. Finally we see if ⊗ distributes over . Consider (a b) ⊗ c = (ab)c = ac bc = (a ⊗ c)(b ⊗ c) = (a ⊗ c) (b ⊗ c) which shows that ⊗ distributes over on the right. Now we check the other direction, c ⊗ (a b) = c ⊗ (ab) = cab = ca cb = ca cb = (c ⊗ a) (c ⊗ b). So ⊗ is distibutive over both directions. We conclude that R is not a ring since ⊗ is not associative. Exercise: 2 Section 5.1 Question: Decide whether the following set with the given operations is a ring. If it is, prove it and decide whether it is commutative and whether it has an identity. If it is not, decide which axioms fail. Let R = Q, with addition x y = x + y + 1, and multiplication x+ ×y = x + y + xy. Solution: We check all the axioms for a ring. It is easy to tell that both operations are binary operations. 217

218

CHAPTER 5. RINGS

Addition is associative : True: (x y) z = (x + y + 1) z = x + y + z + 2 = x (y + z + 1) = x (y z). Addition is commutative : Evident, since + is commutative. Addition has an identity : True. The identity is −1. x (−1) = x + (−1) + 1 = x. Addition has inverses : True. For all x ∈ Q, x (−x − 2) = x − x − 2 + 1 = −1. Multiplication is associative : True: (x+ ×y)+ ×z = (x + y + xy)+ ×z = x + y + z + xy + xz + yz + xyz = x+ ×(y + z + yx) = x+ ×(y+ ×z). Multiplication is Commutative : Obvious. (We checked this first so that we only have to verify one side distributivity later.) Multiplication has an identity : 0 works as an identity because for all x ∈ Q, we have x+ ×0 = x. Distributivity : True: x+ ×(y z) = x+ ×(y + z + 1) = x + y + z + 1 + x(y + z + 1) = 1 + 2x + y + z + xy + xz and (x+ ×y) (x+ ×z) = (x + y + xy) (x + z + xz) = 2x + y + z + xy + xz + 1. These two are equal so we have distributivity. We have checked all the axioms of a ring and established that (Q, , × +) is a commutative ring with an identity.

Exercise: 3 Section 5.1 Question: Let R = R3 with addition as vector addition and multiplication as the cross product. Solution: It is clear that for any a, b ∈ R we have a + b = c ∈ R and by the definition of the cross product we also have a × b = c ∈ R. So that both operations are binary operations on R. Now we check that (R, +) is an abelian group. Associativity Since vector addition is defined component-wise, and since + is associative in R, we can conclude that ~a + (~b + ~c) = (~a + ~b) + ~c for all ~a, ~b, ~c ∈ R3 . Identity Let ~0 be the zero vector. Since zero is the additive identity in R and vector addition is component wise, it is clear that for any ~a ∈ R3 we have ~a + ~0 = ~a = ~0 + ~a. So (R, +) has an identity. Inverses For any ~a ∈ R3 , we know that −~a ∈ R3 and ~a + (−~a) = ~0 = (−~a) + ~a. So that (R, +) has inverses Commutativity Since vector addition is defined component wise and addition is commutative in R, we know that ~a + ~b = ~b + ~a. This shows that the pair (R, +) is an abelian group. Now we will show that the cross product, or ×, is not associative. We will exploit the fact that for any vector ~a ∈ R3 , ~a × ~a = ~0. Let ~i, ~j, and ~k represent three standard basis vectors in R3 . Consider the expression (~i × ~i) × ~j = ~0 × ~j = ~0. However, grouping differently we find ~i × (~i × ~j) = ~i × ~k = −~j 6= ~0. This is an example of how the cross product is not associative in R3 . From the properties of the cross product, we know that (~a + ~b) × ~c = ~a × ~c + ~b × ~c and ~c × (~a + ~b) = ~c × ~a + ~c × ~b. So R is not a ring with the cross product failing to be associative. Exercise: 4 Section 5.1 Question: Let S be any set and consider R = P(S) with 4 as the addition operation and ∩ as the multiplication. Solution: Since the symmetric difference and the intersection of any two sets from P(S) can only result in a set that contains elements of S and so is an element of P(S), we must have that the symmetric difference and intersection are binary operation on P(S). We show that the pair (P(S), 4) is an abelian group.

5.1. INTRODUCTION TO RINGS

219

Associativity No matter how we place the parentheses, A4B4C consists of the elements that occur in an odd number of the sets A, B, and C. So 4 is associative. Identity The operation has an identity, namely ∅, because for all A ∈ P(S), A4∅ = A = ∅4A. Inverses For all A ∈ P(S), we have A4A = ∅ so every element is its own inverse. Commutativity The symmetric difference is commutative. Since A4B = A∪B−A∩B = B∪A−B∩A = B4A. Hence (P(S), 4) is an abelian group. Now, we show that ∩ is associative. Let A, B, C ∈ P(S) then A ∩ (B ∩ C) = (A ∩ B) ∩ C since both sides of the equation represent a set that contains all the element that exist in A, B, and C. So ∩ is associative. We know from Exercise 1.1.9 that A ∩ (B4C) = (A ∩ B)4(A ∩ C). Since ∩ is commutative, we know that A ∩ (B4C) = (B4C) ∩ A and that (A ∩ B)4(A ∩ C) = (B ∩ A)4(C ∩ A) which implies that (B4C) ∩ A = (B ∩ A)4(C ∩ A) so that ∩ also distributes from the right over 4. So we have shown that the triple (P(S), 4, ∩) is a ring. We know that ∩ is commutative and for any A ∈ R, A ∩ S = S ∩ A = A. Which shows that S is the multiplicative identity for R. Exercise: 5 Section 5.1 Question: Let S be any set and consider R = P(S) with ∪ as the addition operation and ∩ as the multiplication. Solution: Obviously, the intersection or union of any two sets in P(S) will only contain elements of S and therefore must be a member of P(S) so that ∩ and ∪ are binary operation. Now we will see if the pair (P(S), ∪) is an abelian group. Associativity We know from the section on set theory the ∪ is associative. Identity The empty set, ∅, functions as the identity since for any A, A ∪ ∅ = ∅ ∪ A = A. Inverses For any nonempty set A, and any other set B, we know that A ⊆ A ∪ B, so that A can never become ∅ by unioning with another set. So (P(S), ∪) does not have inverses. Commutativity Taking the union of two sets is commutative. Since the inverse axiom fails, the pair (P(S), ∪) is not a group. We will continue to check the other axioms. We know from Chapter 1 that ∩ is associative and that it distributes both left and right over ∪. So the only axiom that fails is that (P(S)) is not a group which shows that R is not a ring. Exercise: 6 Section 5.1 Question: Let S be any set and consider R = P(S) with 4 as the addition operation (defined as A4B = A4B) and ∪ as the multiplication. Solution: The operation 4 is a binary operation. For associativity, consider A, B, C ∈ P(S). A Venn diagram will show that A4(B4C) = (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C). This is symmetric in A, B and C, so since 4 is commutative, we deduce that (A4B)4C = C4(A4B) = A4(B4C). For the identity, note that A4S = A4S = Ā = A. Hence, the whole context set S is the identity for 4. For inverses, we see that A4A = A4A = ∅ = S. Hence, A is its own inverse with respect to 4. We already observed that since 4 is commutative, then so is 4. We know from earlier sections that ∪ is associative in P(S) and we also know that ∅ is the identity for ∪. We just need to check distributivity of ∪ over 4. We have A ∪ (B4C) = A ∪ (B ∪ C) ∩ B ∩ C =A∪B∪C ∪B∩C = A ∪ (B ∩ C) ∪ (B ∩ C),

220

CHAPTER 5. RINGS

and (A ∪ B)4(A ∪ C) = (A ∪ B ∪ C) ∩ (A ∪ B) ∩ (A ∪ C) = (A ∩ B ∩ C) ∪ A ∪ (B ∩ C) = ((A ∪ A) ∩ ((B ∩ C) ∪ A)) ∪ (B ∩ C) = (S ∩ ((B ∩ C) ∪ A)) ∪ (B ∩ C) = (B ∩ C) ∪ A ∪ (B ∩ C) = A ∪ (B4C).

Exercise: 7 Section 5.1 Question: Let R be the set of finite unions of bounded intervals in R (possibly empty or singletons {a}). Let A, B ∈ R. Define the symmetric difference A4B on R as the addition and the convex hull of A ∪ B as the multiplication. (We define the convex hull of a subset S of R as the smallest bounded interval containing S.) Solution: We know from other exercises (see Exercise 3.2.12) that (P(R), 4) is an abelian group. We first need to check that R, as a subset of P(R) is a subgroup. R is obviously nonempty. Let A, B ∈ R. Then A4B −1 = A4B, since the inverse of a set under 4 is itself. Suppose that A=

m [

Ii

and

n [

B=

i=1

Jj

j=1

as finite unions of intervals. Then  A4B = (A ∪ B) − (A ∩ B) = 

m [

Ii ∪

=

m [

i=1

Ii ∪

n [ j=1





Jj  − 

n m [ [





Jj  − 

j=1

i=1



n [

m [

! Ii

i=1

 ∩

n [

 Jj 

j=1

 (Ii ∩ Jj ) .

i=1 j=1

So A4B is a finite union of intervals from which are removed a finite union of bounded intervals. The result is a subset of R that consists of fewer or equal to (mn + 1)(m + n) intervals. In particular, A4B ∈ R. By the One-Step Subgroup Criterion, R is a subgroup of P(S). This establishes the first 4 axioms of a ring. Now consider the operation A · B of taking the convex hull of A ∪ B. With minimal geometric reasoning, we can observe that A · (B · C) is the convex hull of A ∪ B ∪ C. Because ∪ is associative and commutative, (A · B) · C is also the convex hull of A ∪ B ∪ C. Hence, · is associative. We need to discuss distributivity of · over 4. We prove that · is not distributive over 4 by giving an example. Let A = [0, 1], B = [2, 4], and C = [3, 5]. Then A · (B4C) = [0, 1] · ([2, 3) ∪ (4, 5]) = [0, 5], whereas (A · B)4(A · C) = [0, 4]4[0, 5] = (4, 5]. The operation · is obviously commutative since ∪ is commutative. The operation does not have an identity. If A is a disjoint union of two intervals, then for any set B, the set A · B is the convex hull of A ∪ B, so contains the convex hull of A. However, when A is a disjoint union of two intervals, it is not the convex hull of any set (since the convex hull consists of a single interval). Exercise: 8 Section 5.1 Question: Let R = Z × Z and define (a, b) + (c, d) = (a + c, b + d) and define also (a, b) × (c, d) = (ad − bc, bd). Solution: Clearly, the + and × operations are binary operations over Z and thus Z × Z. It is known that + is associative and commutative over Z, that 1 is the identity of this operation in Z, and that the inverse of any element a ∈ Z is −a. Thus (Z × Z, +) is an abelian group. We will now see if the triple (Z × Z, +, ×) is a ring. We find that × is not associative in Z × Z because for any (a, b)(c, d), (e, f ) ∈ Z × Z,

5.1. INTRODUCTION TO RINGS

221

((a, b) × (c, d)) × (e, f ) = (ad − bc, bd) × (e, f ) = (adf − bcf − bde, bdf ) 6= (adf − bcf + bde, bdf ) = (a, b) × (cf − de, df ) = (a, b) × ((c, d) × (e, f )) As × is not associative, this is not a ring. Exercise: 9 Section 5.1 Question: Let R be a ring, let r, s ∈ R, and let m, n ∈ Z. Prove the following formulas with the · notation. a) m · (r + s) = (m · r) + (m · s). b) (m + n) · r = (m · r) + (n · r). Solution: Let R be a ring, let r, s ∈ R, and let m, n ∈ Z. a) First assume that m is positive. m times

times m times }| { z m }| z { z }| { m · (r + s) = (r + s) + (r + s) + · · · + (r + s) = r + r + · · · + r + s + s + · · · + s = m · r + m · s,

where the middle equality holds because of commutativity of +. Furthermore, if m is positive then the negative number (−m) satisfies (−m) · (r + s) = −(m · (r + s)) = −(m · r + m · s) = −(m · r) + (−(m · s)) = (−m) · r + (−m) · s. b) This property is just the power rule in the group (R, +). Exercise: 10 Section 5.1 Question: Let R be a ring, let r, s ∈ R, and let m, n ∈ Z. Prove the following formulas with the · notation. a) m · (rs) = r(m · s) = (m · r)s. b) (mn) · r = m · (n · r). Solution: a) First we assume that m is positive. m · (rs) = rs + rs + . . . + rs(m times) = r(s + s + . . . + s(m times)) = r(m · s) In the same way, m · (rs) = rs + rs + . . . + rs(m times) = (r + r + . . . + r(m times))s = (m · r)s Now if m is positive then the negative number −m satisfies, (−m) · (rs) = −(m · (rs)) = −(r(m · s)) = r(−m · s), where the last step is taken by Proposition 5.1.3 part 2. We now show the other half also holds with −m, (−m) · (rs) = −(m · (rs)) = −(m · r)s = (−m · r)s. b) We first assume that mn is positive. (mn) · r = r + r + . . . + r(mn times) = (r + r + . . . + r(n times)) + (r + r + . . . + r(n times)) . . . + (r + r + . . . + r(n times))(m times) = m · (r + r + . . . + r(n times)) = m · (n · r)

222

CHAPTER 5. RINGS Now if mn is positive we will show that this works for −mn, (−mn) · r = −(mn · r) = −(m · (n · r)) = −m · (n · r) = m · (−n · r) So the formula holds.

Exercise: 5 Section 5.11 Question: Prove that in C 0 ([a, b], R) the composition operation ◦ is right distributive over +. Solution: We want to show that for f, g, h ∈ C 0 ([a, b], R), (f + g) ◦ h = f ◦ h + g ◦ h. Let x ∈ R; we will show that this statement holds for any x in the domain. Then ((f + g) ◦ h)(x) = (f + g)(h(x)) = f (h(x)) + g(h(x)) = (f ◦ h)(x) + (g ◦ h)(x) = (f ◦ h + g ◦ h)(x). Therefore ◦ is right distributive over + as desired. Exercise: 12 Section 5.1 Question: Let I be an interval of real numbers. Prove that the zero divisors in Fun(I, R) are nonzero functions f (x) such that there exists x0 ∈ I such that f (x0 ) = 0. Prove that all the elements in Fun(I, R) are either 0, a zero divisor, or a unit. Solution: Suppose f ∈ Fun(I, R) is a zero divisor, so f (x) is not everywhere 0. Then there exists g ∈ Fun(I, R) such that (f g)(x) = 0 for all x ∈ I and such that g is not the zero function. Therefore g(x0 ) 6= 0 for at least one x0 ∈ I; to be the least restrictive we shall say this happens for only one x0 ∈ I. Define g(x) = 0 for x 6= x0 and g(x) = k for x = x0 , where k is any nonzero real number. Then the function f for which f (x0 ) = 0 is a zero divisor because (f g)(x) = f (x) × 0 = 0 for x 6= x0 and (f g)(x0 ) = 0k = 0. This characterizes all zero divisors in the ring because if f (x) 6= 0 for any x ∈ I, then g must be the zero function, but then f would not be a zero divisor. Suppose, then, that f ∈ Fun(I, R) is not a zero divisor. Then either f (x) = 0 for all x ∈ I or f (x) is never 0 for any x ∈ I. If f (x) = 0 for all x ∈ I, then f is the zero function. If f (x) is never 0 on I, then there exists a function g ∈ Fun(I, R), defined by g(x) = 1/f (x), such that g is the inverse of f in Fun(I, R) because (f g)(x) = f (x) × 1/f (x) = 1, so f g is the identity function. Exercise: 13 Section 5.1 Question: Prove (carefully) that the nonzero elements in (C 0 ([a, b], R), +, ×) that are neither zero divisors nor units are functions for which there exists an x0 ∈ [a, b] and an ε > 0 such that f (x0 ) = 0 and for which f (x) 6= 0 for all x such that 0 < |x − x0 | < ε. Solution: Let f be an element in (C 0 ([a, b], R), +, ×) that is neither a zero divisor nor a unit. Functions that are invertible are continuous functions that have no zeros. (Otherwise, 1/f is a continuous function.) Hence, f has a zero. A zero divisor is a nonzero continuous function f such that f g = 0 for some other function g. Since f 6= 0, there’s an element x1 with f (x1 ). Since f is continuous, there is an interval I around x1 in which f is nonzero. Then in order for f g = 0, we must have g(x) = 0 for x ∈ I. The reasoning is symmetric so a function is a zero divisor if and only if it is 0 on a subinterval of its domain. Hence, a function f that is neither a unit nor a zero divisor has a zero but there does not exist an interval such that f is zero on that interval. Exercise: 14 Section 5.1 Question: Prove that multiplication in H is associative. Solution: Let α = α1 + α2 i + α3 j + α4 k, β = β1 + β2 i + β3 j + β4 k, and γ = γ1 + γ2 i + γ3 j + γ4 k be three elements in H. Then αβ = (α1 β1 − α2 β2 − α3 β3 − α4 β4 ) + (α1 β2 + α2 β1 + α3 β4 − α4 β3 )i + (α1 β3 − α2 β4 + α3 β1 + α4 β2 )j + (α1 β4 + α2 β3 − α3 β2 + α4 β1 )k.

5.1. INTRODUCTION TO RINGS

223

Then we find that (αβ)γ = (α1 β1 γ1 − α2 β2 γ1 − α3 β3 γ1 − α4 β4 γ1 − α1 β2 γ2 − α2 β1 γ2 − α3 β4 γ2 + α4 β3 γ2 − α1 β3 γ3 + α2 β4 γ3 − α3 β1 γ3 − α4 β2 γ3 − α1 β4 γ4 − α2 β3 γ4 + α3 β2 γ4 − α4 β1 γ4 ) + (α1 β1 γ2 − α2 β2 γ2 − α3 β3 γ2 − α4 β4 γ2 + α1 β2 γ1 + α2 β1 γ1 + α3 β4 γ1 − α4 β3 γ1 + α1 β3 γ4 − α2 β4 γ4 + α3 β1 γ4 + α4 β2 γ4 − α1 β4 γ3 − α2 β3 γ3 + α3 β2 γ3 − α4 β1 γ3 )i + (α1 β1 γ3 − α2 β2 γ3 − α3 β3 γ3 − α4 β4 γ3 − α1 β2 γ4 − α2 β1 γ4 − α3 β4 γ4 + α4 β3 γ4 + α1 β3 γ1 − α2 β4 γ1 + α3 β1 γ1 + α4 β2 γ1 + α1 β4 γ2 + α2 β3 γ2 − α3 β2 γ2 + α4 β1 γ2 )j + (α1 β1 γ4 − α2 β2 γ4 − α3 β3 γ4 − α4 β4 γ4 + α1 β2 γ3 + α2 β1 γ3 + α3 β4 γ3 − α4 β3 γ3 − α1 β3 γ2 + α2 β4 γ2 − α3 β1 γ2 − α4 β2 γ2 + α1 β4 γ1 + α2 β3 γ1 − α3 β2 γ1 + α4 β1 γ1 )k. Similarly for βγ we calculate βγ = (β1 γ1 − β2 γ2 − β3 γ3 − β4 γ4 ) + (β1 γ2 + β2 γ1 + β3 γ4 − β4 γ3 )i + (β1 γ3 − β2 γ4 + β3 γ1 + β4 γ2 )j + (β1 γ4 + β2 γ3 − β3 γ2 + β4 γ1 )k. Then we have α(βγ) = (α1 β1 γ1 − α1 β2 γ2 − α1 β3 γ3 − α1 β4 γ4 − α2 β1 γ2 − α2 β2 γ1 − α2 β3 γ4 + α2 β4 γ3 − α3 β1 γ3 + α3 β2 γ4 − α3 β3 γ1 − α3 β4 γ2 − α4 β1 γ4 − α4 β2 γ3 + α4 β3 γ2 − α4 β4 γ1 ) + (α1 β1 γ2 + α1 β2 γ1 + α1 β3 γ4 − α1 β4 γ3 + α2 β1 γ1 − α2 β2 γ2 − α2 β3 γ3 − α2 β4 γ4 + α3 β1 γ4 + α3 β2 γ3 − α3 β3 γ2 + α3 β4 γ1 − α4 β1 γ3 + α4 β2 γ4 − α4 β3 γ1 − α4 β4 γ2 )i + (α1 β1 γ3 − α1 β2 γ4 + α1 β3 γ1 + α1 β4 γ2 − α2 β1 γ4 − α2 β2 γ3 + α2 β3 γ2 − α2 β4 γ1 + α3 β1 γ1 − α3 β2 γ2 − α3 β3 γ3 − α3 β4 γ4 + α4 β1 γ2 + α4 β2 γ1 + α4 β3 γ4 − α4 β4 γ3 )j + (α1 β1 γ4 + α1 β2 γ3 − α1 β3 γ2 + α1 β4 γ1 + α2 β1 γ3 − α2 β2 γ4 + α2 β3 γ1 + α2 β4 γ2 − α3 β1 γ4 − α3 β2 γ3 + α3 β3 γ2 − α3 β4 γ1 + α4 β1 γ1 − α4 β2 γ2 − α4 β3 γ3 − α4 β4 γ4 )k. Though it takes some time to check, we do indeed find that (αβ)γ = α(βγ). Exercise: 15 Section 5.1 Question: Let α = 1 + 2i + 3j + 4k and β = 2 − 3i + k in H. Calculate the following operations: a) α + β; b) αβ; c) βα; d) αβ −1 ; e) β 2 . Solution: We calculate the products: a) α + β = 3 − i + 3j + 5k. b) Calculating: αβ = (1 + 2i + 3j + 4k)(2 − 3i + k) = 2 − 3i + k + 4i + 6 − 2j + 6j + 9k + 3i + 8k − 12j − 4 = 4 + 4i − 8j + 18k c) Calculating: βα = (2 − 3i + k)(1 + 2i + 3j + 4k) = 2 + 4i + 6j + 8k − 3i + 6 − 9k + 12j + k + 2j − 3i − 4 = 4 − 2i + 20j d) Calculating: 1 (1 + 2i + 3j + 4k)(2 + 3i − k) 4+9+1 1 = (2 + 3i − k + 4i − 6 + 2j + 6j − 9k − 3i + 8k + 12j + 4) 14 1 2 10 1 = (4i + 20j − 2k) = i + j − k 14 7 7 7

αβ −1 =

224

CHAPTER 5. RINGS

e) Calculating: β 2 = (2 − 3i + k)(2 − 3i + k) = 4 − 6i + 2k − 6i − 9 + 3j + 2k − 3j − 1 = −6 − 12i + 4k

Exercise: 16 Section 5.1 Question: Let α, β ∈ H be arbitrary. Decide whether any of the operations αβ −1 , β −1 α, βα−1 , or α−1 β are equal. Solution: Recall that (a + bi + cj + dk)−1 =

1 (a − bi − cj − dk). a2 + b2 + c2 + d2

Due to the denominator, th only possible equalities would be between αβ −1 and β −1 α or between βα−1 and α−1 β. However, if the property that αβ −1 = β −1 α for all α and |beta is tantamount to the multiplication being commutative on nonzero elements but we know that this is not the case. Hence, there is no general equality between any of these four expressions. (As for specific equalities, there may be some but that is a more complicated problem that requires methods from Chapter 12.) Exercise: 17 Section 5.1 Question: Let R = {a + bi + cj + dk ∈ H | a, b, c, d ∈ Z}. Prove that R is a subring of H and prove that U (R) = Q8 , the quaternion group. Solution: The subset R is nonempty since 0 ∈ R. Also, it is obvious from the formulas for addition and multiplication that if α, β ∈ R, then the components of α − β and or αβ have integer components, so are in R. Thus R is a subring of H. Now consider the group of units U (R). An element α = a + bi + cj + dk ∈ R is in U (R) if and only if α−1 =

1 (a − bi − cj − dk) ∈ R. a2 + b2 + c2 + d2

This is equivalent to requiring that all of the coefficients be in Z. Now for all positive integers n, we have n2 > n unless n = 1. In that case, we have equality n2 = n. Furthermore, for all positive integers n and nonnegative integers m, we have n2 + m > n unless n = 1 and m = 0. Applied to α−1 , we see that a ∈Z a2 + b2 + c2 + d2 if and only if a = 0 and b, c, d are arbitrary or a = ±1 and b = c = d = 0. Since all components of α−1 are in Z, the only quaternions in U (R) are ±1, ±i, ±j, ±k. Furthermore, it is easy to check that U (R) ∼ = Q8 , especially since H was designed so that the subgroup ({±1, ±i, ±j, ±k}, ×) is Q8 . Exercise: 18 Section 5.1 Question: Fix an integer n ≥ 2. Let R(n) be the set of symbols a + ib where a, b ∈ Z/nZ. Define + and × on R like addition and multiplication in C. a) Prove that R(n) is a ring b) Set n = 6. Identify all the zero divisors in R(6). Solution: For simplicity, we will write a + ib ∈ R(n) understanding that we mean a + ib. a) For two elements a + ib, c + id ∈ R(n), we know that (a + ib) + (c + id) = a + c + i(b + d). Since a + c, b + d ∈ Z/nZ, we know that a + c + i(b + d) ∈ R(n) so that + is a binary operation. Similarily, (a + ib) × (c + id) = ac − bd + i(bc + ad). Since we know that ac − bd, bc + ad ∈ Z/nZ it follows that ac − bd+i(bc + ad) ∈ R(n) and × is also a binary operation. We can further see that ac − bd+i(bc + ad) = ca − db + i(cb + da) = (c + id) × (a + ib) so that × is commutative. Now we check if the pair (R(n), +) is an abelian group.

5.1. INTRODUCTION TO RINGS

225

Commutativity For any two elements a + ib, c + id ∈ R(n), (a + ib) + (c + id) = a + c + i(b + d) = c + a + i(d + b) = (c + id) + (a + ib). This shows that + is commutative in R(n). Associativity Since addition is defined like addition in C, we know that it will be associative. Identity We know that 0 + i0 ∈ R(n). For any c + id ∈ R(n), (c + id) + (0 + i0) = c + 0 + i(d + 0) = c + id. Since addition is commutative, we know that this will hold when the elements are flipped so that 0 + i0 functions as the identity in our group. Inverses For any a + ib ∈ R(n), we know that −a + i(−b) ∈ R(n). So then (a + ib) + (−a + i(−b)) = a − a + i(b − b) = 0 + i0. Since + is abelian, we know that (−a + i(−b)) + (a + ib) = 0 + i0. So R(n) has inverses. This shows that the pair (R(n), +) is an abelian group. For any a + ib, c + id, e + if ∈ R(n), (a + ib) × ((c + id) × (e + if )) = (a + ib) × (ce − df + i(de + cf )) = ace − adf − bde − bcf + i(bce + bdf + ade − acf ) = (ac − bd)e − (ad + bc)f + i((ad + bc)e + (bd − ac)f ) = (ac − bd + i(ad + bc)) × (e + if ) = ((a + ib) × (c + id)) × (e + if ) Then × is associative as well. We show that × distributes over +. (a + ib) × ((c + id) + (e + if )) = (a + ib) × (c + e + i(d + f )) = (ac + ae − bd − bf ) + i(bc + be + ad + af ) = (ac − bd) + i(bc + ad) + (ae − bf ) + i(be + af ) = (a + ib) × (c + id) + (a + ib) × (e + if ) = (c + id) × (a + ib) + (e + if ) × (a + ib) = ((c + id) + (e + if )) × (a + ib). This shows that × distributes over + both ways and we conclude that R(n) is a ring. b) From part a, we seek elements of the form a + ib and c + id where ac − bd = 0 and ad + bc = 0. The zero-divisors are 2, 3, 4, 2i, 3i, 4i, 2 + i2, 2 + i4, 3 + i3, 4 + i2, 4 + i4. Exercise: 19 Section 5.1 Question: Define Hom(V, W ) as the set of linear transformations from a real vector space V to an other real vector space W . Prove that Hom(V, V ), equipped with + and ◦ (composition) is a ring. Solution: We first prove that the pair (Hom(V, V ), +) is an abelian group. Associativity Let f, g, h ∈ Hom(V, V ) and ~v ∈ V be any vector and consider (f + (g + h))(~v ) = f (~v ) + (g + h)(~v ) = f (~v ) + g(~v ) + h(~v ) = (f + g)(~v ) + h(~v ) = ((f + g) + h)(~v ). This shows that + is associative in Hom(V, V ). Identity Consider the function e where e(~v ) = ~0 ∈ V . Then e(~v + c · w) ~ = 0 = 0 + c · 0 = e(~v ) + c · e(w), ~ so e is a linear transformation and e ∈ Hom(V, V ). Then for any f ∈ Hom(V, V ) and any ~v ∈ V , (f + e)(~v ) = f (~v ) + e(~v ) = f (~v ) + 0 = f (~v ) and (e + f )(~v ) = e(~v ) + f (~v ) = 0 + f (~v ) = f (~v ) which shows that e functions as the identity.

226

CHAPTER 5. RINGS

Inverses For any f ∈ Hom(V, V ) we know that −f ∈ Hom(V, V ) so that for any ~v ∈ V we have (f − f )(~v ) = f (~v ) − f (~v ) = 0 = e(~v ) and (−f + f )(~v ) = −f (~v ) + f (~v ) = 0 = e(~v ) which shows that f − f = e = −f + f . So Hom(V, V ) has inverses. Commutativity Since addition in the vector space is commutative, for any f, g ∈ Hom(V, V ) and ~v ∈ V we have (f + g)(~v ) = f (~v ) + g(~v ) = g(~v ) + f (~v ) = (g + f )(~v ) so that addition is commutative in Hom(V, V ). This shows that the pair (Hom(V, V ), +) is an abelian group. Next we show that ◦ is associative. Let f, g, h ∈ Hom(V, V ) and consider (f ◦ (g ◦ h))(~v ) = f ((g ◦ h)(~v )) = f (g(h(~v ))) = (f ◦ g)(h(~v )) = ((f ◦ g) ◦ h)(~v ), which shows us that ◦ is associative. Now we will show that ◦ distributes both right and left over +. For any f, g, h ∈ Hom(V, V ) and ~v ∈ V consider (f ◦ (g + h))(~v ) = f ((g + h)(~v )) = f (g(~v ) + h(~v )) = f (g(~v )) + f (h(~v )) = (f ◦ g)(~v ) + (f ◦ h)(~v ) = ((f ◦ g) + (f ◦ h))(~v ). Similarily, ((g + h) ◦ f )(~v ) = (g + h)(f (~v )) = g(f (~v )) + h(f (~v )) = (g ◦ f )(~v ) + (h ◦ f )(~v ) = ((g ◦ f ) + (h ◦ f ))(~v ). This shows that ◦ distributes both right and left over +. So we have satisfied all the axioms and know that the triple (Hom(V, V ), +, ◦) is a ring. Furthermore, the identity linear transformation i(~v ) = ~v functions as the multiplicative identity in our ring. However, multiplication is not commutative in our ring. Exercise: 20 Section 5.1 Question: Let R1 and R2 be rings with nonzero identity elements. Prove that U (R1 ⊕ R2 ) ∼ = U (R2 ) ⊕ U (R2 ). Prove the equivalent result for a finite number of rings R1 , R2 , . . . , Rn . Solution: We prove that U (R1 ⊕ R2 ⊕ · · · ⊕ Rn ) ∼ = U (R1 ) ⊕ U (R2 ) ⊕ · · · ⊕ U (Rn ) since this will subsume the case with n = 2. If Ri are rings with identities all designated by 1, then the identity in R1 ⊕R2 ⊕· · ·⊕Rn is (1, 1, . . . , 1). Hence, an element (r1 , r2 , . . . , rn ) ∈ U (R1 ⊕ R2 ⊕ · · · ⊕ Rn ) if and only if there exists (s1 , s2 , . . . , sn ) ∈ R1 ⊕ R2 ⊕ · · · ⊕ Rn such that (r1 , r2 , . . . , rn )(s1 , s2 , . . . , sn ) = (r1 s1 , r2 , s2 , . . . , rn sn ) = (1, 1, . . . , 1) if and only if ri ∈ U (Ri ) for all i = 1, . . . , n. Thus, U (R1 ⊕ R2 ⊕ · · · ⊕ Rn ) ∼ = U (R1 ) ⊕ U (R2 ) ⊕ · · · ⊕ U (Rn ). Exercise: 21 Section 5.1 Question: Prove that the characteristic char(R) of an integral domain R is either 0 or a prime number. [Hint: By contradiction.] Solution: Let R be an integral domain and suppose that char(R) is finite. Then for some n, we have n · 1 = 0. Assume that n = mk is a composite number for some integers m, k ≥ 2. Then by definition we have 0 = mk · 1 = 1 + 1 + . . . + 1(mk times) = (1 + 1 + . . . + 1(m times)) + (1 + 1 + . . . + 1(m times)) + . . . . . . + (1 + 1 + . . . + 1(m times))(k times) = (1 + 1 + . . . + 1(m times))(1 + 1 + . . . + 1(k times)) = (m · 1)(k · 1).

5.1. INTRODUCTION TO RINGS

227

However, since we are in an integral domain, either (m·1) or (k ·1) must be zero. This contradicts the assumption that char(R) was n. This shows that any composite number cannot be the characteristic of an integral domain. So char(R) must be 0 or a prime number. Exercise: 22 Section 5.1 Question: Consider the ring Z ⊕ Z and consider the subset R = {(x, y) | x − y = 0}. Prove that R is a subring. Decide if R is an integral domain. Solution: Let (x1 , y1 ) and (x2 , y2 ) be in R. Then (x1 , y1 ) − (x2 , y2 ) = (x1 − x2 , y1 − y2 ) satisfies (x1 − x2 ) − (y1 − y2 ) = (x1 − y1 ) − (x2 − y2 ) = 0 − 0 = 0. Hence, by the One Step Subgroup Criterion, (R, +) is a subgroup of Z ⊕ Z with +. Also, x1 x2 − y1 y2 = y1 x2 − y1 y2 = y1 (x2 − y2 ) = 0. Hence R is closed under the component-wise multiplication in Z ⊕ Z. Thus R is a subring of Z ⊕ Z. Exercise: 23 Section 5.1 Question: Prove that a finite integral domain is a field. Solution: Let R be a finite integral domain with n elements and consider any nonzero element x ∈ R. Let A = {xi |1 ≤ i ≤ n + 1} be a set of the first n + 1 powers of x. By the pigeon-hole principle, we must have two of the same elements in A. This implies that for some powers j, k where we will assume that j < k, we have xj = xk ⇒ xk − xj = 0 ⇒ xj (xk−j − 1) = 0. Now since we are in an integral domain, one of these elements must equal zero. We know that xj 6= 0 since this would imply that we have zero-divisors. So we must have xk−j − 1 = 0 ⇒ xk−j = 1 ⇒ x(xk−j−1 ) = 1. So any nonzero element x is a unit. Since integral domains are already commutative and every nonzero element has a multiplicative inverse, R is a field. Exercise: 24 Section 5.1 Question: Let R1 and R2 be rings. Prove that R1 ⊕ R2 is an integral domain if and only if R1 is an integral domains and R2 = {0} or vice versa. Solution: (=⇒) Let R1 ⊕ R2 be an integral domain. First we show that at least one of the rings must be {0}. Assume that R1 ⊕ R2 is the direct sum of two rings which each have two or more elements. Then each ring has at least one nonzero element. Let a ∈ R1 and b ∈ R2 be nonzero elements and consider the nonzero elements (a, 0), (0, b) ∈ R1 ⊕ R2 . Then we see that (a, 0) × (0, b) = (a · 0, 0 · b) = (0, 0) so that R1 ⊕ R2 has zero divisors which is a contradiction. So at least one of the rings be trivial ({0}). Now we will show that if R2 is trivial, then R1 must be an integral domain. Since we’re in an integral domain, we know that (a, 0)(b, 0) = (b, 0)(a, 0) = (ba, 0). This implies that ab = ba which implies that R1 is commutative. For some element, (d, 0), we have (a, 0)(d, 0) = (d, 0)(a, 0) = (a, 0) for any element (a, 0). This implies that ad = a for all elements a ∈ R1 , which implies that R1 contains an identity element (in our case, d). Additionally, since (a, 0)(b, 0) = (ab, 0) 6= (0, 0) for all nonzero (a, 0), (b, 0) ∈ R1 ⊕ R2 , this implies that for all nonzero elements a, b ∈ R1 , ab 6= 0. So R1 is commutative, has an identity, and no zero divisors. So R1 is an integral domain. The argument is symmetric for when R1 is trivial. (⇐=) Let R1 be an integral domain and R2 = {0}. Then for all (a, 0), (b, 0) ∈ R1 ⊕ R2 , we have (a, 0)(b, 0) = (ab, 0) = (ba, 0) = (b, 0)(a, 0) so that multiplication is commutative. For any nonzero (a, 0), (b, 0) ∈ R1 ⊕R2 this implies that a and b are nonzero and it follows that the product ab is nonzero. This shows that for any nonzero elements, (a, 0)(b, 0) = (ab, 0) 6= (0, 0) so that R1 ⊕ R2 doesn’t contain any zerodivisors. Since 1 ∈ R1 , we know that (a, 0)(1, 0) = (1, 0)(a, 0) = (a, 0) for any (a, 0). So R1 ⊕ R2 is an integral domain. The argument is symmetric for when R1 is trivial and R2 is an integral domain.

228

CHAPTER 5. RINGS

Exercise: 25 Section 5.1 Question: Let R be a ring and suppose that x and y commute in R. Prove that for all positive integers n, n

(x + y) =

n X n i=0

i

xn−i y i .

Solution: Let n be a positive integer and x, y ∈ R with x and y commuting. Because of commutativity, when we distribute (x + y)n , every product can be written in the form xa y b . Furthermore, because of multiplications, every term will contain a total of n of the symbols x or y. Hence, a + b = n. In other words, every term in the distributed expression of (x + y)n is of the form xn−i y i for some i with 0 ≤ i ≤ n. For each i, we need to determine how many terms of the form xn−i y i . There are n expressions (x + y) in (x + y)n = (x + y)(x + y) · · · (x + y) We get a unique term xn−i y i by selecting i of the (x + y) parentheses from which to take the y element (and we select the x element from the n − i remaining expressions). There are ni ways to make this selection, because we are selecting i of the (x + y) expressions from the n of them. Hence, there are ni terms of the form xn−i y i in the distributed expression of (x + y)n . The binomial formula follows. Exercise: 26 Section 5.1 Question: Let R be a ring and suppose that x and y commute in R. Prove that for all positive integers n, xn − y n = (x − y)(xn−1 + xn−2 y + xn−3 y 2 + · · · + y n−1 ).

Solution: We will start with the right hand side of the equation. (x − y)(xn−1 + xn−2 y + xn−3 y 2 + · · · + y n−1 ) = (x − y)(

n−1 X

xn−1−i y i )

i=0

= x(

n−1 X

n−1 X

xn−1−i y i ) − y(

i=0

i=0

n−1 X

n−1 X

=(

xn−1−i+1 y i ) − (

i=0 n−1 X

=(

xn−1−i y i ) xn−1−i y i+1 )

i=0 n−1 X

xn−i y i ) − (

i=0

xn−1−i y i+1 )

i=0

= (xn +

n−1 X

xn−i y i ) − (y n +

i=1

n−2 X

xn−1−i y i+1 ).

i=0

Now we reindex the series on the left by i → i + 1 by replacing i with i + 1 everywhere in the terms and setting our counters to one lower, (xn +

n−2 X i=0

xn−(i+1) y i+1 ) − (y n +

n−2 X

xn−1−i y i+1 ) = (xn +

i=0

n−2 X

xn−1−i y i+1 ) − (y n +

i=0

= xn − y n +

xn−1−i y i+1 )

i=0 n−2 X

xn−i−1 y i−1 −

i=0 n

n−2 X

n

=x −y .

Exercise: 27 Section 5.1 Question: Prove that if a ∈ R is idempotent, then an = a for all positive integers n.

n−2 X i=0

xn−1−i y i+1

5.1. INTRODUCTION TO RINGS

229

Solution: We will prove this using induction. For our base step we will use the definition that a2 = a. Now, suppose that for all integer less than or equal to k we have ak = a. Now consider, ak+1 = (ak )a = (a)a = a2 = a. So the induction step holds and we have proven that an = a for all positive integers n. Exercise: 28 Section 5.1 Question: In the ring M2 (Z), a) find two nilpotent elements; b) find two idempotent elements that are not the identity. Solution: a) There are many nilpotent elements, here are two, when squared each becomes the zero matrix: 0 0 0 1 , . 1 0 0 0 b) There are many idempotent elements, here are two: 1 0 0 , 1 0 0 .

1 . 1

Exercise: 29 Section 5.1 Question: Prove that if R contains an identity, then all idempotent elements that are not the identity are zero divisors. Solution: Suppose that r is an idempotent element in a ring R with identity 1. Then r2 = r. Thus r2 − r = 0 so r(r − 1) = 0. Thus, if r 6= 0, then it is a zero divisor. Exercise: 30 Section 5.1 Question: Prove that if A ∈ Mn (R) is nilpotent then all of its eigenvalues are 0. [Remark: The converse is also true but follows from the Jordan Canonical Form of a matrix. See Section 10.3] Solution: Let A be a nilpotent element of Mn (R) where Ak = 0 and assume that A has a nonzero eigenvalue λ. Then for some vector ~v , we know that A~v = λ~v and more generally Ai~v = λi~v . Consider Ak~v = λk~v . However, since Ak = 0, this should be Ak~v = 0~v = ~0, which is a contradiction since λ is nonzero. So our assumption is wrong and all of A’s eigenvalues must be 0. Exercise: 31 Section 5.1 Question: Let R be a commutative ring. a) Prove that the set of nilpotent elements, N (R), is closed under addition. [Hint: Binomial formula. See Exercise 5.1.25] b) Prove that N (R) is closed under multiplication. c) Conclude that N (R) is a subring of R. Solution: a) Let x, y ∈ N (R) be any two nilpotent elements where xm = y n = 0. Now consider the element (x + y) we will show that this is nilpotent using the binomial formula from Exercise 5.1.25: m+n X m + n m+n (x + y) = xm+n−i y i . i i=0 Now, in every term of that sum either xm+n−i = 0 when m + n − i ≥ m and when m + n − i < m this implies that i ≥ n which implies that y i = 0. So every term becomes zero and we get (x + y)m+n = 0 which implies (x + y) ∈ N (R). So N (R) is closed under addition.

230

CHAPTER 5. RINGS

b) Let x, y ∈ N (R) be any two nilpotent elements where xm = y n = 0. We will assume that n ≤ m. Consider (xy)n = xn y n = xn (0) = 0. This shows that xy ∈ N (R) and that N (R) is closed under multiplication of elements. c) We need to check that the pair (N (R), +) is a group. It contains 0, inherits associativity and commutativity from R, and since it is closed under multiplication, we know that for any x ∈ N (R), −x ∈ N (R) so that x + (−x) = (−x) + x = 0. So we know that (N (R), +) is an abelian group and N (R) is closed under multiplication so it is a subring. Exercise: 32 Section 5.1 Question: Let R be a commutative ring with an identity 1 6= 0. Let x ∈ N (R). Prove that 1 − x is a unit. Solution: Let n be the integer where xn = 0. We will construct an element that is the multiplicative inverse of 1 − x. Let blog2 (n)c

u=

Y

i

(1 + x2 ).

i=0

Now, when we do the product (1 − x) × u this is equivalent to k

(1 − x)(1 + x)(1 + x2 )(1 + x4 ) . . . (1 + x2 ) where 2k is the closest power of two less than or equal to n. So then (1 − x) × u = 1 − x2 1 − 0 = 1. So 1 − x is a unit.

k+1

= 1 − xn x(2

k+1

−n)

=

Exercise: 33 Section 5.1 Question: Let R = Z/81Z. Determine the elements in N (R). In particular determine the cardinality of N (R). Solution: Suppose that for some positive integer n and x̄ ∈ Z/81Z we had x̄n ≡ 0̄. This would imply that for some multiple, k, of 81 we had xn = 81k. Now, 81 = 34 . So for any number of the form 3k for any other natural number k, we have (3k)4 = 34 k 4 = 81k 4 . So the nilpotent elements are all multiples of 3 and zero. So |N (R)| = 27. Exercise: 34 Section 5.1 Question: Let R = Z/700Z. Determine the elements in N (R). In particular determine the cardinality of N (R). Solution: If an element x̄n = 0 this implies that for some integer k, we have xn = k700. So for any prime p|700, p|x. Since 700 = 22 52 7, this implies that any multiple of 2·5·7 = 70 will be a zero divisor since (70k)2 = 4900k 2 = 7k 2 (700). So the zero divisors are all multiples of 70 and 0. So N (R) = {0, 70, 140, 210, 350, 420, 490, 560, 630} and |N (R)| = 10. Exercise: 35 Section 5.1 Question: A Boolean ring is a ring R in which r2 = r for all r ∈ R. a) Prove that the characteristic of a Boolean ring with an identity is 2. b) Prove that every Boolean ring is commutative. Solution: a) For 1 ∈ R we have 1 = (−1)2 = −1 ⇒ 2 = 0. Which implies that 2 · x = 0 · x = 0 for all x ∈ R. So the characteristic of any Boolean ring is 2. b) For x, y ∈ R, consider x + y = (x + y)2 = x2 + xy + yx + y 2 = x + xy + yx + y,

5.1. INTRODUCTION TO RINGS

231

Equating the initial expression with the last we have x + y = x + xy + yx + y ⇒ 0 = xy + yx ⇒ 2 · xy = xy + yx ⇒ xy + xy = xy + yx ⇒ xy = yx. So every Boolean ring is commutative. Exercise: 36 Section 5.1 Question: Let R be a ring and suppose that a and b are two elements such that a3 = b3 and a2 b = b2 a. Can a2 + b2 be a unit? [This exercise appeared in modified form as problem A2 on the 1991 Putnam Mathematics Competition.] Solution: We will construct an element to show that a2 + b2 is a zero divisor. Consider, (a2 + b2 )(a − b) = a3 − b3 − a2 b + b2 a = a3 − a3 − a2 b + a2 b = 0. By Proposition 5.1.12, we know that the set of zero divisors and units are mutually exclusive so that the element a2 + b2 cannot be a unit. Exercise: 37 Section 5.1 Question: Let R be a ring such that x3 = x for all x ∈ R. Prove that R is commutative. Solution: Let R be a ring such that x3 = x for all x ∈ R. Now let r, s ∈ R be arbitrary. Claim 1: For all r, s ∈ R, we have 2 · (sr − rs) = 0. We have r + s = (r + s)3 =⇒ r + s = r + r2 s + rsr + rs2 + sr2 + srs + s2 r + s =⇒ r2 s + rsr + rs2 + sr2 + srs + s2 r = 0. We then also have −r2 s − rsr + rs2 − sr2 + srs + s2 r = 0. Adding these two together, we obtain 2 · (rs2 + srs + s2 r) = 0. Then 0 = s(2 · (rs2 + srs + s2 r)) − (2 · (rs2 + srs + s2 r))s = 2 · (srs2 + s2 rs + sr − rs − srs2 − s2 rs) = 2 · (sr − rs), for all r, s ∈ R. Claim 2: For all x ∈ R, we have 6 · x = 0. From 2 · x = (2 · x)3 = 8 · x3 = 8 · x, we deduce that 6 · x = 0 for all x ∈ R. Claim 3: For all x ∈ R, we have 3 · (x + x2 ) = 0. Since x + x2 = (x + x2 )3 = x3 + 3 · x4 + 3 · x5 + x6 = x + 3 · x2 + 3 · x + x2 , after subtracting x + x2 , we have 3 · (x + x2 ) = 0. For all r, s ∈ R, by Claim 3, we have 0 = 3 · ((r + s) + (r + s)2 ) = 3 · (r + r2 + s + s2 + rs + sr) = 3 · (r + r2 ) + 3 · (s + s2 ) + 3 · (rs + sr) = 3 · (rs + sr). By Claim 2, 6 · rs = 0 so 0 = 3 · (rs + sr) − 6 · (rs) = 3 · (sr − rs). Since 2 · (sr − rs) = 0, subtracting these two we deduce that sr − rs = 0. This establishes that rs = sr for all r, s ∈ R. Hence, R is commutative. Exercise: 38 Section 5.1 Question: Prove that nk | n, k ∈ Z with k odd is a subring of Q. Solution: Let A = nk | n, k ∈ Z with k odd . We first prove that A is a subgroup of Q under addition. Let a b k , l ∈ A, then a b al − bk − = . k l kl

232

CHAPTER 5. RINGS

Now, al − bk ∈ Z and since k and l are both odd, kl is also odd so that al−bk ∈ A. So by the one step subgroup kl criteria, A is a subgroup under addition. Now we show that A is closed under multiplication. ab a b · = ∈ A. k l kl So A is a subring of Q. Exercise: 39 Section 5.1 Question: Prove that {a + bi | a, b ∈ Z} is a subring of C. Solution: Let R0 = {a + bi | a, b ∈ Z} and H = (Z, +). We will start by showing that H is a subgroup of (C, +) using the onestep subgroup criteria. Let (a + bi), (c + di) ∈ H, then (a + bi) − (c + di) = (a − c) + (b − d)i. Then since a − c ∈ Z and b − d ∈ Z, we know that (a − c) + (b − d)i ∈ H so by the one step subgroup criteria, H is a subgroup of (C, +). Now we check that R0 is closed under multiplication. (a + bi) × (c + di) = (ac − bd) + (ad + bc)i ∈ R0 . So R0 is a subring. Exercise: 40 Section 5.1 Question: Let R be any ring and let n be a positive integer. Prove that {n · r | r ∈ R} is a subring of R. Solution: Let R0 = {n · r | r ∈ R}. We will show that it is a subgroup of R by the one step subgroup criteria. Let n · r, n · q ∈ R0 . Then (n · r) − (n · q) = (n · r) + (n · −q) = n · (r − q) ∈ R0 . The last step is justified by Exercise 5.1.9 part b. So R0 is an abelian subgroup. Now we check that it is closed under multiplication. (n · r) × (n · q) = (n · r) × (q + q + . . . + q(n times)) = ((n · r) × q) + ((n · r) × q) + . . . + ((n · r) × q)(n times) = (n · (rq)) + (n · (rq)) + . . . + (n · (rq))(n times) = n · (n · (rq)) ∈ R0 . The second to last step was justified by the results of Exercise 5.1.10 part a. We can conclude that it exists in R0 since (n · (rq)) = s ∈ R so that the last line becomes n · s ∈ R0 . So R0 is closed under multiplication and we conclude that R0 is a subring. Exercise: 41 Section 5.1 Question: Determine with proof, which of the following subsets are subrings of Z ⊕ Z. a) {(a, b) ∈ Z ⊕ Z | 2a + b = 0}. b) {(a, b) ∈ Z ⊕ Z | a = b}. c) {(a, b) ∈ Z ⊕ Z | a + b is even }. d) {(a, b) ∈ Z ⊕ Z | ab = 0}. Solution: a) We find a quick counter example. Since 2 · 1 + (−2) = 0 we have that (1, −2) exists in our set. If our set was a subring, then it would be closed under multiplication. Consider (1, −2)(1, −2) = (1, 4). However, 2 · 1 + 4 = 6 6= 0. So our set is not closed under multiplication and is not a subring. b) Let A denote our set. We first show that it is a subgroup under addition. Let (a, a), (b, b) ∈ A, then (a, a) − (b, b) = (a − b, a − b) ∈ A. By the one step subgroup criteria, (A, +) is a subgroup. Now we show that it is closed under multiplication. (a, a) × (b, b) = (ab, ab) ∈ A. So A is a subring of Z ⊕ Z. c) Let B denote our set. We first show that it is a subgroup under addition. Let (a, b), (c, d) ∈ B so that a+b ≡ 0 mod 2 and c+d ≡ 0 mod 2. Then (a, b)−(c, d) = (a−c, b−d) and a−c+b−d = a+b−(c+d) ≡ 0 − 0 = 0 mod 2 so that (a − c, b − d) ∈ B. By the one step subgroup criteria, (B, +) is a subgroup. Now we show that it is closed under multiplication. (a, b) × (c, d) = (ac, bd). We consider all possibilities. In our

5.1. INTRODUCTION TO RINGS

233

set, any pair is composed of two even numbers or two odd numbers. If either pair is composed of two even numbers, then the product is two even numbers which when added together will be even and so will be a member of the set. So the only case we must check is when both pairs are composed of two odd numbers. Then the result is two odd numbers who’s sum is even and therefore is in our set. So our set is closed under multiplication of elements. So B is a subring. d) Let C denote our set. We will provide a quick counter example. We know that (0, 1), (1, 0) ∈ C. Then consider the sum (0, 1) + (1, 0) = (1, 1) 6∈ C. So C cannot be a subring. Exercise: 42 Section 5.1 Question: Determine with proof, which of the following subsets are subrings of Q. a) Fractions, which when written in reduced form, have an odd denominator. b) Fractions, which when written in reduced form, have an even denominator. c) Fractions of the form k2n , where k is odd and n ∈ Z. n d) Fractions, which when written in reduced form, are 2k . Solution: In order to check if a subset is a subring, we need to check that it is closed under subtraction and closed under multiplication. For reference c ad − bc a − = b d bd

and

a c ac × = . b d bd

a) Let R be the subset of fractions, which when written in reduced form have an odd denominator. Suppose that b and d are odd, then the denominator of the difference or product of two fractions is bd is odd and when rewritten in reduced form, the denominator is still odd. Hence R is a subring. b) Let S be the set of fractions which when written in reduced form have an even denominator. Notice that 1 1 1 6 + 6 = 3 . We see that S is not closed under addition. Hence S is not a subring. c) Let T be the set of fractions of the form k2n , where n ∈ Z. These are fractions with denominators that are powers of 2. Note that if b = 2m and d = 2n , then any divisor of bd is a divisor of 2m+n so is still a power of 2. Thus T is closed under subtraction and multiplication. Thus T is a subring. n d) Let U be the set of fractions, which when written in reduced form, are 2k . Note that 2 2 24 + = . 5 7 35 Hence U is not closed under addition. Hence U is not a subring of Q. Exercise: 43 Section 5.1 Question: Prove that the set of periodic real-valued functions of period p is a subring of Fun(R, R). [Note: Functions that are periodic with period p satisfy f (x + p) = f (x) for all x. Such functions may be periodic with a lower period or even constant.] Solution: We denote our subring by S. Let f and g ∈ S and let x be any member of R. We show S is closed under addition, (f − g)(x) = f (x) − g(x) = f (x + p) − g(x + p) = (f − g)(x + p). Then (f − g) ∈ S and by the one step subgroup criterion, S is closed under addition. Now we examine multiplication, (f · g)(x) = f (x) · g(x) = f (x + p) · g(x + p) = (f · g)(x + p). Then (f · g) ∈ S and S is closed under multiplication. So S is a subring of Fun(R). Exercise: 44 Section 5.1 Question: Let C n ([a, b], R) be the set of real-valued functions on [a, b] whose first n derivatives exist and are continuous. Prove that C n+1 ([a, b], R) is a proper subring of C n ([a, b], R). Solution: We first prove by induction that C k ([a, b], R) is a ring for all k ≥ 0. The case k = 0 was treated in the section. Now suppose that C n ([a, b], R) is a ring for all n with 0 ≤ n ≤ k. By differentiation rules, we

234

CHAPTER 5. RINGS

know that if f, g ∈ C k+1 ([a, b], R), then (f − g)(k+1) exists and is equal to f (k+1) − g (k+1) . Furthermore, as a difference of two continuous functions, f (k+1) − g (k+1) is continuous. Thus f − g ∈ C k+1 ([a, b], R). Also, by differentiation formulas, (f g)(k+1) will be a sum of products of functions of the form f (n) or g (n) with 0 ≤ n ≤ k. Since these exist and are continuous in the ring C k ([a, b], R) by the induction hypothesis, then (f g)(k+1) exists and is continuous. Hence, C k+1 ([a, b], R) is a ring as a subring of Fun([a, b], R). By induction, C k ([a, b], R) is a ring for all k ≥ 0. By rules of differentiation, if a function f is n + 1 times differentiable with f (n+1) continuous, then f (n) exists and is continuous. Thus, C n+1 ([a, b], R) ⊆ C n ([a, b], R). To see that C n+1 ([a, b], R) is a proper subring of C n ([a, b], R), consider the function f define over [−1, 1] by ( −xn+1 if x ≤ 0 f (x) = xn+1 if x > 0. Polynomials have derivatives of all orders so the only concern about continuity occurs at x = 0. For all 0 ≤ k ≤ n + 1, we have ( (n+1)! − (n+1−k)! xn+1−k if x ≤ 0 (k) f = (n+1)! n+1−k if x > 0. (n+1−k)! x For 0 ≤ k ≤ n, these functions are continuous at 0. However, if k = n + 1, then ( −(n + 1)! if x > 0 (n+1) f = (n + 1)! if x > 0 so is not differentiable or continuous at 0. Thus f ∈ C n ([a, b], R) but not it C n+1 ([a, b], R). Hence, C n+1 ([a, b], R) ( C n ([a, b], R). Exercise: 45 Section 5.1 Question: Let R be any ring and let {Si }i∈I be a collection of subrings (not necessarily finite or countable). Prove that the intersection \ Si i∈I

is a subring of R. T Solution: The additive identity 0 is in every subring Si so 0 is in S = i∈I Si . Hence, S is nonempty. Let x, y ∈ S. Then x, y ∈ Si for all i ∈ I. Since each Si is a subring of R, then x − y ∈ Si and also xy ∈ S. Thus x − y ∈ S and xy ∈ S. Hence, by definition of a subring, S is a subring of R. Exercise: 46 Section 5.1 Question: Let R be a ring and let R1 and R2 be subrings. Prove by a counterexample that R1 ∪ R2 is in general not a subring. Solution: Consider the ring R = Z with the usual + and ×. Let R1 = 2Z and R2 = 3Z. The elements 2 and 3 are in R1 ∪ R2 but 2 + 3 = 5 is not in R1 ∪ R2 . Thus, R2 ∪ R2 is not closed under + so it cannot be a subring of Z. Exercise: 47 Section 5.1 Question: Let R be a ring and let a be a fixed element of R. Define C(a) = {r ∈ R | ra = ar}. Prove that C(a) is a subring of R. Solution: Let R be a ring and let a be a fixed element of R. Define C(a) = {r ∈ R | ra = ar}. The set C(a) is not empty since 0 ∈ C(a) because a0 = 0 = 0a. Let x, y ∈ C(a). Hence, ax = xa and ay = ya. Then (x − y)a = xa − ya = ax − ay = a(x − y) and also (xy)a = xay = axy. Thus x − y ∈ C(a) so (C(a), +) is a subgroup of (R, +) by the One-Step Subgroup Criterion and xy ∈ C(a). Thus, C(a) is a subring of R. Exercise: 48 Section 5.1 Question: Let R be a ring and let a be a fixed element of R. 1. Prove that the set {x ∈ R | ax = 0} is a subring of R.

5.2. RINGS GENERATED BY ELEMENTS

235

2. With R = Z/100Z, and a = 5, find the subring defined in part (a). Solution: Let R be a ring and let a be a fixed element of R. 1. Call S = {x ∈ R | ax = 0}. The set S is nonempty since 0 ∈ S. Let x, y ∈ S. Then ax = 0 = ay. Then a(x − y) = ax − ay = 0 − 0 = 0. Hence, by the One-Step Subgroup Criterion, (S, +) is a subgroup of R. Also, a(xy) = (ax)y = 0y = 0. Thus, S is also closed under multiplication. Hence, S is a subring of R. 2. Let R = Z/100Z and a = 5. The set S = {x ∈ Z/100Z | 5x = 0} consists of elements that are “multiples” of 20, i.e., S = {0, 20, 40, 60, 80}.

5.2 – Rings Generated by Elements Exercise: 1 Section 5.2 Question: Prove that Z[i] is an integral domain. Solution: From Example 5.2.2, we do know that the Gaussian integers are a ring. So we just need to check that multiplication is commutative, it contains a one, and it has no zero divisors. Commutative For a + bi, c + di ∈ Z[i] then (a + bi)(c + di) = (ac − bd) + (ad + bc)i = (ca − db) + (da + cb)i = (c + di)(a + bi). So multiplication is commutative. Identity For any a + bi ∈ Z[i] we have (a + bi)(1 + 0i) = (1 + 0i)(a + bi) = (1a − 0b) + (0a + 1b)i = a + bi. So Z[i] has an identity. Zero-Divisors Suppose that (a + bi)(c + di) = (ac − bd) + (ad + bc)i = 0 + 0i. This implies that ac − bd = 0 ⇒ ac = bd and since ad + bc = 0 this implies that ad = −bc. First we show that these equations cannot hold for all nonzero values. By the first equation, if a and c have the same sign or different signs, then it follows that b and d have the same sign or different signs respectively. By the second equation, however, if a and c have the same sign or different signs, then it follows that b and d must have different signs or the same sign respectively. This contradicts the relations between the signs of the pairs of integers from the first equation. So no nonzero integers can satisfy both equations due to sign arguments. Now we consider what happens when one of the terms is zero. We will not consider the case when both are zero since we are looking for nonzero elements. If a is zero and b is nonzero, this forces both c and d to be zero, which contradicts the fact that both element must be nonzero. If b is zero and a is nonzero, this forces both c and d to be zero again. The argument is identical for c and d. So we have shown that no elements from the integers can satisfy both equations, so there are no zero-divisors. We have shown that Z[i] is an integral domain. Exercise: 2 Section 5.2 √ √ Question: Prove that Z[ 2, 5] consists of the following subring of R, √ √ √ {a + b 2 + c 5 + d 10 ∈ R | a, b, c, d ∈ Z}.

236

CHAPTER 5. RINGS

√ √ √ √ √ √ Write out the multiplication between a + b 2 + c 5 + d 10 and a0 + b0 2 + c0 5 + d0 10 and collect like terms. √ √ √ √ √ 10. √ It also must Solution: Since Z[ 2, 5] is closed under multiplication, it must contain 1, 2, 5, and √ 2, 5] contains contain all integer multiples of these square roots. Since it is closed under addition, then Z[ √ √ √ elements of the form a + b 2 + c 5 + d 10. The set √ √ √ S = {a + b 2 + c 5 + d 10 ∈ R | a, b, c, d ∈ Z} is closed under subtraction with √ √ √ √ √ √ √ √ √ (a + b 2 + c 5 + d 10) − (a0 + b0 2 + c0 5 + d0 10) = (a − a0 ) + (b − b0 ) 2 + (c − c0 ) 5 + (d − d0 ) 10. The product of two elements in S is √ √ √ √ √ √ (a + b 2 + c 5 + d 10)(a0 + b0 2 + c0 5 + d0 10)

√ = (aa0 + 2bb0 + 5cc0 + 10dd0 ) + (ab0 + ba0 + 5cd0 + 5dc0 ) 2 √ √ = +(ac0 + ca0 + 2bd0 + 2db0 ) 5 + (ad0 + da0 + bc0 + cb0 ) 10. √ √ This shows that S is closed under multiplication. Hence S is a subring and therefore is Z[ 2, 5].

Exercise: 3 Section 5.2 1 for some integer m. Question: Let r1 , r2 , . . . , rn ∈ Q. Prove that Z[r1 , r2 , . . . , rn ] = Z m a 1 Solution: First we show that if b ∈ Z[r1 , r2 , . . . , rn ] then b ∈ Z[r1 , r2 , . . . , rn ]. We will assume that ab is reduced to where gcd(a, b) = 1. Now, since gcd(a, b) = 1, we know that na + mb = 1 for some integers n and m. Then, 1 na + mb = b b na mb = + b b a = n · + m ∈ Z[r1 , r2 , . . . , rn ]. b Now, for each ri we will write ri = abii . We will show that Z[r1 , r2 , . . . , rn ] = Z[ b1 ,b21,...,b ]. Since r1 · r2 · · · · · rn = a1 ·a2 ·····an 1 b1 ·b2 ·····bn ∈ Z[r1 , r2 , . . . , rn ], from our earlier result we know that b1 ·b2 ·····bn ∈ Z[r1 , r2 , . . . , rn ] which implies that 1 1 Z[ b1 ·b2 ·····bn ] is a subring of Z[r1 , r2 , . . . , rn ]. Now, in Z[ b1 ·b2 ·····bn ], for any ri , ai bi b1 · b2 . . . · bi−1 · ai · bi+1 . . . bn = b1 · b2 · · · · · bn

ri =

= (b1 · b2 . . . · bi−1 · ai · bi+1 . . . bn ) ·

1 1 ∈ Z[ ]. b1 · b2 · · · · · bn b1 · b2 · · · · · bn

So that the set S = {r1 , r2 , . . . rn } is contained in the ring Z[ b1 ·b21·····bn ]. This implies that Z[r1 , r2 , . . . , rn ] is a subring of Z[ b1 ·b21·····bn ]. Since both contain one another, we must have equivalence. So letting b1 · b2 · · · · · bn = m 1 we have Z[r1 , r2 , . . . , rn ] = Z[ m ] Exercise: 4 Section 5.2 Question: Let p be a prime number. Consider the subset R of Q of fractions, which, when written in reduced form, has a denominator that is not divisible by p. Prove that R is a subring of Q. Prove that R cannot be written as Z[S] for any finite set S ⊆ Q. Solution: We first prove that the pair (R, +) is an abelian subgroup of (Q, +) using the one step subgroup criteria. Let ab , dc ∈ R and consider a c ad − bc − = . b d bd

5.2. RINGS GENERATED BY ELEMENTS

237

We know that this indeed exists in Q and we must just show that it cannot be written with a denominator of p. Now, since both b and d are not divisible by p, this implies that the product bd is not divisible by p. If the fraction reduces at all, bd can only reduce to a divisor, which shows that the fraction can not be written in ∈ R. So R is a subgroup of Q under reduced form with a denominator divisible by p, which implies that ad−bc bd addition. Now we show that it is closed under multiplication. a c ac · = ∈ R. b d bd We can see that this element is a member of R by the exact same reasoning as before. So R is closed under multiplication and is a subring of Q. Assume that for some finite set S ⊆ Q, we have R = Z[S]. We know 1 ] for some integer m from Exercise 5.2.2. Let q be a prime which does not divide m and is that Z[S] = Z[ m not p. Consider 1q which should exist in our ring since it’s denominator cannot be written in terms of p. If αk 1 a 1 α2 m = pα βk for some integer a and any combination of 1 p2 . . . pk , then everything in Z[ m ] looks like, β1 β2 p1 p2 ...pk

nonnegative integers β1 , β2 . . . βk . Since the denominator can never reduce or multiply to be q, we know that 1 1 q 6∈ Z[ m ], which is a contradiction. This implies that R 6= Z[S] for any finite set S. Exercise: 5 Section 5.2 √ Question: Prove that for all primes p, the ring Q[ p] is a field. √ √ √ Solution: Let R = Q[ p]. We note that for all even powers, p2n = ( p2 )n = pn . For all odd powers, √ 2n+1 √ 2n √ √ √ p = ( p )( p) = pn p. So every element can be written in the form a + b p for some rationals a and b. Commutativity We check that R is commutative. √ √ √ (a + b p)(c + d p) = (ac + bdp) + (ad + bc) p √ = (ca + dbp) + (da + cb) p √ √ = (c + d p)(a + b p). So R is commutative. Identity We check that R has an identity. √ √ √ (a + b p)(1 + 0 p) = (a1 + b0p) + (b1 + a0) p √ = a + b p. The identity works in both directions since multiplication is commutative. √ √ √ √ Inverses For any element, a + b p, since a − b p ∈ R, then (a + b p)(a − b p) = a2 − b2 p ∈ Q. So then √ a−b p the element a2 −b2 p will function as an inverse as long as a2 − b2 p 6= 0. We show that this cannot happen. Assume that a2 − b2 p = 0. We will write a = aa21 and b = bb21 . Then a1 2 b1 2 p − a2 b2 ⇒ a21 b22 = a22 b21 p.

a2 − b2 p =

1 2α2 Now we write a21 b22 = p2α . . . pk2αk for primes pi to even powers since the integers are squared. We also 1 p2 write a22 b21 = q12β1 q22β2 . . . qn2βn for primes qi to even powers since the integers are squared. So 1 2α2 k p2α . . . p2α = q12β1 q22β2 . . . qn2βn p. 1 p2 k

It’s easy to see that the integer on the left cannot possibly equal the integer on the right. The integer on the right has a prime that is raised to an odd power where as the integer on the left only has primes raised to an even power. Since prime factorizations are unique, they cannot be equal. This√is a contradiction and a−b p so our assumption must have been wrong and a2 − b2 p 6= 0. Then the element a2 −b2 p works as an inverse √ for any element a + b p. √ We conclude that the ring Q[ p] is a field for any prime p. Exercise: 6 Section 5.2 √ Question: Consider the ring Q[ 3 2] as a subring of R.

238

CHAPTER 5. RINGS

√ √ a) Prove that it consists of elements of the form a + b 3 2 + c( 3 2)2 with a, b, c ∈ Q. √ b) Prove that every element of the form a + b 3 2 with (a, b) 6= (0, 0) is a unit. √ Solution: Consider the ring Q[ 3 2] as a subring of R. √ √ √ √ a) The ring Q[ 3 2] must contain all the rationals and 3 2. By multiplying 3 2 with itself, we do obtain 3 2 and √ 2 3 2 , but all higher powers become powers of 2 times one of these. By multiplying by rationals we get all √ √ 2 the expressions a,√b 3 2, and c 3 2 . Since the ring is closed under addition, we see that every real number √ √ 3 of the form a + b 3 2 + c( 3√ 2)2 with a, b, c ∈ Q is in Q[ 2]. However, it is easy to see that the difference of √ √ √ 3 3 3 3 2 2 2 + c( 2) is again of the same form. Also, consider a + b 2 + c ( 2) and numbers of the form a + b 1 1 1 √ √ a2 + b2 3 2 + c2 ( 3 2)2 with coefficients in Q, we have √ √ √ √ 3 3 3 3 (a1 + b1 2 + c1 ( 2)2 )(a2 + b2 2 + c2 ( 2)2 ) √ √ 2 3 3 = (a1 a2 + 2b1 c2 + 2c1 b2 ) + (a1 b2 + b1 a2 + 2c1 c2 ) 2 + (a1 c2 + b1 b2 + c1 a2 ) 2 . √ Hence, the coefficients of the powers of 3 2 are all rationals. Hence, √ √ the set of elements of the form √ a + b 3 2 + c( 3 2)2 with a, b, c ∈ Q is a ring and thus this must be Q[ 3 2]. √ √ √ 2 b) Consider elements of the form α = a + b 3 2. Associated to it is β = a2 − ab 3 2 + b2 3 2 . By usual identities, we have √ √ √ 2 3 3 3 (a + b 2)(a2 − ab 2 + b2 2 ) = a3 − 2b3 So

√ √ 2 1 1 3 3 √ = 3 (a2 − ab 2 + b2 2 ). 3 3 a − 2b a+b 2

3 There is no pair of rationals (a, b) √ such that a3 − 2b = 0. Hence, the real number on the right consists of √ 3 3 rational numbers times powers of 2. Thus, a + b 2 is a unit.

Exercise: 7 Section 5.2 Question: Consider the ring (Z/2Z)[x]. Let α(x) = x3 + x + 1 and β(x) = x2 + 1. Calculate: a) α(x) + β(x); b) α(x)β(x); c) α(x)2 . Solution: a) α(x) + β(x) = x3 + x + 1 + x2 + 1 = x3 + x2 + x. b) α(x)β(x) = (x3 + x + 1)(x2 + 1) = x5 + 2x3 + x2 + x + 1 = x5 + x2 + x + 1. c) α(x)2 = (x3 + x + 1)2 = x6 + 2x4 + 2x3 + x2 + 2x + 1 = x6 + x2 + 1.

Exercise: 8 Section 5.2 Question: Consider the ring (Z/6Z)[x]. Let α(x) = 2x3 + 3x + 1 and β(x) = 2x2 + 5. Calculate: a) α(x) + β(x); b) α(x)β(x); c) α(x)3 . Solution: We calculate: a) (2x3 + 3x + 1) + (2x2 + 5) = 2x3 + 2x2 + 3x b) (2x3 + 3x + 1)(2x2 + 5) = 4x5 + 4x3 + 0x3 + 3x + 2x2 + 5 = 4x5 + 4x3 + 2x2 + 3x + 5.

5.2. RINGS GENERATED BY ELEMENTS

239

c) For α(x)3 , we have (2x3 + 3x + 1)3 = (2x3 + 3x + 1)(4x6 + 3x2 + 4x3 + 1) = 2x9 + 2x6 + 2x3 + 3x3 + 3x + 4x6 + 4x3 + 3x2 + 1 = 2x9 + 3x3 + 3x2 + 3x + 1

Exercise: 9 Section 5.2 Question: Calculate (2x + 3)n for all n in Z/6Z. Repeat the same question but in Z/12Z. Solution: a) We calculate some powers of 2x + 3: (2x + 3)1 = 2x + 3 (2x + 3)2 = 4x2 + 3 (2x + 3)3 = 2x3 + 3 (2x + 3)4 = 4x4 + 3. We observe that in general n

(2x + 3) =

2xn + 3 4xn + 3

: n is odd : n is even

b) We consider some powers of 2x + 3: (2x + 3)1 = 2x + 3 (2x + 3)2 = 4x2 + 9 (2x + 3)3 = 8x3 + 6x + 3 (2x + 3)4 = 4x4 + 9 (2x + 3)5 = 8x5 + 6x + 3 (2x + 3)6 = 4x6 + 9. We can see that in general:   2x + 3 4xn + 9 (2x + 3)n =  8xn + 6x + 3

:n=1 : n is even : n is odd

Exercise: 10 Section 5.2 Question: Suppose that R is a ring with identity 1 6= 0 of characteristic n. Prove that R[x] is also of characteristic n. Solution: Let a(x) = ak xk + ak−1 xk−1 + . . . + a0 ∈ R[x] be any polynomial. By Exercises 5.1.9 and 5.1.10 we know that n · (a(x)) = n · (ak xk + ak−1 xk−1 + . . . + a0 ) = n · ak xk + n · ak−1 xk−1 + . . . + n · a0 = (n · ak )xk + (n · ak−1 )xk−1 + . . . + (n · a0 ) = 0xk + 0xk−1 + . . . + 0 = 0. Which shows that R[x] has characteristic less than or equal to n. However since R is embedded in R[x] as the polynomials of degree zero this implies that n is the smallest number where n · 1 = 0. So we must have that the characteristic of R[x] is exactly n.

240

CHAPTER 5. RINGS

Exercise: 11 Section 5.2 Question: Let p be a prime number. Prove that for all a ∈ Z/pZ, the following identity holds in the ring (Z/pZ)[x], (x + a)p = xp + a. Prove that n is prime if and only if (x + a)n = xn + a in Z/nZ[x]. Solution: By Fermat’s Little Theorem, ap−1 ≡ 1 (mod p) for all prime numbers p and p - a. Multiplying by a, we get ap ≡ a when p - a. However, if p | a, then a ≡ 0 and ap ≡ 0. Hence, ap ≡ a (mod p) for all a. In (Z/pZ)[x], we have p X p p−n n p (x + a) = x a . n n=0 p! and since p, as a prime, does not appear in the factorial n! or (p − n)!. Thus p remains However, np = n!(p−n)! in the factorization of the binomial coefficient np . Thus np ≡ 0 (mod p). Hence, (x + a)p = xp + ap = xp + a, where the second equality holds because of the previous comment. Let n be a composite number with n = pα m where p is the largest prime in the prime decomposition of n and pα+1 does not divide n. Then n(n − 1)(n − 2) · · · (n − p + 1) n n! = . = p!(n − p)! p(p − 1)(p − 2) · · · 1 p In the list n, (n − 1), (n − 2), . . . , (n − p + 1), only n is divisible by p. There is a single p in the prime factorization of p! so pα−1 | np but pα - np . In particular, n - np so in the binomial expansion of (x + a)n , the xp term is not 0 for all a ∈ Z/nZ. Exercise: 12 Section 5.2 Question: Let G = S3 and let R = Z/3Z. In the group ring Z/3Z[S3 ] consider α = 1 + (1 2) + 2(1 3) + (2 3) and β = 2 + 2(1 3) + (1 2 3). Calculate: a) α + 2β; b) αβ; and c) βα. Solution: a) ; α + 2β = 1 + (1 2) + 2(1 3) + (2 3) + 2(2 + 2(1 3) + (1 2 3)) = 1 + (1 2) + 2(1 3) + (2 3) + 4 + 4(1 3) + 2(1 2 3)) = 2 + (1 2) + (2 3) + 2(1 2 3). b) αβ = (1 + (1 2) + 2(1 3) + (2 3))(2 + 2(1 3) + (1 2 3)) = 1(2 + 2(1 3) + (1 2 3)) + (1 2)(2 + 2(1 3) + (1 2 3)) + 2(1 3)(2 + 2(1 3) + (1 2 3)) + (2 3)(2 + 2(1 3) + (1 2 3)) = 2 + 2(1 3) + (1 2 3) + 2(1 2) + 2(1 3 2) + (2 3) + 4(1 3) + 4 + 2(1 2) + 2(2 3) + 2(1 2 3) + (1 3) = 6 + 4(1 2) + 7(1 3) + 3(2 3) + 3(1 2 3) + 2(1 3 2) = (1 3) + (1 2) + 2(1 3 2) c) βα = (2 + 2(1 3) + (1 2 3))(1 + (1 2) + 2(1 3) + (2 3)) = 2(1 + (1 2) + 2(1 3) + (2 3)) + 2(1 3)(1 + (1 2) + 2(1 3) + (2 3)) + (1 2 3)(1 + (1 2) + 2(1 3) + (2 3)) = 2 + 2(1 2) + 4(1 3) + 2(2 3) + 2(1 3) + 2(1 2 3) + 4 + 2(1 3 2) + (1 2 3) + (1 3) + 2(2 3) + (1 2) = 6 + 3(1 2) + 7(1 3) + 4(2 3) + 3(1 2 3) + 2(1 3 2) = (1 3) + (2 3) + 2(1 3 2).

5.2. RINGS GENERATED BY ELEMENTS

241

Exercise: 13 Section 5.2 Question: Show that the element (1 2) + (1 3) + (2 3) is in the center of the group ring Z/3Z[S3 ]. Solution: Let τ = (1 2) + (1 3) + (2 3). We first show that for any 1 · σ ∈ Z/3Z[S3 ], (1 · σ)(τ )(1 · σ −1 ) = τ . Recall that the conjugacy classes of S3 are the cycle types. Since τ is the sum of all members of a single conjugacy class of S3 , we know that under conjugation by any element σ, all members will be present after the conjugation somewhere in the sum. σ will act as a bijection on the conjugacy class and so will permute the elements of τ . Since all the coefficient in the sum are 1 and addition is abelian, we can simply reorder the elements after permutation so that τ will ramain the same. So, σ(τ )σ −1 = σ(1 2)σ −1 + σ(1 3)σ −1 + σ(2 3)σ −1 = (1 2) + (1 3) + (2 3) = τ. Which shows that for any σ, στ = τ σ. Note that τ also commutes with any member of Z/3Z since we are in a group ring. Now consider an arbitrary element an σn + an−1 σn−1 + . . . + a0 σ0 ∈ Z/3Z[S3 ]. Then (an σn + an−1 σn−1 + . . . + a0 σ0 )τ = (an σn )τ + (an−1 σn−1 )τ + . . . + (a0 σ0 )τ = an (σn τ ) + an−1 (σn−1 τ ) + . . . + a0 (σ0 τ ) = an (τ σn ) + an−1 (τ σn−1 ) + . . . + a0 (τ σ0 ) = τ (an σn ) + τ (an−1 σn−1 ) + . . . + τ (a0 σ0 ) = τ (an σn + an−1 σn−1 + . . . + a0 σ0 ). Which shows that τ commutes with any arbitrary element so that (1 2) + (1 3) + (2 3) is in the center of the group ring. Exercise: 14 Section 5.2 Question: In the ring Z[Q8 ], find (i + j)n for all positive integers n. Solution: We begin by calculating some of the powers and hope to find a pattern. We point out that in the group ring Z[Q8 ], (−1)k and (−k) (and similarly for other pairs) are different elements because k and (−k) are different elements in Q8 . (i + j)1 = i + j (i + j)2 = i2 + ij + ji + j 2 = 2(−1) + k + (−k) − 1 = 2(−1) + k + (−k) (i + j)3 = (i + j)(2(−1) + k + (−k)) = 2(−i) + (−j) + j + 2(−j) + i + (−i) = i + 3(−i) + j + 3(−j) = (i + j) + 3((−i) + (−j)) 4

(i + j) = (i + j)(i + 3(−i) + j + 3(−j)) = (i + j)2 + 3(i + j)((−i) + (−j)) = 2(−1) + k + (−k) + 3(2 + (−k) + k) = 6 + 2(−1) + 4k + 4(−k) = 6 + 2(−1) + 4(k + (−k)) 5

(i + j) = (i + j)(6 + 2(−1) + 4k + 4(−k)) = 6i + 2(−i) + 4(−j) + 4j + 6j + 2(−j) + 4i + 4(−i) = 10i + 6(−i) + 10j + 6(−j) = 10(i + j) + 6((−i) + (−j)) 6

(i + j) = (i + j)(10i + 6(−i) + 10j + 6(−j)) = 10(−1) + 6 + 10k + 6(−k) + 10(−k) + 6k + 10(−1) + 6 = 12 + 20(−1) + 16k + 16(−k) = 12 + 20(−1) + 16(k + (−k)) We make a few remarks on the patterns. • (i + j)((−i) + (−j)) = 2 + (k + (−k)); • (i+j)(k+(−k)) = (−j)+j +i+(−i) = (i+j)+((−i)+(−j)) and also (k+(−k))(i+j) = j +(−i)+(−j)+i = (i + j) + ((−i) + (−j)). • (i + j)2 ((−i) + (−j)) = 3(i + j) + ((−i) + (−j)).

242

CHAPTER 5. RINGS • all the elements are in a subring described by R = {a + b(−1) + c(i + j) + d((−i) + (−j)) + e(k + (−k)) | a, b, c, d, e ∈ Z}; • this ring R is commutative. • If n is odd, then (i + j)n = a(i + j) + b((−i) + (−j)). Furthermore, (i + j)n+2 = a(i + j)3 + b(3(i + j) + ((−i) + (−j))) = a((i + j) + 3((−i) + (−j))) + b(3(i + j) + ((−i) + (−j))) = (a + 3b)(i + j) + (3a + b)((−i) + (−j)). • So this leads to a recursive formula. If we write (i + j)2n+1 = an (i + j) + bn ((−i) + (−j)), then a0 = 1, b0 = 0 and an+1 = an + 3bn and bn+1 = 3an + bn . • Having a formula for (i + j)2n+1 gives us a formula for even powers of (i + j).

We can take this further by using techniques from linear algebra. We notice that n an 1 3 1 = . bn 3 1 0 This transition matrix has two eigenvalues of 4 and −2, with eigenvectors of 11 and the transition matrix diagonalizes to −1 1 3 1 −1 4 0 1 −1 = . 3 1 1 1 0 −2 1 1

−1 1

respectively. Hence

In particular, 1 3

3 1

n

n −1 1 −1 4 0 1 −1 = 1 1 0 (−2)n 1 1 n n 1 1 1 4 −(−2) = 4n (−2)n 2 −1 1 1 4n + (−2)n 4n − (−2)n = . 2 4n − (−2)n 4n + (−2)n

This gives

an bn

1 = 2

n 4 + (−2)n 4n − (−2)n

4n − (−2)n 4n + (−2)n

2n−1 1 4n + (−2)n 1 2 − (−2)n−1 = = 0 22n−1 + (−2)n−1 2 4n − (−2)n

This gives us, for all the odd powers, (i + j)2n+1 = (22n−1 − (−2)n−1 )(i + j) + (22n−1 + (−2)n−1 )((−i) + (−j)). For the even powers of (i + j), we can now calculate (i + j)2n = (i + j)(i + j)2n−1 = (i + j) (22n−3 − (−2)n−2 )(i + j) + (22n−3 + (−2)n−2 )((−i) + (−j)) = (22n−3 − (−2)n−2 )(i + j)2 + (22n−3 + (−2)n−2 )(i + j)((−i) + (−j)) = (22n−3 − (−2)n−2 )(2(−1) + (k + (−k))) + (22n−3 + (−2)n−2 )(2 + (k + (−k))) = (22n−2 − (−2)n−1 ) + (22n−2 + (−2)n−1 )(−1) + 22n−2 (k + (−k))

Exercise: 15 Section 5.2 Question: In the subring {a + bi + cj + dk ∈ H | a, b, c, d ∈ Z} of H, find (i + j)n for all positive integers n.

5.2. RINGS GENERATED BY ELEMENTS

243

Solution: Let R = {a + bi + cj + dk ∈ H | a, b, c, d ∈ Z}. We calculate the first few powers of the element i + j. We have i+j =i+j (i + j)2 = i2 + ij + ji + j 2 = −1 + k − k − 1 = −2 (i + j)3 = −2i − 2j (i + j)4 = 4. From (i + j)2 = −2, it is obvious that (i + j)2n = (−2)n . Then for all odd powers, (i + j)2n+1 = (−2)n (i + j). Exercise: 16 Section 5.2 Question: Let R be a commutative ring and G a group. Prove that R[G] is commutative if and only if G is abelian. Solution: (=⇒) Let R[G] be commutative. Then for any g, h ∈ G, we must have (1 · g)(1 · h) = (1 · h)(1 · g) ⇒ (1 · gh) = (1 · hg) ⇒ gh = hg. Since this must hold for all g, h ∈ G, we find that G is abelian. (⇐=) Let G be abelian. Consider the arbitrary elements α, β ∈ R[G] and suppose that together α and β only use the finite set of elements {g1 , g2 , . . . , gn } ⊆ G so that they can be written as α=

n X

ai gi

and

β=

n X

bj gj

j=1

i=1

with ai , bi ∈ R. Recall that R must be a commutative ring. Then  ! n n X X αβ = ai gi  bj gj  i=1

j=1



  n n X X ai gi  = bj gj  i=1

j=1

n X n X = (ai bj )(gi gj ) i=1 j=1 n X n X = (bj ai )(gj gi ) i=1 j=1

=

n X j=1

bj gj

n X

!! ai gi

i=1

  ! n n X X   = bj gj ai gi j=1

i=1

= βα So R[G] is commutative. This proves both directions so that for a commutative ring R, R[G] is commutative if and only if G is abelian.

Exercise: 17 Section 5.2 Question: Let R be a commutative ring with an identity 1 6= 0. Prove that G is subgroup of U (R[G]). Find an example of a ring R and a group G in which G is a strict subgroup of U (R[G])

244

CHAPTER 5. RINGS

Solution: For any g ∈ R[G] we know that g −1 ∈ R[G] as well so that (gg −1 ) = 1 = (g −1 g). Which shows that G is a subset of U (R[G]). It is obviously a subgroup so, G ≤ U (R[G]). Consider that group ring Z/3Z[S3 ]. G is a strict subgroup of U (R[G]) since 2̄ ∈ U (R[G]). Exercise: 18 Section 5.2 Question: Let R be a commutative ring and let G be a group. Prove that α is in the center of R[G] if and only if gα = αg for all g ∈ G. Solution: Let R be a commutative ring, G a group, and α an element of R[G]. (=⇒) Let α be in the center of R[G]. For any g ∈ R[G], this implies that gα = αg. Pn (⇐=) Let gα = αg for all g ∈ G. Consider an arbitrary element of R[G], i=0 ri gi . Then n n n n X X X X ( ri gi )α = (ri gi )α = ri (gi α) = ri (αgi ) i=0

i=0

=

i=0

n X

α(ri gi ) = α(

i=0

i=0

n X

ri gi ).

i=0

This shows that α is in the center of R[G]. This completes both directions of the if and only if proof. Exercise: 19 Section 5.2 Question: Let R be a commutative ring. We denote by R[[x]] the set of formal power series ∞ X

an xn

n=0

with coefficients in R. In R[[x]], we do not worry about issues of convergence. Addition of power series is performed term by term and for multiplication ! ∞ ! ∞ ∞ n X X X X X n n an x bn x = cn xn where cn = ak bn−k = ai bj . n=0

n=0

n=0

i+j=n

k=0

a) Prove that R[[x]] with the addition and the multiplication defined above is a commutative ring. b) Suppose that R has an identity 1 6= 0. Prove that 1 − x is a unit. P∞ c) Prove that a power series of n=0 an xn is a unit if and only if a0 is a unit. Solution: a) Since R is a ring, it is clear that all the coefficient will remain in R under multiplication and addition so that addition and multiplication defined on R[[x]] are binary operations. We now check that (R[[x]], +) is an abelian group. Associative Consider the sum, ∞ X n=0

an xn + (

∞ X

bn x n +

n=0

∞ X

cn xn ) =

n=0

= = =

∞ X

an xn +

n=0 ∞ X

(bn + cn )xn

n=0

(an + (bn + cn ))xn

n=0 ∞ X

((an + bn ) + cn )xn

n=0 ∞ X

(an + bn )xn +

n=0 ∞ X

=(

n=0

So addition is associative on R[[x]].

∞ X

∞ X

cn xn

n=0

an xn +

∞ X n=0

bn xn ) +

∞ X n=0

cn xn .

5.2. RINGS GENERATED BY ELEMENTS

245

Commutative Consider the sum, ∞ X

an xn +

n=0

∞ X

bn xn =

n=0

∞ X

∞ X

(an + bn )xn =

n=0

∞ X

(bn + an )xn =

n=0

∞ X

bn xn +

n=0

a n xn .

n=0

So addition is commutative in R[[x]]. Identity Consider the sum, ∞ X

an xn +

n=0

∞ X

∞ X

0xn =

n=0

(an + 0)xn =

n=0

∞ X

an xn .

n=0

Since addition is commutative, the identity works both ways. P∞ P∞ Inverses For any n=0 an xn , we claim that n=0 −an xn is the inverse element. We will only check one direction since we already know that addition is commutative in R[[x]]. ∞ X

an xn +

n=0

∞ X

∞ X

−an xn =

n=0

(an + −an )xn =

n=0

∞ X

(0)xn .

n=0

So R[[x]] has additive inverses. This shows that the pair (R[[x]], +) is an abelian group. Now we check that multiplication is associative. ! ∞ ∞ ∞ ∞ ∞ X X X X X n n cn xn ) dn xn ) · ( cn x n ) = ( bn x ) · ( an x )( ( n=0

n=0

n=0

n=0

n=0

=

∞ X

en xn .

n=0

Where dn =

X

X

ai bj and en =

i+j=n

(

X

X

ai bj )cl =

k+l=n i+j=k

X

(ai bj cl )

k+l=n i+j=k

X

=

(ai bj cl )

i+j+l=n

X

=

ai

i+r=n

X

(bj cl ).

j+l=r

Plugging this formula for the coefficients back into the product we get,   ∞ ∞ X X X X  ai (bj cl ) xn en xn = n=0

n=0

i+r=n

j+l=r

 ! ∞ ∞ X X X an xn ·  (bj cl )xn 

=

n=0

=

∞ X

n=0 j+l=n ∞ X

n

an x ·

n=0

n

bn x ·

n=0

∞ X

! cn x

n

n=0

So that multiplication is associative. Now we show that multiplication is commutative. We know that ! ∞ ! ∞ ∞ X X X X X n n an x bn x = cn x n where cn = ai bj = bj ai n=0

n=0

⇒

n=0 ∞ X

i+j=n

cn x n =

n=0

=

∞ X

∞ X n=0

(

X

bj ai )xn

n=0 j+i=n

! bn x

n

∞ X n=0

! an x

n

.

j+i=n

246

CHAPTER 5. RINGS So multiplication is commutative. Now we show that multiplication distributes over addition from the left and it follows, since multiplication is commutative, that it distributes over from the right. ! ∞ ∞ ∞ ∞ ∞ X X X X X n n n ( an x ) · ( bn x ) + ( cn x ) = ( an xn ) · ( (bn + cn )xn ) n=0

n=0

n=0

n=0

=

=

=

n=0

∞ X X

(ai (bj + cj ))xn

n=0 i+j=n ∞ X X n=0 i+j=n ∞ X X

(ai bj + ai cj )xn (ai bj )xn +

∞ X X

(ai cj )xn

n=0 i+j=n n=0 i+j=n ∞ ∞ ∞ ∞ X X X X =( an xn ) · ( bn x n ) + ( an xn ) · ( cn xn ). n=0 n=0 n=0 n=0

This shows that multiplication distributes over addition both ways and we conclude that R[[x]] with addition and multiplication defined as above is a commutative ring. P∞ b) Let n=0 anP xn be the element where a0 = 1 and a1 = −1 and an = 0 for all other n. We will show that ∞ the element n=0 xn is it’s inverse. Note that we only need to show it functions as an inverse on one side since multiplication is commutative. So ∞ X

(

an xn ) · (

n=0

Now

∞ X

xn ) =

n=0

∞ X X ( ai 1)xn n=0 i+j=n

P

i+j=n ai equals 1 when n = 0 and 0 when n ≥ 1. So that

(1 − x)(

∞ X

xn ) = 1.

n=0

P∞ P∞ P∞ c) (=⇒) P Suppose that P xn is a unit. This implies that for some n=0 bn xn , we have ( n=0 an xn ) · n=0 anP ∞ ∞ n ( n=0 bn xn ) = = 1. In particular, for the zeroth term this implies that n=0 ( i+j=n ai bj )x a0 b0 = 1 which shows that a0 is a unit. −1 P (⇐=) Suppose that a0 is a unit. Let b0 = a−1 0 and let bn = −a0 ( i+j=n−1 ai+1 bj ) for n ≥ 1. We will prove this works as an inverse. We check that the zeroth coefficient of the product of the two series is the identity 1, a0 · b0 = a0 · (a−1 0 ) = 1. Now we show that the nth coefficient is 0 for n ≥ 1. The formula for the nth coefficient is X X ai bj = a0 bn + ai+1 bj i+j=n

i+j=n−1

X

= a0 (−a−1 0

i+j=n−1

X

=−

X

ai+1 bj ) +

ai+1 bj +

i+j=n−1

ai+1 bj

i+j=n−1

X

ai+1 bj = 0.

i+j=n−1

P∞ n So P∞that alln other coefficients are zero. This shows that n=0 bn x is the inverse element and that a x is a unit. This completes both directions of the if and only if statement. n=0 n Exercise: 20 Section 5.2 Question: Consider the power series ring Q[[x]]. (See Exercise 5.2.19.) P∞ a) Suppose that c0 is a nonzero square element. Prove that that there exists a power series n=0 an xn such that !2 ∞ ∞ X X n an x = cn xn . n=0

n=0

5.2. RINGS GENERATED BY ELEMENTS

247

b) Find a recurrence relation for the terms an such that !2 ∞ X n an x = 1 + x. n=0

Solution: Consider the power series ring Q[[x]]. a) By using the product expression of power series, if ∞ X

!2 an x

n

n=0

then

=

∞ X

cn xn .

n=0

( P (n−1)/2 2 k=0 an−k ak cn = an−k ak = Pn/2−1 an/2 + 2 k=0 an−k ak k=0 n X

if n is odd if n is even.

Suppose that ci are arbitrary with the one condition that c0 is a nonzero square of a rational number. We wish to solve for the ai in terms of the ci . The condition that c0 is a square, allows us first to solve the √ a0 = ± c0 . Then c1 = 2a1 a0 . Since a0 6= 0, then a1 = c1 /(2a0 ). We prove by induction that ai exists and is well-defined for all integers i. We have already established a bases case. Now suppose that ai exists and is well-defined for all i < n. If n is odd then   (n−1)/2 X 1  an−k ak  . cn − 2 an = 2a0 k=1

If n is even then

 an =

1  cn − an/2 − 2 2a0

n/2−1



X

an−k ak  .

k=1

Thus, an exists andP is well-defined for all n. Note that the sign choice for a0 leads to two different ∞ n possibilities for the n=0 an x power series. However, the choice of sign uniquely determines all the remaining values of an . Exercise: 21 Section 5.2 Question: Let R be a ring and let X be a set. Prove that the set Fun(X, R) of functions from X to R is a ring with the addition and multiplication of functions defined by (f1 + f2 )(x) (f1 f2 )(x)

def

= f1 (x) + f2 (x)

def

(5.1)

= f1 (x)f2 (x).

Solution: We see from the definitions that the addition and multiplication of two functions remains a function from X to R so that both operations are binary operations. We will prove that (Fun(X, R), +) is an abelian group under addition. Associativity (f1 + (f2 + f3 ))(x) = f1 (x) + (f2 + f3 )(x) = f1 (x) + (f2 (x) + f3 (x)) = (f1 (x) + f2 (x)) + f3 (x) = (f1 + f2 )(x) + f3 (x) = ((f1 + f2 ) + f3 )(x). Commutativity We rely on the fact that addition is abelian within R. (f1 + f2 )(x) = f1 (x) + f2 (x) = f2 (x) + f1 (x) = (f2 + f1 )(x).

248

CHAPTER 5. RINGS

Identity Let i ∈ Fun(X, R) be the function where i(x) = 0 for every x ∈ X. Then for any f ∈ Fun(X, R), (f + i)(x) = f (x) + i(x) = f (x) + 0 = f (x). The identity holds for both directions since addition is commutative. Inverses Let f ∈ Fun(X, R) and define g : X → R to be g(x) = −f (x) for all x ∈ X. It is obvious that g ∈ Fun(X, R). So, (f + g)(x) = f (x) + g(x) = f (x) − f (x) = 0 = i(x). The inverse holds for both sides because of commutativity. So Fun(X, R) has inverses. This shows that the pair (Fun(X, R), +) is an abelian group. Now we check associativity for multiplication. So (f (gh))(x) = f (x)(gh)(x) = f (x)(g(x)h(x)) = (f (x)g(x))h(x) = (f g)(x)h(x) = ((f g)h)(x). So multiplication is associative. Now we check if multiplication is left distributive over addition, (f (g + h))(x) = f (x)(g + h)(x) = f (x)(g(x) + h(x)) = f (x)g(x) + f (x)h(x) = (f g)(x) + (f h)(x) = (f g + f h)(x), and right distributive over addition, ((f + g)h)(x) = (f + g)(x)h(x) = (f (x) + g(x))h(x) = f (x)h(x) + g(x)h(x) = (f h)(x) + (gh)(x) = (f h + gh)(x). So both hold and we find that Fun(X, R) is a ring. Exercise: 22 Section 5.2 Question: Let R be a ring and let X be a set. The support of a function f ∈ Fun(X, R) is the subset Supp(f ) = {x ∈ X | f (x) 6= 0}. Consider the subset Funf s (X, R) of functions in Fun(X, R) that are of finite support, i.e., that are 0 except on a finite subset of X. Prove that Funf s (X, R) is a subring of Fun(X, R) as defined in Exercise 5.2.21. Solution: We first show that Funf s (X, R) is a subgroup under addition of Fun(X, R). We will use the one step subgroup criteria. Let f, g ∈ Funf s (X, R) and let K and H be the finite subsets that f and g are nonzero on respectively. First we make sure that −g ∈ Funf s (X, R). It is clear that −g is only nonzero on H. So −g ∈ Funf s (X, R). Now consider the sum (f − g). Now, everywhere that f and −g are both zero, then (f − g) is also zero. The contrapostitive statement is that f − g can only be nonzero on either of the sets where f or −g are nonzero. The set that (f − g) is nonzero on is a subset of H ∪ K (which is finite as well). This implies that (f − g) ∈ Funf s (X, R). By the one step subgroup criteria, Funf s (X, R) ≤ Fun(X, R). Now we show that Funf s (X, R) is closed under multiplication. Consider (f g)(x) = f (x)g(x). Now, if either f or g is zero at x, then (f g)(x) = 0. This implies that (f g) is nonzero exactly on the set K ∩ H which must be finite since both K and H are finite. This implies that (f g) ∈ Funf s (X, R). So it is closed under multiplication and we see that Funf s (X, R) is a subring of Fun(X, R).

5.3 – Matrix Rings Exercise: 1 Section 5.3 Question: In M2 (Z/4Z), consider the matrices 1 2 0 A= , B= 2 3 1

1 , 1

C=

2 3

2 1

Perform the following calculations, if they are defined: 1) A + BC; 2) ABC; 3) B n for all n ∈ N; 4) C −1 ; 5) A−1 B. Solution: (We omit the bars over top of the integers and understand that the notation represents congruence classes.) Matrix calculations:

5.3. MATRIX RINGS

249

a) A + BC: A + BC =

1 2

2 3 + 3 1

1 2

2 3 3 1

1 4 = 3 3

3 2

b) ABC:

1 1 = 3 1

3 3

c) For B n , we calculate a few powers and hope to find a pattern 0 1 0 1 1 2 B = = 1 1 1 1 1 0 1 1 1 1 B3 = = 1 1 1 2 2 0 1 1 2 2 B4 = = 1 1 2 3 3 0 1 2 3 3 B5 = = 1 1 3 1 1 0 1 3 1 1 B6 = = 1 1 1 0 0

1 2 2 3 3 1 1 0 0 1

ABC =

We now see that B is in the finite group U (M2 (Z/4Z)) and has order 6. Every power of B will correspond to one of these, depending on the remainder of n when divided by 6. d) We notice that det C = 0 so C is not invertible and C −1 does not exist. e) A−1 B: 0 1 2 1 2 3 −1 −1 3 2 A B=3 =3 = 2 1 1 1 1 3 3 1 Exercise: 2 Section 5.3 Question: Repeat Exercise 5.3.2, but with the ring of coefficient in Z/5Z. Solution: Matrix calculations: a) A + BC: 1 2 3 1 4 3 A + BC = + = 2 3 0 3 2 1 b) ABC:

1 3 = 3 1

2 1

c) For B n , we calculate a few powers and hope to find a pattern 0 1 0 1 1 2 B = = 1 1 1 1 1 0 1 1 1 1 B3 = = 1 1 1 2 2 0 1 1 2 2 B4 = = 1 1 2 3 3 0 1 2 3 3 B5 = = 1 1 3 0 0

1 2 2 3 3 0 0 3

ABC =

1 2

2 3 3 0

Now, with the powers of B that we have we can describe all possible powers. We note that since B 5 = 3I, this implies that B 25 = (B 5 )5 = (3I)5 = 35 I = I and B is a unit. Every time we reach a power that is a multiple of 5, we can multiply the scalar factor by 3 and start over with our powers. This implies that n

B n = 3b 5 c B n̄ mod 5 .

250

CHAPTER 5. RINGS

d) We notice that det C = 1 so C is invertible. Then 1 −1 C =1· 2

3 2

1 = 2

3 . 2

e) A−1 B: A−1 B = 4−1

3 3

3 1

0 1

1 1

=4

3 1

1 2 = 4 4

4 1

Exercise: 3 Section 5.3 Question: Find the inverse of the matrix in Example 5.3.5. Solution: The matrix in question is in R = Z/6Z and is  2 3 A = 1 0 4 2

 5 3 . 1

There are two strategies taught in linear algebra for how to find the inverse of a matrix. One uses a cofactor formula, whereas the other is an application of the reduced row echelon form. We use the latter. Recall that if A is invertible, then rref(A I) = (I A−1 ). Using technology that relies on R, we calculate     1 0 0 −6/31 7/31 9/31 2 3 5 1 0 0 rref 1 0 3 0 1 0 = 0 1 0 11/31 −18/31 −1/31 . 0 0 1 2/31 8/31 −3/31 4 2 1 0 0 1 All of the operations performed in the Gauss-Jordan elimination live entirely inside a ring, except possibly when it is necessary to divide by a nonunit. However, the above calculation shows that even in the integers,    2 3 5 −6 7 9 1 0 3  11 −18 −1 = 31I. 4 2 1 2 8 −3 Reducing this equality modulo 6, we find that in Z/6Z (where we did not write the bars over the numbers to indicate congruence classes)    0 1 3 2 3 5 1 0 3 5 0 5 = I. 2 2 3 4 2 1 Thus the inverse of A with coefficients in the ring R is the second matrix in the equality. Exercise: 4 Section 5.3 Question: Consider the following matrices in M2 (H): 1 i i+j A= , B= j k i

k . j

Calculate: (a) A + B; (b) AB; (c) B 3 . Solution: a) For A + B, we simply have A+B =

1+i+j i+j

i+k . j+k

b) For AB, we have AB =

1 j

i k

i+j i

k j

=

i + j + ik j(i + j) + ki

k + ij ji + kj

=

i −1 + j − k

2k . −i − k

5.3. MATRIX RINGS

251

c) For B 3 , we first calculate i+j k i+j 2 B = i j i

k j

=

(i + j)2 + ki i(i + j) + ji

(i + j)k + kj ik + j 2

−2 + j = −1

−j , −1 − j

and then 3

i+j i

−1 − 2i − 2j −2i − j + k

B = =

k j

−2 + j −1

−j −1 − j 1 + i − 2k . 1−j−k

(i + j)(−2 + j) − k = i(−2 + j) − j

−(i + j)j − k(1 + j) −ij − j(1 + j)

Exercise: 5 Section 5.3 Question: Let S be a subring of R. Prove that Mn (S) is a subring of Mn (R). Solution: Let S be a subring of a ring R. Let A, B ∈ Mn (S) ⊂ Mn (R). Then the entries of C = A − B are cij = aij − bij . Since S is closed under subtraction, then cij ∈ S so C ∈ Mn (S). This shows that (Mn (S), +) is a subgroup of (Mn (R), +). Furthermore, if D = AB, then the components are dij =

n X

aik bkj .

k=1

Now S is closed under multiplication, so for all i, j, k, the product aik bkj ∈ S. Since S is closed under addition, then dij ∈ S for all i, j. Thus D ∈ M( S). Hence, Mn (S) is closed under multiplication. Therefore, Mn (S) is a subring of Mn (R). Exercise: 6 Section 5.3 Question: Prove that the subset of upper triangular matrices in Mn (R) is a subring. Solution: Let H = {A = (aij ) ∈ Mn (R) aij = 0 when i < j} denote the set of all upper trianglar matrices. We first prove that H is a subgroup under addition, consider the two arbitrary matrices, A = (aij ), B = (bij ) ∈ H. Then A − B = A + (−B) = (aij − bij ). Now, when i < j this implies that aij − bij = 0 − 0 = 0 which implies that A − B ∈ H. So by the one step subgroup criteria, (H, +) is a subgroup of (Mn (R), +). Now we show that it is closed under multiplication. The (i, j)th entry of AB is defined as n X

aik bkj .

k=1

Assume that i < j. Then while k < j, we know that bkj = 0. When k ≥ j, then k > i which implies that aik = 0. So the (i, j)th term of AB whenever i < j is n X k=1

aik bkj =

j−1 X k=1

aik · 0 +

n X

0 · bkj = 0 + 0 = 0.

k=j

We note that when j = 1, the lefthand summation goes away since it is undefined, however the sum still equals zero since the righthand summation covers all of the sums and products. This implies that AB ∈ H and H is closed under multiplication and so is a subring. Exercise: 7 Section 5.3 Question: Prove that the subset of diagonal matrices in Mn (R) is a subring.

252

CHAPTER 5. RINGS

Solution: Let H = {A = (aij ) ∈ Mn (R) aij = 0 when i 6= j} denote the set of all diagonal matrices. We first prove that H is a subgroup under addition, consider the two arbitrary matrices, A = (aij ), B = (bij ) ∈ H. Then A − B = A + (−B) = (aij − bij ). Now, when i 6= j this implies that aij − bij = 0 − 0 = 0 which implies that A − B ∈ H. So by the one step subgroup criteria, (H, +) is a subgroup of (Mn (R), +). Now we show that it is closed under multiplication. The (i, j)th entry of AB is defined as n X

aik bkj .

k=1

Let i 6= j and assume that for some k, we have aik bkj 6= 0. This implies that aik 6= 0 and bkj 6= 0. This means that i = k and k = j which implies that i = j and contradicts the fact that i 6= j. So whenever i 6= j we know that the product aik bkj = 0 for every k and the (i, j)th entry is n X

aik bkj = 0.

k=1

This implies that AB ∈ H. So H is closed under multiplication and so is a subring. Exercise: 8 Section 5.3 Question: Consider the ring Mn (Z). Consider the subsets S of upper triangular matrices A = (aij ) in which 2j−i divides aij for all indices (i, j) with j ≥ i. Prove that S is a subring of Mn (Z). Solution: Let S = {(aij ) where 2j−i |aij when i ≤ j otherwise aij = 0} be a subset of the upper triangular matrices. By Exercise 5.3.7, we know that any combination of sums and products of matrices from this set will remain in the set of upper triangular matrices since it is a subring. This implies that we only need to check that the extra properties of divisibility hold under sums and products. We will show that K is a subgroup under addition. Let B = (bij ), A = (aij ) ∈ S and consider the sum A + (−B) = (aij − bij ). Now, when i ≤ j, then we know that aij = 2j−i k and bij = 2j−i l for some integers l and k. Then the (i, j)th entry of A + (−B) is 2j−i k − 2j−i l = 2j−i (k − l) and is divisible by 2j−i . We know that it will also remain upper triangular by Exercise 5.3.7 so that A + (−B) ∈ S. By the one step subgroup criteria, S is a subgroup under addition of Mn (Z). Now we check that S is closed under multiplication. The (i, j)th term of AB is defined as n X

aik bkj .

k=1

Now, we can split this sum up into three sums, n X

aik bkj =

k=1

i−1 X

aik bkj +

k=1

j X

n X

aik bkj +

k=i

aik bkj .

k=j+1

In the first summation, we have that k < i, so aik = 0. Similarily, in the third sum, we have that j < k, so that bkj = 0. In the middle sum, we have i ≤ k ≤ j, so that 2k−i |aik and 2j−k |akj so we can rewrite the product aik bkj as 2k−i maik 2j−k lbkj for some integers maik and lbkj which are indexed by the terms they came from. So we rewrite the (i, j)th term as i−1 X k=1

aik bkj +

j X k=i

aik bkj +

n X

aik bkj =

k=j+1

i−1 X

0bkj +

k=1

=

j X

j X

2k−i maik 2j−k lbkj +

k=i

2k−i+j−k maik lbkj

k=i

=

j X

2j−i maik lbkj

k=i

= 2j−i

j X k=i

maik lbkj .

n X k=j+1

aik 0

5.3. MATRIX RINGS

253

We can now see that 2j−i divides the (i, j)th term of AB when i ≤ j. We know that the product AB will still be upper triangular from the previous exercise. So AB satisfies all the properties and the product AB ∈ S. So S is closed under multiplication and is a subring of Mn (Z). Exercise: 9 Section 5.3 Question: (Multivariable calculus required) Recall the Hessian matrix of a real-valued function f (x, y) defined on an open set D ⊆ R2 . In what algebraic structure does the Hessian matrix of f exist? Solution: The Hessian matrix of a real-valued function f defined over an open set D ⊆ R2 is fxx (x, y) fxy (x, y) . fyx (x, y) fyy (x, y) (The determinant of the Hessian is used in the Second Derivative Test for functions over R2 .) The assumption is that these second derivatives are continuous so the Hessian matrix is an element in M2 (C 0 (D, R)). Exercise: 10 Section 5.3 Question: Suppose that R is a ring with identity 1 6= 0. Prove that the center Z(Mn (R)) = {aIn | a ∈ Z(R)} and In is the n × n identity matrix. Give an example where this result fails if R does not have an identity. Solution: Let C represent a matrix that we believe to be in the center of Mn (R). Let Aij represent a matrix with zeroes in every entry but the (i, j)th entry which holds 1. Then we examine the (i, d)th entry (where d 6= j) of both Aij C and CAij . Since C is in the center, the entries should be the same. So we should have n X

aik ckd =

k=1

n X

cik akd

k=1

Since all the akd terms on the righthand side are 0 and most of the terms on the left are zero we end up with cjd = 0 where j 6= d. Since the choices are arbitrary, we know that for any j, d where j 6= d we need cjd = 0 so that C can only have elements on the diagonal. Consider B be a matrix which has an arbitrary element on the (d, d)th entry and zeroes every where else. We consider the (d, d)th entry of BC and CB which should be equal since C is in the center. n X

bdk ckd =

k=1

n X

cdk bkd

k=1

⇒ bdd cdd = cdd bdd . Now, since the element bdd could be any element of R and this equality must hold, this forces cdd to be in the center of R for any element along the diagonal. So we have shown that it is necessary for a matrix in the center of Mn (R) to only have elements from the center of R along the diagonal. Now we will show that every element along the diagonal must be the same. Let C be a matrix which only has elements from the center of R along the diagonal. We will examine the (i, j)th entry of Aij C and CAij which must be equal. Then, n X

aik ckj =

k=1

n X

cik akj

k=1

Since, whenever k 6= j on the left hand side, ckj = 0 and likewise whenever k 6= i on the righthand side we are left with the equation, aij cjj = cii aij ⇒ cii = cjj . So all the element along the diagonal must be equal. Let C be a matrix of the form cI where c ∈ Z(R). We will now show that C is in the center. Let A be any arbitrary matrix. We will show that the (i, j)th entry of AC and CA are the same. n X k=1

aik ckj =

n X

cik akj

k=1

⇒ aij cjj = cii aij ⇒ caij = caij .

254

CHAPTER 5. RINGS

So the (i, j)th entries are equal for any i and j which implies that AC = CA so that C is in the center. This shows that the conditions are both necessary and sufficient and completes the proof. Now we give an example where the result P fails when R does not have an identity. Let R = {0, a} have charn acteristic two and a2 = 0. Then the sum k=1 aik bkj is always zero. This implies that for any two matrices, A, B ∈ Mn (R) we have AB = 0 = BA. So that every possible matrix is in the center of the ring. Exercise: 11 Section 5.3 Question: Let R be a commutative ring. Prove that the determinant is “linear by row.” In other words prove that 

a11  a 21   ..  .  det  rai1 + sa0i1  ..   . an1

a12 a22 .. . rai2 + sa0i2 .. . an2

··· ··· .. . ··· .. . ···

  a1n a11   a21 a2n     . ..   . .   . = r det   rain + sa0in   ai1   ..   ..   . . ann an1

a12 a22 .. . ai2 .. . an2

··· ··· .. . ··· .. . ···

  a1n a11   a21 a2n    . ..    . .   .  + s det  0 ain   ai1   ..   ..  . .  ann an1

a12 a22 .. . a0i2 .. . an2

··· ··· .. . ··· .. . ···

 a1n a2n   ..   .  . a0in   ..  .  ann

Solution: Call A0 the matrix where the ith row of A has a0ij instead and call B the matrix where the ith row of A, instead of being aij is written as raij + sa0ij . Then X det B = (sign σ)a1σ(1) a2σ(2) · · · (raij + sa0ij ) · · · anσ(n) σ∈Sn

so by distributivity X = (sign σ)(a1σ(1) a2σ(2) · · · (raij ) · · · anσ(n) + a1σ(1) a2σ(2) · · · (sa0ij ) · · · anσ(n) σ∈Sn

by commutativity with r and s and separating the sums X X = (sign σ)ra1σ(1) a2σ(2) · · · aij · · · anσ(n) + (sign σ)sa1σ(1) a2σ(2) · · · a0ij · · · anσ(n) σ∈Sn

σ∈Sn

then by distributivity ! =r

X

(sign σ)a1σ(1) a2σ(2) · · · aij · · · anσ(n)

! +s

σ∈Sn

X

(sign σ)a1σ(1) a2σ(2) · · · a0ij · · · anσ(n)

σ∈Sn

= r(det A) + s(det A0 ). This shows that the determinant function is linear in each row. Exercise: 12 Section 5.3 Question: Let R be a commutative ring with an identity 1 6= 0. Let σ ∈ Sn and defined the matrix Eσ as the n × n matrix with entries (eij ) such that ( 1 if j = σ(i) eij = 0 otherwise. Prove that det Eσ = sign σ. Solution: We use the definition of determinant given in the text. X det Eσ = (sign τ )e1τ (1) e2τ (2) . . . enτ (n) . τ ∈Sn

Now for any τ ∈ Sn where τ 6= σ, this implies that for some 1 ≤ i ≤ n , we have τ (i) 6= σ(i), which implies that aiτ (i) = 0 and that entire term becomes 0 in the summation. Since this holds for all permutations that are not sigma we see that X det Eσ = (sign τ )e1τ (1) e2τ (2) . . . enτ (n) τ ∈Sn

= (sign σ)e1σ(1) e2σ(2) . . . enσ(n) .

5.3. MATRIX RINGS

255

Since every term in this product is one, we get det Eσ = sign σ. Exercise: 13 Section 5.3 Question: In the ring M2 (Q[x]), prove that the following matrix is invertible and find the inverse: x+1 x−2 . x+6 x+3 Solution: Let M represent our matrix. We first calculate the determinant to see if it belongs to U (Q[x]). So det M = sign(1)a11 a22 + sign(1 2)a12 a21 = 1(x + 1)(x + 3) − 1(x − 2)(x + 6) = x2 + 4x + 3 − x2 − 4x − 12 = −9 ∈ U (Q[x]). So it is invertible. Now we calculate the (i, j)th entries of the inverse using the formula. (i, j) = (detM )−1 (−1)i+j det(Mj i) −1 −x − 1 (1, 1) = (1)(x + 1) = 9 9 −1 x+6 (1, 2) = (−1)(x + 6) = 9 9 −1 x−2 (2, 1) = (−1)(x − 2) = 9 9 −1 −x − 3 (2, 2) = (1)(x + 3) = . 9 9 So the inverse is M

−1

−x−1 =

9 x−2 9

x+6 9 −x−3 9

.

Exercise: 14 Section 5.3 Question: Let a, b ∈ Mn (R), where R is a ring. a) Prove that (AB)> = B > A> if R is commutative. b) Prove that this identity does not necessarily hold if R is not commutative. Solution: a) Let R be a commutative ring. Let A and B be matrices. Consider the (i, j)th entry of the transpose of the product, (AB)> which is the (j, i)th entry of the product, n X

ajk bki .

k=1

Now we examine the (i, j)th entry of the product of the transposes, or B > A> , we denote the entries of B > and A> with b0 and a0 respectively, n X

b0ik a0kj =

k=1

=

n X k=1 n X

bki ajk ajk bki .

k=1

We see that the last sum is equivalent to the (i, j)th entry of (AB)> which shows that B > A> = (AB)> .

256

CHAPTER 5. RINGS

b) If R is not commutative consider the matrices B and Aij which has zeroes everywhere except the (i, j)th entry which is an arbitrary element. Then the (j, i)th entry of (AB)> is the (i, j)th entry of the product which is n X

aik bkj = aij bjj .

k=1

Now the (j, i)th element of B > A> is n X

b0jk a0ki =

k=1

n X

bkj aik

k=1

= bjj aij . Now in order for (AB)> to equal B > A> we need the (j, i)th entries to be equal, or bjj aij = aij bjj , which does not hold for some elements if R is not commutative. So the identity does not hold. Exercise: 15 Section 5.3 Question: Recall that the trace Tr : Mn (R) → R of a matrix is the sum of its diagonal elements. Let R be a commutative ring. Prove that Tr(AB) = Tr(BA) for all matrices A, B ∈ Mn (R). Solution: Let R be a commutative ring. Let A ∈ Mn (R), we will define T r(A) =

n X

aii .

i=1

Then T r(AB) =

n n X X aik bki ) ( i=1 k=1

n X n X = ( aik bki )

=

k=1 i=1 n n X X

(

bki aik )

k=1 i=1

= T r(BA).

Exercise: 16 Section 5.3 Question: Let R be any ring. Suppose that A ∈ Mn (R) is strictly upper triangular (upper triangular but with zeros on the diagonal). Prove that A is nilpotent. Prove that if B ∈ GLn (R), then BAB −1 is also nilpotent. Solution: The matrix A = (aij ) has the property that aij = 0 if j ≤ i. We prove by induction on k that if Ak = (cij ), then cij = 0 if j ≤ i + k − 1. By definition of strictly upper triangular, this statement is true for the base case of k = 1. Suppose that the hypothesis is true for a given k ≥ 1. Then Ak+1 = Ak A and so its ijth entry is n X dij = ci` a`j . `=1

Consider the entries dij with j ≤ i + k. By the induction hypothesis, if ` ≤ i + k − 1, then ci` = 0 so ci` a`j = 0. Thus n X dij = ci` a`j . `=i+k

But a`j = 0 if j ≤ `. Since the summation has i + k ≤ `, we deduce that when j ≤ i + k, then j ≤ ` so a`j = 0 and thus dij = 0. Hence, we have shown that dij = 0 for j ≤ i + k. By induction, we have shown that if Ak = (cij ), then cij = 0 if j ≤ i + k − 1. In particular, since j ≤ i + n − 1 for all i ≥ 1, we conclude that An = 0. This proves that A is nilpotent.

5.3. MATRIX RINGS

257

Note that if Ak = 0, the zero matrix, then (BAB −1 )k = BAk B −1 = B0B −1 = 0. Therefore, any matrix similar to a nilpotent matrix is again nilpotent.

Exercise: 17 Section 5.3 Question: Prove Proposition 5.3.12. Solution: We wish to prove the following proposition. Let R be a commutative ring with an identity 1 6= 0. A matrix A ∈ Mn (R) is a unit if and only if det A ∈ U (R). Furthermore, the (i, j)th entry of the inverse matrix A−1 is (det A)−1 (−1)i+j det(Aji ). Suppose that A is invertible with inverse A−1 . By Proposition 5.3.11, det(A) det(A−1 ) = det I = 1 so det A is a unit in R. The converse is not so obvious. Given a matrix A, consider the matrix C whose ijth entry is cij = (−1)i+j det(Aji ). Note that C ∈ Mn (R). The product AC will have for ijth entry

dij =

n X

aik ckj =

k=1

n X

aik (−1)k+j det(Ajk )

k=1

From Theorem 5.3.8, we see that if i = j, then dii = det A. If i 6= j, then dij represents (up to a possible sign change) the Laplace expansion formula about the jth row for the determinant of the matrix A but changed so that the jth row is replaced with the ith row. But then dij represents the determinant of a matrix with a repeated row. This determinant is therefore 0. Hence, we have found that AC = det(A)I. By a nearly identical reasoning, CA = det(A)I. Thus, if det(A) ∈ U (R), then A is invertible with inverse A−1 = (det A)−1 C and this gives the formula for the inverse as described in Proposition 5.3.12.

Exercise: 18 Section 5.3 Question: Find the number of units in M2 (Z/4Z). Solution: The units in M2 (Z/4Z) are the matrices who have inverse, or equivalently, who have determinants that are in U (4) = {1, 3}. So we examine the determinant of a matrix,

a det c

b = ad − bc. d

We could have one of two situations. If ad is 0 or 2, then bc must be 1 or 3. If ad is 1 or 3, then bc must be 0 or 2. So we will figure out how many ways two number can multiply to 1 or 3, {1 · 1, 3 · 3, 1 · 3, 3 · 1}, so we have 4 ways to do so. Then the remaining 42 − 4 = 12 products must multiply to be 0 or 2. Examining our cases we see that we have 12 · 4 + 4 · 12 = 96 possible ways to make our determinant in U (4). So there are 96 units in M2 (Z/4Z).

Exercise: 19 Section 5.3 Question: Prove that a function F : Mn (R) → R that is linear in each row and linear in each column satisfies the alternating property (described in Proposition 5.3.10) if and only if F (A) = 0 for every matrix A that has a repeated row or a repeated column. Solution: Let F : Mn (R) → R be a function that is linear in each row and linear in each column satisfies the alternating property. Suppose that F satisfies the alternating property. If A has a repeated column (respectively row), then the transposition τ that interchanges the equal columns (resp. rows) leads to F (A) = F (A0 ) = sign(τ )F (A) = −F (A). Hence, F (A) = 0. Conversely, suppose that F is such that F (A) = 0 for every matrix A that has a repeated row. Let B be an

258

CHAPTER 5. RINGS

arbitrary matrix and let (i, j) with i 6= j. By linearity of F    b11 b11 b12 ··· b1n    . .. .. . . .. ..  ..   . .     bi1 bi1 + bj1 bi2 + bj2 · · · bin + bjn      .   . . . .. .. .. .. F  = F  .. .     bi1 bi1 + bj1 bi2 + bj2 · · · bin + bjn      .   .. .. . . . .  ..   . . . . bm1 bm1 b12 ··· b1n  b11  ..  .   bj1   = F  ...   bi1   .  .. bm1

b12 .. . bi2 .. . bi2 .. . b12 b12 .. . bj2 .. . bi2 .. . b12

··· .. . ··· .. . ··· .. . ··· ··· .. . ··· .. . ··· .. . ···

  b11 b1n  .. ..   . .     bi1  bin   ..  + F  ..  . .     bj1 bin     . ..   ..  .

b12 .. .

bm1  b11 b1n  .. ..   . .     bj1 bjn    ..  + F  ..  .  .    bj1  bin    .  ..  ..  .

b12

b1n



b1n

bm1

bi2 .. . bj2 .. . b12 .. . bj2 .. . bj2 .. . b12

··· .. . ··· .. . ··· .. . ··· ··· .. . ··· .. . ··· .. . ···

 b1n ..  .   bin   ..  .   bjn   ..  .  b1n  b1n ..  .   bjn   ..  .   bjn   ..  .  b1n

By the condition on F , the left hand side is 0, as are the first and fourth matrix on the right hand side. Hence,     b11 b12 · · · b1n b11 b12 · · · b1n  ..  .. .. ..  .. ..  .. ..  .  . . . . .  . .       bj1 bj2 · · · bjn   bi1 bi2 · · · bin       .. ..  = −F  .. .. ..  . .. .. F  ...   . . . .  . .     .  bi1 bi2 · · · bin   bj1 bj2 · · · bjn       .  . .. ..  .. ..  .. ..  ..  .. . . . .  . .  bm1 b12 · · · b1n bm1 b12 · · · b1n Consequently, if B 0 is the matrix B but with rows permuted by a transposition, then F (B 0 ) = −F (B). Any permutation σ ∈ Sn can be written as a product of transpositions σ = τ1 , τ2 , . . . , τk . Then if A0 is a matrix obtained from a matrix A by permuting the rows according to σ, then F (A0 ) = (−1)k F (A) = (sign σ)F (A). The same reasoning happens for the columns. This proves that if F (A) = 0 for all matrices with repeated row or repeated columns, then F satisfies the alternating property. Exercise: 20 Section 5.3 Question: Let F : Mn (R) → R be a function that is linear in each row and linear in each column of an input matrix A. Prove that F (A> ) = F (A) for all A ∈ Mn (R) if and only if the function f in Proposition 5.3.9 satisfies f (σ −1 ) = f (σ) for all σ ∈ Sn . Solution: Let F : Mn (R) → R be a function that is linear in each row and linear in each column of an input matrix A. By Proposition 5.3.9, there exists a function f : Sn → R such that X F (A) = f (σ)a1σ(1) a2σ(2) · · · anσ(n) . σ∈Sn >

Suppose that F (A ) = F (A) for all A ∈ Mn (R). Note that X > > F (A> ) = f (σ)a> 1σ(1) a2σ(2) · · · anσ(n) σ∈Sn

=

X

f (σ)aσ(1)1 aσ(2)2 · · · aσ(n)n

σ∈Sn

=

X σ∈Sn

f (σ)a1σ−1 (1) a2σ−1 (2) · · · anσ−1 (n) ,

5.4. RING HOMOMORPHISMS

259

where we permuted the terms in each product by σ −1 so that the first indices each each are counted 1, 2, . . . , n. Since there is a bijection on Sn of σ ↔ σ −1 , this is equal to X F (A> ) = f (σ −1 )a1σ(1) a2σ(2) · · · anσ(n) σ −1 ∈Sn

=

X

f (σ −1 )a1σ(1) a2σ(2) · · · anσ(n) .

(5.2)

σ∈Sn

Since F (A> ) = F (A) for all A ∈ Mn (R), we have X X f (σ)a1σ(1) a2σ(2) · · · anσ(n) . f (σ −1 )a1σ(1) a2σ(2) · · · anσ(n) = σ∈Sn

σ∈Sn

Now, for each σ ∈ Sn , we can choose an appropriate A such that F (A) = f (σ). Thus, f (σ) = f (σ −1 ). Conversely, if f (σ −1 ) = f (σ) for all σ ∈ Sn , then from (5.2) and Proposition 5.3.9, we immediately see that F (A> ) = F (A) for all A ∈ Mn (R).

5.4 – Ring Homomorphisms Exercise: 1 Section 5.4 Question: Prove that there are only two ring homomorphisms f : Z → Z. Solution: Suppose that ϕ is a ring homomorphism such that ϕ(1) = c. Then ϕ(−1) = −ϕ(1) = −c and for any positive number k, we have ϕ(k) = ϕ(1 + 1 + . . . + 1) = ϕ(1) + ϕ(1) + . . . + ϕ(1) = c + c + ... + c = c · k. Likewise for −k we have ϕ(−k) = −ϕ(k) = −ck = c(−k). Now ϕ(ab) = c(ab) but ϕ(a)ϕ(b) = cacb = c2 (ab). This implies that c2 = c. There are only two numbers where this holds true, c = 0 which associates with the trivial ring homomorphism (ϕ(x) = 0 for all x). c = 1 is associated with the identity automorphism (ϕ(x) = x for all x) which we know is a ring homomorphism since it can be labeled an inclusion function which is a ring homomorphism by Example 5.4.3. Exercise: 2 Section 5.4 Question: Find all positive integers n and k such that the function Z → Z/nZ defined by f (a) = ka is a homomorphism. Solution: Let n ∈ Z be any nonzero number. For every k it always holds that ϕ(a + b) = k(a + b) = ka + kb = ka + kb = ϕ(a) + ϕ(b). However we also need to have ϕ(ab) = kab be equal to ϕ(a)ϕ(b) = kakb = k 2 ab. Letting a and b be 1 shows us 2 that we need k = k ⇒ k(k − 1) = 0. So for every n, letting k be any number congruent to 1 or 0 mod n always produces a ring homomorphism. For any prime number p, those will be the only two types of homomorphisms since we know that Fp is a field and if another k worked this would imply that k(k − 1) = 0̄ and Fp would have αn 1 α2 zero-divisors. For any composite number n = pα 1 p2 . . . pn made up of two or more distinct primes we have α1 α2 αn other options however. Let A = {p1 , p2 , . . . , pn } be the set of all prime divisors with their powers. Consider any partition of A into two nonempty sets, A1 and A2 . Now take all the numbers in A1 and multiply them, call that number p. Do the same for A2 and call that number q. Now, we know that gcd(p, q) = 1. This implies that their exist some integers, c and d, where cp + dq = 1. Since p and q are both positive, we know that only one of c and d is negative. Without loss of generality we will rewrite the equation as cp − dq = 1, where c and d are now

260

CHAPTER 5. RINGS

both positive numbers. Then we reorder so that we see that cp−1 = dq. We will claim that for k = cp will always work. First we show that k 2 ≡ k mod n by showing the equivalent statement, that k 2 − k = k(k − 1) ≡ 0 mod n. k(k − 1) = cp(cp − 1) = cp(dq) = (cd)(pq) = (cd)n ≡ 0 mod n Next we show that k = cp is not the same as k = 0 or k = 1. We can tell that cp 6≡ 0 mod n since q does not divide cp but q does divide any multiple of n. If cp ≡ 1 mod n this would imply that cp − 1 = dq ≡ 0 mod n but this is also not true since p does not divide dq but it must divide any multiple of n. This shows that k = cp will produce another ring homomorphism. Exercise: 3 Section 5.4 Question: Prove that the “projection” function π1 : R ⊕ R → R given by π1 (a, b) = a is a ring homomorphism and determine the kernel of it. Solution: For any (a, b), (c, d) ∈ R ⊕ R, π1 (a + c, b + d) = a + c = π1 (a, b) + π1 (c, d) and π1 (ac, bd) = ac = π1 (a, b)π1 (c, d) so that π1 is a ring homomorphism. For all, d ∈ R, we know that π1 (0, d) = 0 so that Ker π1 = {(0, d)∀d ∈ R}. Exercise: 4 Section 5.4 Question: Prove that the function f : Z ⊕ Z → Z given by f (m, n) = m − n is not a homomorphism. Solution: Consider (1, 0), (0, 1) ∈ Z ⊕ Z, then f ((1, 0) · (0, 1)) = f (1, 1) = 1 − 1 = 0 but f (1, 0) · f (0, 1) = 1 · −1 = −1 which shows that f is not a ring homomorphism. Exercise: 5 Section 5.4 Question: Let D be an integer not divisible by a square integer (i.e., square-free). Prove that the function √ f : Z[ D] → M2 (Z) defined by √ a b f (a + b D) = Db a √ is an injective ring homomorphism. Deduce that Z[ D] is ring-isomorphic to Im f . Solution: We check the homomorphism properties of f . For the addition √ √ √ a+c b+d f ((a + b D) + (c + d D)) = f ((a + c) + (b + d) D) = D(b + d) a + c √ √ a b c d a+c b+d f ((a + b D) + f ((c + d D)) = + = . Db a Dd c D(b + d) a + c These two are equal so f satisfies the homomorphism property for +. For multiplication, √ √ √ ac + Dbd ad + bc f ((a + b D)(c + d D)) = f ((ac + Dbd) + (ad + bc) D) = D(ad + bc) ac + Dbd √ √ a b c d ac + bdD ad + bc . = f ((a + b D)f ((c + d D)) = Db a Dd c bcD + adD bdD + ac Again, we observe that these two are the same so f also satisfies the homomorphism property for multiplication. Thus f is a ring homomorphism. √ √ Furthermore, f (a + b D) is the 0 matrix if and only if a = b = 0, i.e., a + b D = 0. Since ker f = {0} then f is injective. Since f is injective, the domain is isomorphic to the image subring Im f in M2 (Z). Exercise: 6 Section 5.4 Question: Let R be a commutative ring and let a ∈ R be a fixed element. Consider the function fa : R[x, y] → R[x] defined by fa (p(x, y)) = p(x, a). Prove that fa is a ring homomorphism. Solution: From Example 5.4.4, we know that evaluation of a single variable polynomial at a fixed value r with coefficients in a commutative ring R is a ring homomorphism. We will rewrite our situation into the situation of Example 5.4.4. We will write p(x, y) = b(y) = bn y n + bn−1 y n−1 + . . . + b1 y + b0 ∈ R[x][y] where bi are polynomials in R[x] which is a ring. Let S = R[x], we know that a ∈ S as a zero-degree polynomial. We know from Proposition 5.2.4 that R[x] or S is in fact a commutative ring since R is commutative. Now we will rewrite

5.4. RING HOMOMORPHISMS

261

from before, fa : S[y] → S defined by fa (b(y)) = b(a). It is clear now that this case is no different from Example 5.4.4 and so is a ring homomorphism. Exercise: 7 Section 5.4 Question: Let R and S be commutative rings and let ϕ : R → S be a homomorphism. Prove that the function ψ : R[x] → S[x] defined by ψ(an xn + · · · + a1 x + a0 ) = ϕ(an )xn + · · · + ϕ(a1 )x + ϕ(a0 ). Prove that ψ is a homomorphism. Solution: Let a(x) and b(x) be polynomials in R[x] with a(x) = am xm + · · · + a1 x + a0 b(x) = bn xn + · · · + b1 x + b0 . For simplicity when proving that ψ has the homomorphism property with addition, assume that m = n, possibly allowing for an or bn to be 0. Then ! n n n X X X k k ψ(a(x) + b(x)) = ψ (ak + bk )x = ϕ(ak + bk )x = (ϕ(ak ) + ϕ(bk ))xk k=0

ψ(a(x)) + ψ(b(x)) =

n X

k=0

! ϕ(ak )xk

+

k=0

n X

k=0

! ϕ(bk )xk

=

k=0

n X

(ϕ(ak ) + ϕ(bk ))xk .

k=0

We see that these are equal. For multiplication,         n n n X X X X X X  ai bj  xk  = ϕ ai bj  xk = ϕ ϕ(ai )ϕ(bj ) xk ψ(a(x)b(x)) = ψ  k=0

ψ(a(x))ψ(b(x)) =

n X k=0

i+j=k

! ϕ(ak )xk

k=0 n X

! ϕ(bk )xk

k=0

=

i+j=k n X

k=0

 ϕ

k=0

i+j=k

 X

ϕ(ai )ϕ(bj ) xk .

i+j=k

We have shown that ψ is a ring homomorphism. Exercise: 8 Section 5.4 Question: Let R be a commutative ring of prime characteristic p. a) Prove that p divides kp for all integers 1 ≤ k ≤ p − 1. b) Prove that the function f : R → R given by f (x) = xp is a homomorphism. In other words, prove that (a + b)p = ap + bp

and

(ab)p = ap bp .

Solution: . Since we know this is an integer, we just need a) Let 1 ≤ k ≤ p − 1 and consider kp = p(p−1)(p−2)...(p−k+1) k! to make sure that p divides it. Now p divides the numerator so if p does not divide the denominator, this implies that p divides kp . Since all number in the denominator are less than p and p is prime, the numbers are all relatively prime to p. This implies that their product is relatively prime to p as well so that p does not divide the denominator and must divide kp . b) Let a, b ∈ R and consider f (a + b) = (a + b)p . By Exercise 5.1.25, p X p p−i i (a + b)p = a b i i=0 p−1 p p 0 p 0 p X p p−i i = a b + a b + a b. 0 p i i=1

262

CHAPTER 5. RINGS Now since all of the binomials in the summation are divisible by p and R is characteristic p, this implies that p−1 p−1 X p p 0 p 0 p X p p−i i p p a b + a b + a b =a +b + 0ap−i bi 0 p i i=1 i=1 = ap + bp = f (a) + f (b). Now we check multiplication. We note that since R is commutative, we can reorder as we feel led. f (ab) = (ab)p = (ab)(ab) . . . (ab)(p times) = ap bp = f (a)f (b). This shows that f (x) is a ring homomorphism.

Exercise: 9 Section 5.4 Question: Given any set S, the triple (P(S), 4, ∩) has the structure of a ring. Let S 0 be any subset of S. Show that ϕ(A) = A ∩ S 0 is a ring homomorphism from P(S) to P(S 0 ). Solution: For any set A, A ∩ S 0 ⊆ S 0 ⇒ A ∩ S 0 ∈ P(S 0 ) so that ϕ is a function from P(S) to P(S 0 ). Let A, B ∈ S. We will utilize results from Exercise 1.1.9. Consider ϕ(A4B) = (A4B) ∩ S 0 = (A ∩ S 0 )4(B ∩ S 0 ) = ϕ(A)4ϕ(B). Next we check that ϕ separates over ∩. ϕ(A ∩ B) = (A ∩ B) ∩ S 0 = (A ∩ S 0 ) ∩ (B ∩ S 0 ) = ϕ(A) ∩ ϕ(B). This shows that ϕ is a ring homomorphism from P(S) to P(S 0 ). Exercise: 10 Section 5.4 Question: Prove that (5Z, +) is group-isomorphic to (7Z, +) but that (5Z, +, ×) is not ring-isomorphic to (7Z, +, ×). Solution: Consider ϕ(a) = a5 · 7 as a function from 5Z to 7Z. Then for any elements 5a, 5b ∈ 5Z we have ϕ(5a + 5b) = ϕ(5(a + b)) = 7(a + b) = 7a + 7b = ϕ(5a) + ϕ(5b) so that ϕ is a homomorphism. For any 7a ∈ 7Z we have ϕ(5a) = 7a so that ϕ is surjective. Furthermore if x5 · 7 = 0 this implies that x = 0 which shows that ϕ is injective. So ϕ is an isomorphism. Now we consider the rings (5Z, +, ×) and (7Z, +, ×). Suppose that some ring isomorphism f existed between the rings. Then for some integer a, we have f (5) = 7a. Then we should have that f (52 ) = f (5)2 = (7a)2 = 49a2 = 7(7a2 ). But at the same time f (52 ) = f (5 + 5 + 5 + 5 + 5) = f (5) + f (5) + f (5) + f (5) + f (5) = 7a + 7a + 7a + 7a + 7a = 7(5a). However these are not equal unless a = 0 which would imply that Ker f is not just 0 and would contradict the fact that f should be injective. So we have a contradiction and no such ring isomorphism can exist. Exercise: 11 Section 5.4 Question: Let U be a set. Show that (Fun(U, Z/2Z), +, ×) is isomorphic to (P(U ), 4, ∩).

5.4. RING HOMOMORPHISMS

263

Solution: Let f ∈ Fun(U, Z/2Z) and define ϕ(f ) = Tf = {u ∈ U |f (u) = 1}. Then Tf ∈ P(U ). From here on out we will refer to f (u) = 1 and u ∈ Tf as equivalent notions. Now we will show that ϕ is a ring homomorphism. Let f, g ∈ Fun(U, Z/2Z) and consider ϕ(f + g) = Tf +g = {u ∈ U |(f + g)(u) = 1} Then if u ∈ Tf XOR u ∈ Tg we have (f + g)(u) = f (u) + g(u) = 1. But if u ∈ Tf and u ∈ Tg then we have (f + g)(u) = f (u) + g(u) = 1 + 1 = 0. And if u is in neither of the sets then (f + g)(u) = f (u) + g(u) = 0 + 0 = 0 and u 6∈ Tf +g . This shows that Tf +g = Tf ∪ Tg − Tf ∩ Tg = Tf 4Tg . This implies that ϕ(f + g) = Tf +g = Tf 4Tg = ϕ(f )4ϕ(g). Now we check that ϕ separates over multiplication. ϕ(f × g) = Tf ×g = {u ∈ U |(f × g)(u) = f (u)g(u) = 1}. u 6∈ Tf ×g if either f (u) = 0 and/or g(u) = 0. So that u ∈ Tf ×g if and only if u ∈ Tf ∩ Tg . This implies that ϕ(f × g) = Tf ×g = Tf ∩ Tg = ϕ(f ) ∩ ϕ(g). This shows that ϕ is a ring homomorphism. We observe that ϕ(i) = ∅ for only one function, i, where i(u) = 0 for all u ∈ U . This implies that ϕ is injective. It is clear that for any A ∈ P(U ) we can define a function f where f (a) = 1 if a ∈ A and f (a) = 0 elsewise. Then ϕ(f ) = A so that ϕ is also surjective. This shows that ϕ is an isomorphism. Exercise: 12 Section 5.4 Question: Show that (Z ⊕ Z)[x] is not isomorphic to Z[x] ⊕ Z[x]. Solution: Idempotence is a property that is preserved under isomorphisms. Assume that ϕ : (Z ⊕ Z)[x] → Z[x] ⊕ Z[x] is an isomorphism. Both rings have precisely three idempotent elements, namely, (1, 0), (0, 1), and (1, 1). Zero divisors are mapped to zero divisors under an isomorphism so ϕ(0, 0) = (0, 0), ϕ(1, 1) = (1, 1), and ϕ may map (1, 0) to (1, 0) or to (0, 1) and vice versa. Without loss of generality, we can suppose that ϕ(1, 0) = (1, 0) and ϕ(0, 1) = (0, 1). Then ϕ maps Z ⊕ Z (as the subring of coefficients of the domain) to Z ⊕ Z as the identity. Now, from properties of a ring homomorphism, once ϕ(x) is determined, the ϕ is determined by isomorphism properties. Still assuming that ϕ is an isomorphism, we must have ϕ(x) = (a(x), b(x)) with a(x) and b(x) not both constant. Suppose that a(x) (and similarly for b(x)) is a constant polynomial. Then if q(x) ∈ (Z ⊕ Z)[x] with q(x) = (cn , dn )xn + · · · + (c1 , d1 )x + (c0 , d0 ), then the image ϕ(q(x)) ϕ(q(x)) = (cn , dn )ϕ(x)n + · · · + (c1 , d1 )ϕ(x) + (c0 , d0 ) has only constants in the first entry of the pair in Z[x] ⊕ Z[x]. For ϕ to be an isomorphism, we must assume that deg a(x) > 0 and deg b(x) > 0. Furthermore, we must have deg a(x) = deg b(x) = 1 because then otherwise, ϕ could not be surjective. But then, with the above q(x), we have ϕ(q(x)) = (cn a(x)n + · · · + c1 a(x) + c0 , dn b(x)n + · · · + d1 b(x) + d0 ). In particular, both polynomials in the pair are of the same degree of q(x). However, pairs in Z[x] ⊕ Z[x] include pairs of polynomials of different degrees. Hence, ϕ is not surjective, which contradicts the assumption that ϕ is bijective. Exercise: 13 Section 5.4 Question: Consider the ring (R, +, ×), where R = Z/2Z × Z/2Z as a set, where + is the component-wise addition but the multiplication is done according to the following table. × (0, 0) (1, 0) (0, 1) (1, 1)

(0, 0) (0, 0) (0, 0) (0, 0) (0, 0)

(1, 0) (0, 0) (1, 1) (1, 0) (0, 1)

(0, 1) (0, 0) (1, 0) (0, 1) (1, 1)

(1, 1) (0, 0) (0, 1) (1, 1) (1, 0)

Prove that (R, +, ×) is a ring. Also prove that (R, +, ×) is not isomorphic to Z/2Z ⊕ Z/2Z. Solution: (At this point) it is obvious that (R, +) is an abelian group. Looking at the multiplication table, we see that × on R − {0} is a commutative operation (with identity (0, 1)). Furthermore, we note that (R − {0}, ×)

264

CHAPTER 5. RINGS

is a group, namely isomorphic to Z3 , generated by the element (1, 0). Considering that (0, 0) multiplied by anything is (0, 0), we conclude that × is associative on R. The primary difficulty involves checking distributivity. Since × on R is commutative, we only need to check left distributivity, i.e., that a × (b + c) = a × b + a × c for all a, b, c ∈ R. At this point, we would need to check the 43 different possibilities for a, b, c. However, if a, b or c is equal to 0, then the distributivity property holds automatically (since 0×r = 0 for all r ∈ R). Also, if a = (0, 1), the multiplicative identity, then the distributivity property holds automatically. With these considerations, we only need to check 2 · 32 = 18 combinations. Also, since + is commutative, we only need to check the distributivity property once if b 6= c. This brings us down to checking 2·6 = 12 calculations. Also, if b = c, then b+c = 0, which leads to a(b+c) = 0 and ab+ac = ab+ab = 0 as well. Hence, we only need to check when b 6= c. This brings us down to only 2 · 3 = 6 calculations to check: (1, 0) × ((1, 0) + (0, 1)) = (1, 0) × (1, 1) = (0, 1) (1, 0) × (1, 0) + (1, 0) × (0, 1) = (1, 1) + (1, 0) = (0, 1) (1, 0) × ((1, 0) + (1, 1)) = (1, 0) × (0, 1) = (1, 0) (1, 0) × (1, 0) + (1, 0) × (1, 1) = (1, 1) + (0, 1) = (1, 0) (1, 0) × ((0, 1) + (1, 1)) = (1, 0) × (1, 0) = (1, 1) (1, 0) × (0, 1) + (1, 0) × (1, 1) = (1, 0) + (0, 1) = (1, 1) (1, 1) × ((1, 0) + (0, 1)) = (1, 1) × (1, 1) = (1, 0) (1, 1) × (1, 0) + (1, 1) × (0, 1) = (0, 1) + (1, 1) = (1, 0) (1, 1) × ((1, 0) + (1, 1)) = (1, 1) × (0, 1) = (1, 1) (1, 1) × (1, 0) + (1, 1) × (1, 1) = (0, 1) + (1, 0) = (1, 1) (1, 1) × ((0, 1) + (1, 1)) = (1, 1) × (1, 0) = (0, 1) (1, 1) × (0, 1) + (1, 1) × (1, 1) = (1, 1) + (1, 0) = (0, 1) We see that these 6 operations done both ways are equal. This suffices to show distributivity of × over +. Exercise: 14 Section 5.4 Question: Prove that Mm (Mn (R)) is isomorphic to Mmn (R). Solution: This exercise follows from viewing a matrix as a block matrix. In particular, for multiplication, a result of matrix multiplication is that if Aij and Bij are n × n matrices, then the ijth entry of    A11 A12 · · · A1n B11 B12 · · · B1n  A21 A22 · · · A2n   B21 B22 · · · B2n      .. .. ..   .. .. ..  .. ..  . . . . .  . . .  An1 is the m × m matrix

An2

···

Ann n X

Bn1

Bn2

···

Bnn

Aik Bkj .

k=1

It is obvious that the process of considering an mn × mn matrix as an n × n matrix of m × m matrices is a bijection. Exercise: 15 Section 5.4 Question: Show that the function ϕ : H → M2 (C) defined by a + bi −c − di ϕ(a + bi + cj + dk) = c − di a − bi is an injective homomorphism. Conclude that H is ring-isomorphic to Im ϕ. Solution: It is obvious that the function ϕ is injective since ϕ(a + bi + cj + dk) = ϕ(a0 + b0 i + c0 j + d0 k) implies that a + bi = a0 + b0 i and −c − di = −c0 − d0 i so (a, b, c, d) = (a0 , b0 , c0 , d0 ). Also, ϕ((a + bi + cj + dk) + (a0 + b0 i + c0 j + d0 k)) = ϕ((a + a0 ) + (b + b0 )i + (c + c0 )j + (d + d0 )k) (a + a0 ) + (b + b0 )i −(c + c0 ) − (d + d0 )i = (c + c0 ) − (d + d0 )i (a + a0 ) − (b + b0 )i 0 a + bi −c − di a + b0 i −c0 − d0 i = + 0 c − di a − bi c − d0 i a0 − b0 i = ϕ(a + bi + cj + dk) + ϕ(a0 + b0 i + c0 j + d0 k).

5.4. RING HOMOMORPHISMS

265

For multiplication, we check first ϕ((a + bi + cj + dk)(a0 + b0 i + c0 j + d0 k)) = ϕ((aa0 − bb0 − cc0 − dd0 ) + (ab0 + ba0 + cd0 − dc0 )i + (ac0 − bd0 + ca0 + db0 )j + (ad0 + bc0 − cb0 + da0 )k) (aa0 − bb0 − cc0 − dd0 ) + (ab0 + ba0 + cd0 − dc0 )i −(ac0 − bd0 + ca0 + db0 ) − (ad0 + bc0 − cb0 + da0 )i = . 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (ac − bd + ca + db ) − (ad + bc − cb + da )i (aa − bb − cc − dd ) − (ab + ba + cd − dc )i

Furthermore, ϕ(a + bi + cj + dk)ϕ(a0 + b0 i + c0 j + d0 k) 0 a + bi −c − di a + b0 i −c0 − d0 i = c − di a − bi c0 − d0 i a0 − b0 i 0 aa − bb0 + ab0 i + ba0 i − cc0 − dd0 + cd0 i − dc0 i = ca0 + db0 + cb0 i − da0 i + ac0 − bd0 − ad0 i − bc0 i

−ac0 + bd0 − ad0 i − bc0 i − ca0 − db0 + cb0 i − da0 i . −cc0 − dd0 − cd0 i + dc0 i + aa0 − bb0 − ab0 i − ba0 i

By inspection, we see that ϕ((a + bi + cj + dk)(a0 + b0 i + c0 j + d0 k)) = ϕ(a + bi + cj + dk)ϕ(a0 + b0 i + c0 j + d0 k). Thus, we see that ϕ satisfies the definition of a homomorphism. Since ϕ is injective and ϕ is surjective on Im ϕ, then H is isomorphic to the subring Im ϕ of M2 (C). Exercise: 16 Section 5.4 Question: Use the reduction homomorphism with n = 5 to show that the polynomial x4 + 3x2 + 4x + 1 has no roots in Z. Solution: Let p(x) = x4 + 3x2 + 4x + 1. Letting n = 5 we see that under the reduction homomorphism, π(p(x)) = x4 + 3x2 + 4x + 1. We will check all the representatives of the equivalence class. π(p(0)) = 1, π(p(1)) = 4, π(p(2)) = 2, π(p(3)) = 1, π(p(4)) = 1. Since π(p(x)) has no roots in Z/5Z, this implies that p(x) has no roots in Z. Exercise: 17 Section 5.4 Question: Use the reduction homomorphism with n = 3 to show that the polynomial x5 − x + 2 has no roots in Z. Solution: Let p(x) = x5 −x+2 and consider the reduction homomorphism to Z/3Z[x], then π(p(x)) = x5 +2x+2. We check all the representatives of the equivalence class. π(p(0)) = 2, π(p(1)) = 2, π(p(2)) = 2. We conclude that p(x) has no roots in Z. Exercise: 18 Section 5.4 Question: Let A be a matrix in M2 (R). Define the function ϕ : R[x] → M2 (R) defined by ϕ(an xn + an−1 xn−1 + · · · + a1 x + a0 ) = an An + an−1 An−1 + · · · + a1 A + a0 I. This looks like plugging A into the polynomial except that the constant term becomes the diagonal matrix a0 I. 1 2 a) Only for this part, take A = . Calculate ϕ(x2 + x + 1) and ϕ(3x3 − 2x). 3 4 b) Show that ϕ is a ring homomorphism and that Im ϕ is a commutative subring of M2 (R). c) For any A ∈ M2 (R), show that the characteristic polynomial fA (x) of A is in Ker ϕ. Solution: a) We calculate 2

2

ϕ(x + x + 1) = A + A + I = 3

3

ϕ(3x − 2x) = 3A − 2A = 3

7 15

37 81

10 1 2 1 0 9 12 + + = , 22 3 4 0 1 18 27 54 1 2 109 158 −2 = . 118 3 4 237 346

266

CHAPTER 5. RINGS

b) Let p(x) = am xm + · · · + a1 x + a0 and q(x) = bm xm + · · · + b1 x + b0 be two polynomials. Then ϕ(p(x) + q(x)) = ϕ((am + bm )xm + · · · + (a1 + b1 )x + (a0 + b0 )) = (am + bm )Am + · · · + (a1 + b1 )A + (a0 + b0 )I = (am Am + · · · + a1 A + a0 I) + (bm Am + · · · + b1 A + b0 I) = ϕ(p(x)) + ϕ(q(x)). (We may use this and assume that so of the apparently leading terms of q(x) are zero and hence generalized to when the degree of p(x) and q(x) are not equal.) If q(x) = bn xn + · · · + b1 x + b0 then p(x)q(x) = cm+n xm+n + · · · + c1 x + c0 , where for all 0 ≤ k ≤ m + n X ck = ai bj , i+j=k

where we assume we only take 0 ≤ i ≤ m and 0 ≤ j ≤ n. Thus, ϕ(p(x)q(x)) =

m+n X

ck Ak = (am Am + · · · + a1 A + a0 )(bn An + · · · + b1 A + b0 ) = ϕ(p(x))ϕ(q(x)),

k=0

where the second equality holds because powers of A commute with powers of A. Thus, we have show that ϕ is a ring homomorphism. By Proposition 5.4.10, Im ϕ is a subring of M2 (R) but it is a commutative subring since R[x] is commutative. c) Let A ∈ M2 (R) and call fA (x) the characteristic polynomial of A. We know that fA (x) = det(xI − A). Then ϕ(fA (x)) = det(A − A) = 0. Thus fA (x) ∈ Ker ϕ. is in Ker ϕ. Exercise: 19 Section 5.4 Question: Suppose that R is a ring with identity 1 6= 0. Let ϕ : R → S be a nontrivial ring homomorphism (i.e., ϕ is not identically 0). a) Suppose that ϕ(1) is not the identity element in S (in particular, if S does not contain an identity element). Prove that ϕ(1) is idempotent. b) Prove that whether or not S has an identity, Im ϕ has an identity, namely 1Im ϕ = ϕ(1). c) Suppose that S contains an identity 1S and that ϕ(1) 6= 1S . Prove that ϕ(1) is a zero divisor. d) Deduce that if S is an integral domain, then ϕ(1) = 1S . e) Suppose that R and S are integral domains. Prove that a nontrivial ring homomorphism ϕ : R → S induces a group homomorphism ϕ× : U (R) → U (S). Solution: Let R be a ring with identity. Let ϕ : R → S be a nontrivial ring homomorphism. a) Since 1 · 1 = 1 in R, then by the homomorphism property, ϕ(1) = ϕ(1)ϕ(1). This shows that ϕ(1) is idempotent. b) For any a ∈ R, we have the following chain of equivalences, ϕ(1)ϕ(a) = ϕ(1a) = ϕ(a) = ϕ(a1) = ϕ(a)ϕ(1) By definition, ϕ(1) functions as the identity for all ϕ(a) ∈ Im ϕ. So 1Im ϕ = ϕ(1). c) If S contains an identity and ϕ(1) 6= 1S , then from ϕ(1) = ϕ(1)ϕ(1). So ϕ(1)1S − ϕ(1)ϕ(1) = 0 =⇒ ϕ(1)(1S − ϕ(1)) = 0. If ϕ(1) 6= 1S , then ϕ(1) = 0 or ϕ(1) is a zero divisor. If ϕ(1) = 0, then for all r ∈ R, ϕ(r) = ϕ(r · 1) = ϕ(r)ϕ(1) = 0. The assumption that ϕ is nontrivial implies that ϕ(1) is a zero divisor. d) Now if S is an integral domain, then since it contains no zero divisors, we see that ϕ(1)(1S − ϕ(1)) = 0 implies that ϕ(1) = 1S . e) Suppose that r ∈ U (R), then there exists some s ∈ R with rs = 1. Then ϕ(r)ϕ(s) = ϕ(1) = 1S . Thus ϕ(r) is invertible with multiplicative inverse ϕ(s). Thus, the ring homomorphism ϕ has ϕ(U (R)) ⊆ U (S). Furthermore, since ϕ(rs) = ϕ(r)ϕ(s), then we see that ϕ is a group homomorphism from U (R) to U (S).

5.4. RING HOMOMORPHISMS

267

Exercise: 20 Section 5.4 Question: Solution: Exercise: 21 Section 5.4 Question: Let R be a commutative ring and let R[x, y] be the polynomial ring on two variables x and y. Consider the function f : R[x, y] → R[x, y] such that f (p(x, y)) = p(y, x). Prove that f is a nontrivial automorphism on R[x, y]. Solution: Let X X aij xi y j , b(x, y) = bij xi y j ∈ R[x, y] a(x, y) = 0≤i,j≤n

0≤i,j≤m

be any two polynomials with coefficients aij , bP ij ∈ R. To simplify, we will assume without losing generality that m ≥ n and extend the summation a(x, y) = 0≤i,j≤m aij xi y j where some coefficients are simply zeroes now. Now consider X X f (a(x, y)b(x, y)) = f (( aij xi y j )( bij xi y j )) 0≤i,j≤m

= f(

X

0≤i,j≤m i j

cij x y )

0≤i,j≤m2

X

=

cij y i xj

0≤i,j≤m2

=(

X

0≤i,j≤m

= f(

X

aij y i xj )(

bij y i xj )

0≤i,j≤m

X

X

i j

aij x y ) · f (

0≤i,j≤m

bij xi y j ).

0≤i,j≤m

We knew we could separate the products because of symmetry. Now we check addition, X X X f (( aij xi y j ) + ( bij xi y j )) = f ( (aij + bij )xi y j ) 0≤i,j≤m

0≤i,j≤m

0≤i,j≤m

X

=

(aij + bij )y i xj

0≤i,j≤m

=(

X

0≤i,j≤m

= f(

X

aij y i xj ) + (

X

bij y i xj )

0≤i,j≤m i j

aij x y ) + f (

0≤i,j≤m

X

bij xi y j ).

0≤i,j≤m

This shows that f is a ring homomorphism from R[x, y] to R[x, y]. To see that it is nontrivial, we simply note that x ∈ R[x, y] and f (x) = y 6= x. Exercise: 22 Section 5.4 Question: Denote by End(R) the set of endomorphisms on R (homomorphisms from R to itself). a) Show that End(R) is closed under the operation of function composition. b) Show that End(R) is closed neither under function addition nor under function multiplication. Solution: a) Let f, g ∈ End(R) and x, y ∈ R where g(x) = x0 , g(y) = y 0 ∈ R. Then, (f ◦ g)(x + y) = f (g(x + y)) = f (g(x) + g(y)) = f (x0 + y 0 ) = f (x0 ) + f (y 0 ) = f (g(x)) + f (g(y)) = (f ◦ g)(x) + (f ◦ g)(y).

268

CHAPTER 5. RINGS And we also check that it separates under multiplication, (f ◦ g)(xy) = f (g(xy)) = f (g(x)g(y)) = f (x0 y 0 ) = f (x0 )f (y 0 ) = f (g(x))f (g(y)) = (f ◦ g)(x)(f ◦ g)(y).

This shows that f ◦ g is still a ring homomorphism from R to R. So End(R) is closed under composition. b) First we show that End(R) is not closed under addition. Consider (f + g)(x)(f + g)(y) = (f (x) + g(x))(f (y) + g(y)) = f (x)f (y) + g(x)f (y) + f (x)g(y) + g(x)g(y). However, (f + g)(xy) = f (xy) + g(xy) = f (x)f (y) + g(x)g(y). So that (f + g) does not sperate over addition of elements and is not a ring homomorphism and End(R) is not closed under multiplication. Now we show that End(R) is not closed under multiplication. (f g)(x + y) = (f g)(x) + (f g)(y) = f (x)g(x) + f (y)g(y). However f (x + y)g(x + y) = (f (x) + f (y))(g(x) + g(y)) = f (x)g(x) + f (y)g(x) + f (x)g(y) + f (y)g(y). Since (f g)(x + y) 6= f (x + y)g(x + y), we see that End(R) is not closed under multiplication. Exercise: 23 Section 5.4 Question: Let R be a commutative ring with identity 1 6= 0 and let G be a finite group. Define the function ψ : R[G] → R by ψ(a1 g1 + a2 g2 + · · · + an gn ) = a1 + a2 + · · · + an . Prove that ψ is a ring homomorphism. This function is called the augmentation map of the group ring R[G]. Solution: Let α, β ∈ R[G] with α = a1 g1 + a2 g2 + · · · + an gn and β = b1 g1 + b2 g2 + · · · + bn gn . Then for addition, ψ(α) + ψ(β) = ψ(a1 g1 + a2 g2 + · · · + an gn ) + ψ(b1 g1 + b2 g2 + · · · + bn gn ) = a1 + a2 + · · · + an + b1 + b2 + · · · + bn = (a1 + b1 ) + (a2 + b2 ) + . . . + (an + bn ) = ψ((a1 + b1 )g1 + (a2 + b2 )g2 + . . . + (an + bn )gn ) = ψ(α + β) and for multiplication ψ(α) · ψ(β) = ψ((a1 g1 + a2 g2 + · · · + an gn ) · (b1 g1 + b2 g2 + · · · + bn gn )) n n X X = ψ( (ak gk bi gi )) = ψ(

k=1 n X

i=1

((ak b1 )gk g1 + (ak b2 )gk g1 + . . . + (ak bn )gk gn )).

k=1

Now since we can pull apart sums, we can apply ψ to each term and keep the summation, so that we get ψ(α) · ψ(β) =

n X

(ak b1 ) + (ak b2 ) + . . . + (ak bn )

k=1 n X

=(

k=1 n X

=(

k=1

ak (

n X

bi ))

i=1 n X ak ) · ( bi ) i=1

= ψ(a1 g1 + a2 g2 + · · · + an gn ) · ψ(b1 g1 + b2 g2 + · · · + bn gn ).

5.4. RING HOMOMORPHISMS

269

This shows that ψ is a ring homomorphism. Exercise: 24 Section 5.4 Question: The augmentation map for a group ring generalizes in the following way. Let R be a commutative ring with identity 1 6= 0 and let G be a finite group. Let f : G → U (R) be a group homomorphism. Prove that the function ψ : R[G] → R defined by ψ(a1 g1 + a2 g2 + · · · + an gn ) = a1 f (g1 ) + a2 f (g2 ) + · · · + an f (gn ), where ai f (gi ) involves a product in the ring R, is a ring homomorphism. Solution: Let α, β ∈ R[G] with α = a1 g1 + a2 g2 + · · · + an gn and β = b1 g1 + b2 g2 + · · · + bn gn . Then for addition Then for addition, ψ(α) + ψ(β) = ψ(a1 g1 + a2 g2 + · · · + an gn ) + ψ(b1 g1 + b2 g2 + · · · + bn gn ) = (a1 + b1 )f (g1 ) + (a2 + b2 )f (g2 ) + . . . + (an + bn )f (gn ) = a1 f (g1 ) + b1 f (g1 ) + a2 f (g2 ) + b2 f (g2 ) + . . . + an f (gn ) + bn f (gn ) = (a1 f (g1 ) + a2 f (g2 ) + · · · + an f (gn )) + (b1 f (g1 ) + b2 f (g2 ) + · · · + bn f (gn )) = ψ(a1 g1 + a2 g2 + · · · + an gn ) + ψ(b1 g1 + b2 g2 + · · · + bn gn ). Now we check multiplication, ψ((a1 g1 + a2 g2 + · · · + an gn ) · (b1 g1 + b2 g2 + · · · + bn gn ))

= ψ(

X

(ai bj )(gi gj ))

1≤i,j≤n

X

=

(ai bj )f (gi gj )

1≤i,j≤n

=

X

(ai bj )f (gi )f (gj )

1≤i,j≤n

= (a1 f (g1 ) + a2 f (g2 ) + · · · + an f (gn )) · (b1 f (g1 ) + b2 f (g2 ) + · · · + bn f (gn )) = ψ(a1 g1 + a2 g2 + · · · + an gn ) · ψ(b1 g1 + b2 g2 + · · · + bn gn ). So ψ is a ring homomorphism. Exercise: 25 Section 5.4 Question: Let R be a commutative ring and let G be a group. Prove that is ϕ : G → G is a group automorphism, then the function Φ : R[G] → R[G] defined by   X X Φ ag g  = ag ϕ(g) g∈G

g∈G

is an automorphism of R[G]. Solution: WePfirst show that Φ is a endomorphism, which requires showing that it is a ring homomorphism. P Let g∈G ag g, g∈G bg g ∈ R[G] be any two elements. Then     X X X Φ ag g + bg g  = Φ  (ag + bg )g  g∈G

g∈G

g∈G

=

X

(ag + bg )ϕ(g)

g∈G

=

X

ag ϕ(g) + bg ϕ(g)

g∈G

=

X

ag ϕ(g) +

g∈G

X

bg ϕ(g)

g∈G

 = Φ

 X

g∈G



ag g  + Φ 

 X

g∈G

bg g  .

270

CHAPTER 5. RINGS

Next we check that Φ separates over multiplication.     X X X Φ ( ag g) · ( bg g) = Φ  (ag bh )(gh) g∈G

g∈G

g,h∈G

X

=

(ag bh )ϕ(gh)

g,h∈G

X

=

(ag bh )ϕ(g)ϕ(h)

g,h∈G

X

=

(ag bh )ϕ(g)ϕ(h)

g,h∈G

=(

X

ag ϕ(g)) · (

g∈G

X

bg ϕ(g)) = Φ(

g∈G

X

ag ϕ(g)) · Φ(

g∈G

X

bg ϕ(g)).

g∈G

This shows that P Φ is a ring homomorphism. Now we show that Φ is injective by showing that Ker Φ = {0}. P Suppose that Φ( a g) = 0 = 0g. This implies that ag = 0 for all g ∈ G. Which implies that g∈G P P g∈G g a g = 0g = 0. This shows that Ker Φ = {0}. Since Φ is injective and an endomorphism, this g∈G g g∈G implies that Φ is bijective as well. So Φ is an automorphism. Exercise: 26 Section 5.4 Question: Let R be a ring with identity 1 6= 0 and let (S, ·) be a monoid (semigroup with an identity e). Prove that the identity for a convolution ring (F, +, ∗) of functions from S to R is the function i : S → R defined by ( 1 if s = e i(s) = . 0 if s 6= e Solution: We will show that i works as the identity for convolution. Let f ∈ F be any element. Then for any element s ∈ S we have X f (s1 )i(s2 ) (f ∗ i)(s) = s1 ·s2 =s

=

X

(f (s1 )i(s2 )) + f (s)i(e)

s1 ·s2 =s|s2 6=e

=

X

(f (s1 )0) + f (s)1

s1 ·s2 =s|s2 6=e

= f (s). Likewise, (i ∗ f )(s) =

X

i(s1 )f (s2 )

s1 ·s2 =s

=

X

(i(s1 )f (s2 )) + i(e)f (s)

s1 ·s2 =s|s1 6=e

=

X

(0f (s1 )) + 1f (s)

s1 ·s2 =s|s1 6=e

= f (s). So i functions as the identity for a convolution ring (F, +, ∗). Exercise: 27 Section 5.4 Question: Let R be a ring and consider the monoid S = ({1, −1}, ×). Prove that as a set Fun(S, R) is in bijection with R × R but that the convolution ring (Fun(R, S), +, ∗) is not isomorphic to R ⊕ R. Solution: The set Fun(S, R) is such that for each of the two elements s ∈ S, there corresponds an element rs . Thus, for each function f ∈ F un(S, R), there corresponds an element (f (1), f (−1)) ∈ R ⊕ R. Furthermore, this

5.4. RING HOMOMORPHISMS

271

is bijective: each pair (r1 , r2 ) ∈ R ⊕ R is mapped to the function f such that f (1) = r1 and f (−1) = r2 . This sets up a bijection ψ : Fun(S, R) → R ⊕ R. Consider the convolution ring (Fun(R, S), +, ∗). If a function f corresponds to (r1 , r−1 ) as described above, then the product satisfies (r1 , r−1 ) ∗ (s1 , s−1 ) = (r1 s1 + r−1 s−1 , r1 s−1 + r−1 s1 ). The bijection ψ satisfies the homomorphism property for addition but it does not satisfy ψ(f ∗ g) = ψ(f )ψ(g)

with product in R ⊕ R.

More generally, assume that there exists an isomorphism ϕ : (Fun(S, R), +, ∗) → (R ⊕ R, +, ·). Also, let us write functions f ∈ Fun(S, R) as a pair (r1 , r−1 ). For the sake of the proof, let R be a commutative ring that contains elements that are not zero divisors, for example Z. Let r be any element in R and suppose that ϕ((r, 0)) = (a, b) and ϕ((0, r)) = (c, d). Then ϕ((r, 0)(0, r)) = ϕ((0, 0)) = (0, 0) = (a, b)(c, d) = (ac, bd). Thus, a, b, c, d are all zero divisors. This is true for all r ∈ R. Let r0 ∈ R, with ϕ((r0 , 0)) = (a0 , b0 ) and ϕ((0, r0 )) = (c0 , d0 ) ϕ((r, r0 )) = ϕ((r, 0) + (0, r0 )) = ϕ((r, 0)) + ϕ((0, r0 )) = (a, b) + (c0 , d0 ) = (a + c0 , b + d0 ) But this element is a zero divisor with (a + c0 , b + d0 )(ca, bd) = (0, 0). Thus Im ϕ only contains zero divisors. If R contains some elements that are not zero divisors, then R ⊕ R contains some elements that are not zero divisors. Hence, ϕ is not surjective. Exercise: 28 Section 5.4 Question: Prove that ring of formal power series (see Exercise 5.2.19) over a commutative ring R is a convolution ring. Solution: Consider the monoid (N, +) and, instead of considering the subring F = Fun(N, R) of Fun(N, R). For each n ∈ N, there are only a finite number of pairs (s1 , s2 ) ∈ N ⊕ N such that s1 + s2 = n. Hence, Fun(N, R) satisfies the convolution condition. Consider the function ψ : Fun(N, R) → R[[x]] defined by ψ(f ) =

∞ X

f (n)xn .

n=0

Let g be another function in Fun(N, R). Then ψ(f + g) =

∞ X

! n

(f (n) + g(n))x =

n=0

X

∞f (n)x

n

! +

n=0

X

∞g(n)x

n

= ψ(f ) + ψ(g).

n=0

Also ψ(f ∗ g) =

∞ X n=0

n

(f ∗ g)(n)x =

∞ X



 X

 n=0

i+j=n

f (i)g(j) xn =

! X n=0

∞f (n)x

n

! X

n

∞g(n)x

= ψ(f )ψ(g).

n=0

This proves that ψ is a homomorphism. It is clear from the flexibility of Fun(N, R) that the function is a bijection. Thus ψ is an isomorphism, which shows that R[[x]] is a convolution ring. Exercise: 29 Section 5.4 Question: Let R be a commutative ring and consider the semigroup Z. Prove that F = {f ∈ Fun(Z, R) | ∃N ∈ Z such that f (n) = 0 for all n < N } is a subring of Fun(Z, R) that satisfies the convolution condition. [Compare to Exercise 5.4.28. If we view Z as isomorphic to the semigroup ({xn | n ∈ Z}, ×), then the convolution ring (F, +, ×) is called the ring of formal Laurent series over R and is denoted by R((x)).]

272

CHAPTER 5. RINGS

Solution: We consider the semigroup (Z, +). Let f1 , f2 ∈ F and suppose that f1 (n) = 0 for all n < N1 and f2 (n) = 0 for all n < N2 . For any n ∈ Z, consider the summation in X (f1 ∗ f2 )(n) = f1 (i)f2 (j) i+j=n

Then, since i = n − j and since f (j) = 0 for j < N2 , then i can be at most n − N2 . Thus, the summation will involve nonzero terms only for i satisfying N1 ≤ i ≤ n − N2 . Thus, there are only max(0, n − N1 − N2 + 1) necessary values of i (and similarly j) in the summation for (f1 ∗ f2 )(n). This number is finite for all n ∈ Z and hence F satisfies the convolution condition.

5.5 – Ideals Exercise: 1 Section 5.5 Question: Prove that Z is not an ideal of Q. Find all the ideals of Q. Solution: It is easy to see that Z is not an ideal since it is not closed under multiplication by ring elements. For instance, 51 · (1) 6∈ Z. It is also easy to notice that any nonzero element is a unit in Q. So by Proposition 5.5.14, any ideal I that is not (0) contains a nonzero element which is a unit, which implies that I = (1). So these are the only two ideals of Q. Exercise: 2 Section 5.5 Question: List all the ideals of Z/12Z. Solution: Recall that every element that is relatively prime to 12 is a unit. Also note that the notions of greatest common divisor and least common multiple still work in Z/12Z so that an ideal generated by two relatively prime elements is the whole ring and an ideal generated by two elements that are not relatively prime is the same as the ideal generated by the greatest common divisor. So all ideals are principal and generated by the divisors of 12, {(0), (1), (2), (3), (4), (6)}. Exercise: 3 Section 5.5 Question: Let R = M2 (Z). Let I be the set of matrices whose entries are multiples of 10. Show that I is an ideal of R. Solution: For some integers a, b, c, and d, every element in I looks like 10a 10b a b = 10 · . 10c 10d c d We first use the one step subgroup criteria to show that I is a subgroup under addition. Consider e f a b e f a b 10 · − 10 · = 10 · − g h c d g h c d e−a f −b = 10 · ∈ I. g−c h−d So I is a subgroup and addition is commutative in I since it’s commutative in M2 (Z). Now we check that it is closed under multiplication on the left and right by any ring elements. So, for any arbitrary matrix of R we have e f a b e f a b 10 · = 10 · g h c d g h c d ae + cf be + f d = 10 · ∈ I. ga + hc gb + hd We also check if it’s closed from multiplication by ring element from the right, a b e f ae + bg f a + bh 10 · = 10 · ∈ I. c d g h ce + dg f c + dh

5.5. IDEALS

273

This shows that I is an ideal of R. Exercise: 4 Section 5.5 Question: Modify Example 5.5.3 to find an ideal in M2 (Z) that is a right ideal but not a left ideal. Prove your claim. Solution: Consider the subset a c I= a, c ∈ Z . a c We will show that this is a right ideal but not a left ideal. For any two elements of I we will use the one step subgroup criteria to show that it is a subgroup under addition. a c b d a−b c−d − = ∈ I. a c b d a−b c−d Next we show that it is closed under multiplication on the right by any element of M2 (Z). e f a b ea + f c be + f d = ∈ I. e f c d ea + f c be + f d Since it is closed under multiplication on the left by any element of M2 (Z), this shows us that I is also a subring. Finally we show that it is not closed under multiplication on the left by an arbitrary element of M2 (Z). a b e f ae + be af + bf = c d e f ce + de cf + df which does not, in general, exist in I. Exercise: 5 Section 5.5 Question: Let R = Z[x]. Which of the following subsets are ideals of R? a) The set of polynomials whose coefficients are even. b) The set of polynomials such that every other coefficient (starting at 0) is even. c) The set of polynomials such that the 0’th coefficient is even. d) The set of polynomials with odd coefficients. e) The set of polynomials whose terms all have even degrees. (Subring but not an ideal.) Solution: Let R = Z[x]. Note that R is commutative so we only need to check whether or not the proposed subset is a left ideal to prove that it is an ideal. a) Let a(x), b(x) be polynomials with even coefficients and let r(x) ∈ Z[x]. The difference a(x) − b(x) will also have even coefficients. Furthermore, suppose that deg a(x) = m and deg r(x) = n. Then   m+n X X  r(x)a(x) = ri aj  xk . k=0

i+j=k

Since each aj is even, then each product ri aj is even, so any sum involving them is even. Hence, r(x)a(x) also has even coefficients. Thus, this set is an ideal. b) Let I2 be the set of polynomials such that every other coefficient is even. In other words, a(x) ∈ I2 if and only 2|ai if 2|i, where ai are the coefficients of a(x). If a(x), b(x) ∈ I2 , then the ith coefficient of c(x) = a(x) − b(x) is ci = ai − bi . If i is even then ai and bi are both even so ci is even. Hence I2 is closed under subtraction. Note that x + 2, x ∈ I2 but x(x + 2) = x2 + 2x + 0 and since the coefficient of x2 is not even, we see that x(x + 2) ∈ / I2 . Hence I2 is not closed under multiplication, so it is not even a subring. c) Let I3 be the set of polynomials such that the 0’th coefficient is even. Let a(x), b(x) ∈ I3 and r(x) ∈ Z[x]. Then a0 and b0 are even. The constant coefficient of a(x) − b(x) is a0 − b0 , which is even. Hence a(x) − b(x) ∈ I3 . Furthermore, the constant coefficient of r(x)a(x) is r0 a0 , which is even since a0 is even. Thus I3 is an ideal. d) The set of polynomials whose coefficients is odd is not closed under addition so it is not a subring of Z[x].

274

CHAPTER 5. RINGS

e) The set of polynomials whose terms only have even degrees is the same as Z[x2 ] so it is a subring of Z[x]. Notice that x ∈ Z[x] and x2 ∈ Z[x2 ] but x · x2 = x3 ∈ / Z[x2 ]. Thus it is not an ideal. Exercise: 6 Section 5.5 Question: Let S = {p(x) ∈ R[x] | p0 (3) = 0}. Prove that S is a subring of R[x] but not an ideal. Solution: We use the one step subgroup criteria to show that S is a subgroup under addition. For p(x), q(x) ∈ S consider (p − q)0 (3) = (p0 − q 0 )(3) = p0 (3) − q 0 (3) = 0 − 0 = 0. So p(x) − q(x) ∈ S and S is a subgroup under addition. Next we show that S is closed under multiplication. (p(x)q(x))0 = p0 (x)q(x) + p(x)q 0 (x) ⇒ p0 (3)q(3) + p(3)q 0 (3) = 0q(3) + p(3)0 = 0. This shows that S is closed under multiplication. So S is a subring. Now let m(x) ∈ R[x] be any polynomial. Then for p(x) ∈ S, (m(x)p(x))0 = m0 (x)p(x)+m(x)p0 (x). Pluggin in 3 we get, m0 (3)p(3)+m(3)p0 (3) = m0 (3)p(3). Letting m(x) = x and p(x) = 1 we see that this can be nonzero which shows that S is not closed under right multiplication by elements of R[x]. It is also not closed under left multiplication by an argument that is symmetric to the one above. So S is a subring but not an ideal. Exercise: 7 Section 5.5 Question: Let S be a set and consider the ring R = (P(S), 4, ∩). Prove that if S 0 ⊂ S, then the subring R0 = (P(S 0 ), 4, ∩) is an ideal. Solution: Recall that in the ring R, a subset A ⊆ S is its own additive inverse. Let A, B ∈ P(S 0 ), then A4B ∈ P(S 0 ). Thus P(S 0 ) is a subgroup of (P(S), 4). Furthermore, for any A ∈ P(S 0 ) and C ∈ P(S), we have A ∩ C ⊆ A ⊆ S 0 . Thus, A ∩ C ∈ P(S 0 ). This shows that P(S 0 ) satisfies the conditions of an ideal. Exercise: 8 Section 5.5 Question: Let I1 , I2 , . . . , In be n ideals in a ring R. Consider the matrix ring Mn (R). a) Prove that the set of matrices in Mn (R) where all the elements of the k’th column are elements of Ik is a left ideal of Mn (R). b) Prove that the set of matrices in Mn (R) where all the elements of the k’th row are elements of Ik is a right ideal of Mn (R). Solution: Let I1 , I2 , . . . , In be n ideals in a ring R. Consider the matrix ring Mn (R). a) Let S = {A = (aij ) ∈ Mn (R) aij ∈ Ij }. We will show that S is an ideal. Let A, B ∈ S and consider A − B = (aij ) − (bij ) = (aij − bij ). Now for any i, j we know that aij , bij ∈ Ij . Then since Ij is an ideal, aij − bij ∈ Ij for any i and j. This implies that A − B ∈ S. By the one step subgroup criteria, S is a subgroup of Mn (R). Now we show that S is closed under left multiplication of Mn (R). Let M = (mij ) be any element of Mn (R). Then we examine the (i, j)th entry of the product M · A and show that it is a member of Ij . (i, j) =

n X

mik akj

k=1

Since akj ∈ Ij for all k, this implies that (mik akj ) ∈ Ij for all k. Then the summation is a sum of elements of the ideal Ij which implies that the (i, j)th entry is an element of the ideal Ij . Since this holds for any j. This implies that M · A ∈ S. So S is a left ideal of Mn (R).

5.5. IDEALS

275

b) Let S = {A = (aij ) ∈ Mn (R) aij ∈ Ii }. We will show that S is an ideal. Let A, B ∈ S and consider A − B = (aij ) − (bij ) = (aij − bij ). Now for any i, j we know that aij , bij ∈ Ii . Then since Ii is an ideal, aij − bij ∈ Ii for any i and j. This implies that A − B ∈ S. By the one step subgroup criteria, S is a subgroup of Mn (R). Now we show that S is closed under right multiplication of Mn (R). Let M = (mij ) be any element of Mn (R). Then we examine the (i, j)th entry of the product A · M and show that it is a member of Ii . (i, j) =

n X

aik mkj

k=1

Since aik ∈ Ii for all k, this implies that (aik mkj ) ∈ Ii for all k. Then the summation is a sum of elements of the ideal Ii which is closed under addition. This implies that the (i, j)th entry is an element of the ideal Ii . Since this holds for any i and j. This implies that A · M ∈ S. So S is a right ideal of Mn (R). Exercise: 9 Section 5.5

0 1 Question: Let R = M2 (Z) and let A = . 0 0 a) Determine all the matrices in RA and all the matrices in AR. b) Obtain the matrices a 0 0 b 0 0 , , , 0 0 0 0 c 0

0 0

0 d

as elements in RAR. c) Deduce that RAR = (A) = M2 (Z). Solution: Let R = M2 (Z) and let A =

1 . 0

0 0

a) Matrices in RA are of the form Matrices in AR are of the form

a c

b 0 d 0

1 0 = 0 0

a . c

1 a 0 c

b c = d 0

d . 0

1 0

0 0

b) From the calculations we just did, we have a 0 0 0 0 0

0 = 0 b 0 = 0 0 0 0 = d d 0 0

0 a 1 0 0 0 0 0 0 0

0 0 0 b 1 0

In order to obtain the remaining desired matrix, we can do 0 0 0 0 0 1 0 = c 0 1 0 0 0 c

0 . 0

c) Now, RAR consists of linear combinations of elements of the form r1 As1 + r2 As2 + · · · rm Asm , where ri , si ∈ M2 (R). By adding the elements we obtained in part (b) as elements of RAR, we deduce that every matrix is in RAR. Exercise: 10 Section 5.5 Question: Let R = M3 (R) and let  A=

n

0 0 0

1 0 0

 0 o 1 . 0

276

CHAPTER 5. RINGS

Determine explicitly RA, AR, RAR and (A). Solution: We examine the product of an arbitrary element of R with A. So      a b c 0 1 0 0 a b d e f  0 0 1 = 0 d e  . g h i 0 0 0 0 g h We see that we can get any real numbers in the right two columns but only zeroes in the left. So   n 0 a b o RA = 0 d e  a, b, d, e, g, h ∈ R 0 g h We examine the product of A with an arbitrary element of R. So     0 1 0 a b c d 0 0 1 d e f  = g 0 0 0 g h i 0

e h 0

 f i . 0

We see that we can get any real numbers in the first two rows but only zeroes in the last. So   n d e f o AR = g h i  d, e, f, g, h, i ∈ R 0 0 0 Now we examine elements in RAR. Since we have an identity matrix, we know that RA ⊆ RAR, AR ⊆ RAR and that (A) = RAR. Now we can show that the identity matrix lives in RAR since   1 0 0 0 0 0 ∈ AR ⊆ RAR 0 0 0   0 0 0 0 1 0 ∈ RA ⊆ RAR. 0 0 1 Then their sum, which is the identity matrix, must also live in RAR. This shows that RAR = (A) = R. Exercise: 11 Section 5.5 Question: Let R be an integral domain. Prove that (a) = (b) in R if and only if a = bu for some unit u ∈ U (R). Solution: First we prove that in an integral domain, every element in a principal ideal (a) looks like ra for some r ∈ R. All elements in (a) looks like r1 as1 + r2 as2 + · · · + rn asn . Since R is commutative we can rewrite this as (r1 s1 )a + (r2 s2 )a + · · · + (rn sn )a = (r1 s1 + r2 s2 + · · · + rn sn )a. Now we will prove both directions of the if and only if statement. (=⇒) If (a) = (b), this implies that ra = b and sb = a for some r, s ∈ R. Then ra = r(sb) = b which implies that (rs)b = (1)b and by the cancellation property rs = 1 so that both r and s are units. Then a = sb = bs for some unit s. (⇐=) If a = bu for some unit u. Then a ∈ (b) so that (a) ⊆ (b). Since for some v ∈ R we have uv = 1, this implies that av = b(uv) = b. Which shows that b ∈ (a) ⇒ (b) ⊆ (a) ⇒ (b) = (a). Exercise: 12 Section 5.5 Question: Let R be a ring. Prove that the subring in M2 (R) of upper triangular matrices is not an ideal of M2 (R). Solution: We will show that the subring is not closed under multiplication for all rings R. Let r, s ∈ R be an r 0 element and consider the upper triangular matrix matrix . Then the product 0 r 0 0 r 0 0 0 = s 0 0 r s·r 0

5.5. IDEALS

277

is in general not an upper triangular matrix. This shows that the subring of upper triangular matrices is not an ideal. Exercise: 13 Section 5.5 Question: Let k be a positive integer. Prove that Mn (kZ) is an ideal in Mn (Z). Solution: We note that each entry in the matrix is a member of the ideal kZ. By the results of Exercise 5.5.8 part a, we know that Mn (kZ) is a left ideal of Mn (Z) and by part b, we know that Mn (kZ) is a right ideal as well. This implies that Mn (kZ) is an ideal in Mn (Z). Exercise: 14 Section 5.5 Question: Prove that if I is an ideal in a ring R and S is a subring that I ∩ S is an ideal of S. Solution: By Exercise 5.1.45, we know that I ∩ S is a subring of both R and also S. So we just need to prove that it is closed under multiplication from the left and right by elements of S. Let i ∈ I and s1 , s2 ∈ S. Since i ∈ I which is an ideal of R and s1 , s2 ∈ R, we know that s1 is2 ∈ I. Since i ∈ S which is a subring, we know that s1 is2 ∈ S. We conclude that s1 is2 ∈ I ∩ S. So I ∩ S is a subring and closed by multiplication on the left and right by elements of S. So I ∩ S is an ideal of S. Exercise: 15 Section 5.5 Question: Let R be a commutative ring and let I be an ideal in R. a) Prove that I[x] = {an xn + · · · + a1 x + a0 | ai ∈ I for all i} is an ideal of R[x]. b) Let a ∈ R be fixed. Prove that {p(x) ∈ R[x] | p(a) ∈ I} is an ideal of R[x]. c) Let k be a nonnegative integer. Prove that {an xn + · · · + a1 x + a0 | ai ∈ I for 0 ≤ i ≤ k} is an ideal of R[x]. d) Prove that {an xn + · · · + a1 x + a0 | a2 ∈ I} is not an ideal of R[x]. Solution: Let R be a commutative ring and let I be an ideal in R. a) Let I1 = I[x] = {an xn + · · · + a1 x + a0 | ai ∈ I for all i}. Let a(x), b(x) ∈ I1 . Then the ith coefficient of a(x) − b(x) is ai − bi , which is in I since I is closed under subtraction. Let r(x) ∈ R[x] and suppose that deg a(x) = n and deg r(x) = m. Then the kth coefficient of c(x) = r(x)a(x) is X ck = ri aj . i+j=k

Since each aj ∈ I, then ri aj ∈ I for any indices i and j so ck ∈ I and thus r(x)a(x) ∈ I1 . Thus I1 is an ideal. b) Let a ∈ R be fixed. Let I2 = {p(x) ∈ R[x] | p(a) ∈ I}. Let p(x), q(x) ∈ I2 . Then p(a) − q(a) = 0, so p(x) − q(x) ∈ I2 . Furthermore, for any polynomial r(x) ∈ R[x], r(a)p(a) = 0, so r(x)p(x) ∈ I2 , which shows that I2 is an ideal. c) Let k be a nonnegative integer. Let I3 = {an xn + · · · + a1 x + a0 | ai ∈ I for 0 ≤ i ≤ k}. Let a(x), b(x) ∈ I3 . Then the ith coefficient of a(x) − b(x) is ai − bi . Furthermore, for all i with 0 ≤ i ≤ k, ai − bi ∈ I since ai , bi ∈ I. Hence I3 is closed under subtraction. Now let r(x) ∈ R[x]. Then the mth coefficient of c(x) = r(x)a(x) is X cm = ri aj . i+j=m

For 0 ≤ m ≤ k, the summation for cm involves only the terms ai with 0 ≤ i ≤ m. Since m ≤ k, then all the involved ai are in I. Thus this above linear combination that gives the cm elements are all in I. Hence c(x) ∈ I3 , which shows that I3 is an ideal. d) Let I4 = {an xn + · · · + a1 x + a0 | a2 ∈ I}. The set I4 is closed under subtraction. Suppose that a(x) ∈ I4 with a1 ∈ / I. Then the x2 coefficient of xa(x) is precisely a1 . Since a1 ∈ / I, then xa(x) ∈ / I4 , so I4 is not closed by multiplication with anything in R. Thus I4 is not an ideal. Exercise: 16 Section 5.5 Question: Let R be a commutative ring. Prove that (a) ⊆ (b) if and only if there exists r ∈ R such that a = rb. Solution: Suppose that (a) ⊆ (b). Since R is commutative, by Proposition 5.5.9, then a ∈ (b) so a = br for some r ∈ R. Conversely, suppose that there exists r ∈ R such that a = br. Then by the definition of ideals, a ∈ (b). But then as the minimal ideal containing a, we conclude that (a) ⊆ (b).

278

CHAPTER 5. RINGS

Exercise: 17 Section 5.5 Question: Show that the ideal (2, x) in Z[x] is not a principal ideal. Conclude that Z[x] is not a PID. Solution: Let I = (2, x) so that every element looks like 2p(x) + xq(x). Now, if I were a principal ideal. It would have to be generated by an element n where there is some r ∈ Z[x] such that r · n = 2. This implies that n must be an integer and a divisor of 2. So n is 1 or -1. However, since these are both units, this would imply that I = Z[x]. We will show that I does not contain all polynomials. Note that for any polynomial in I, it is of the form 2p(x) + xq(x). Then any polynomial solved at 0 is 2p(0) + 0q(0) = 2p(0) an even number. This implies that x + 1 6∈ I. So I cannot be all of Z[x]. So I cannot be a principal ideal. Since I is an ideal of Z[x]. We conclude that Z[x] is not a PID. Exercise: 18 Section 5.5 Question: Consider the ideal I = (13x + 16y, 11x + 13y) in the ring Z[x, y]. a) Prove that I = (x − 2y, 3x + y). [Hint: by mutual inclusion.] b) Prove that (7x, 7y) ⊆ I but prove that this inclusion is strict. Solution: a) Note that 6(3x+y)+(−5)(x−2y) = (18−5)x+(6+10)y = 13x+16y and −4(x−2y)+5(3x+y) = 11x+13y. Since both generators exist in (x − 2y, 3x + y) we see that I ⊆ (x − 2y, 3x + y). Now we look for the other inclusion. We find that (−5)(13x + 16y) + (6)(11x + 13y) = x − 2y and (−4)(13x + 16y) + (5)(11x + 13y) = 3x + y. So that (x − 2y, 3x + y) ⊆ I which implies that I = (x − 2y, 3x + y). b) First we show that (7x, 7y) is a subset of I by showing the generators are elements of I. 11(13x + 16y) + (−13)(11x + 13y) = 7y and (−13)(13x + 16y) + (16)(11x + 13y) = 7x. Now, we notice that all members of (7x, 7y) look like n(7x) + m(7y) for some m, n ∈ Z[x, y]. But then, every element looks like 7(nx + my) and all coefficients are divisible by 7. This is not the case for the generators of I so we conclude that the inclusion, (7x, 7y) ⊆ I, is strict. Exercise: 19 Section 5.5 Question: Prove that Q[x, y] is not a PID. Solution: Consider the ideal I = (x, y). Elements in I must look like h(x, y)x + g(x, y)y. So every element (besides the zero element) is of degree one or greater. Assume that I = (p(x, y)) is a principle ideal. Then for some g(x, y) we have g(x, y)p(x, y) = x and we must have that p(x, y) has no y’s in it. In the same way, for some h(x, y) we must have h(x, y)p(x, y) = y so that p(x, y) has no x’s in it. Then p(x, y) must be of degree zero. However, the only zero degree polynomial in I is 0 and this cannot generate I so we have a contradiction. Therefore I must not be a principle ideal. This shows that Q[x, y] is not a PID. Exercise: 20 Section 5.5 Question: In the ring R[x, y], let I = (ax + by − c, dx + ey − f ) where a, b, c, d, e, f ∈ R. a) Prove that if the lines ax+by = c and dx+ey = f intersect in a single point (r, s), prove that I = (x−r, y−s). b) Prove that if the lines ax + by = c and dx + ey = f are parallel, then I = R[x, y]. c) Prove that if the lines ax + by = c and dx + ey = f are the same, then I = (ax + by − c). Solution: In the ring R[x, y], let I = (ax + by − c, dx + ey − f ) where a, b, c, d, e, f ∈ R. Note that the ideal I must contain d(ax + by − c) − a(dx + ey − f ) = (bd − ae)y − (dc − af ) and e(ax + by − c) − b(dx + ey − f ) = (ea − bd)x − (ec − bf ). a) Suppose that the two lines intersection in the single point (r, s). Then (r, s) solve ( ax + by = c dx + ey = f. Since there is a single solution, then ae − bd 6= 0. In particular, I contains, af − dc 1 ((bd − ae)y − (dc − af )) = y − =y−s −ae + bd ae − bd

5.5. IDEALS

279

by Cramer’s Rule and also 1 ec − bf ((ea − bd)x − (ec − bf )) = x − = x − r. ae − bd ae − bd In particular, we have show that x − r ∈ I and y − s ∈ I, so (x − r, y − s) ⊆ I. Furthermore, a(x − r) + b(y − s) = ax + by − c

and

d(x − r) + e(y − s) = dx + ey − f.

This shows that ax + by − c and dx + ey − f are in (x − r, y − s) so I ⊆ (x − r, y − s). We conclude that I = (x − r, y − s). b) Suppose that the lines are parallel. Then ae − bd = 0 and so d(ax + by − c) − a(dx + ey − f ) = −(dc − af ) 6= 0. Since this real number is not 0, the ideal must also contain −(dc − af )

−1 = 1. dc − af

Since the ideal contains 1, then it contains every element of R[x, y]. c) Suppose that the line equations trace the same line. Then dx + ey − f is a nonzero multiple of ax + by − c so the ideal is simply I = (ax + by − c). Exercise: 21 Section 5.5 Question: Consider the ideal I = (x, y) in the ring R[x, y]. Find a generating set for I k and show that I k requires a minimum of k + 1 generators. Solution: Elements in I 2 are of the form (a(x, y)x + b(x, y)y)(c(x, y)x + d(x, y)y) = a(x, y)c(x, y)x2 + (a(x, y)d(x, y) + b(x, y)c(x, y))xy + b(x, y)d(x, y)y 2 for arbitrary a(x, y), b(x, y), c(x, y), d(x, y) ∈ R[x, y]. This immediately shows that I 2 ⊆ (x2 , xy, y 2 ) but because the polynomials are arbitrary, we can show that x2 , xy, and y 2 are in I so the reverse inclusion holds. Thus a generating set for I 2 is {x2 , xy, y 2 } and similarly, a generating set for I k is {xk , xk−1 y, xk−2 y 2 , . . . , y k }. Call n + m the total degree of a term axn y m . Having determined the generating set of I k , we see that every polynomial in I k must consist of terms of total degree of k or greater. In particular, I k does not contain any polynomials with terms of total degree less than k, so no generators of I k can be such. Since {xk , xk−1 y, xk−2 y 2 , . . . , y k } ⊆ I k , we can consider the set of coefficients of these terms as appearing in polynomials in I k . It is vector space of dimension k + 1 isomorphic to Rk+1 . We can denote these coefficients a (k + 1)-tuple (c0 , c1 , . . . , ck ). If {p1 , p2 , . . . , p` } is any set of polynomials with ` ≤ k, then the set of coefficients of {xk , xk−1 y, xk−2 y 2 , . . . , y k } in r1 (x, y)p1 (x, y) + · · · + r` (x, y)p` (x, y) k+1

is a vector subspace of R of dimension `. Thus the set of combinations of the pi (x, y) cannot be all of I k . k Thus, I requires a minimum of k + 1 generators. Exercise: 22 Section 5.5 Question: Let ϕ : R → S be a ring homomorphism. a) Show that if J is an ideal in S, then ϕ−1 (J) is an ideal in R. b) Show that if I is an ideal in R, then ϕ(I) is not necessarily an ideal in S. c) Show that if ϕ is surjective and I is an ideal of R, then ϕ(I) is an ideal of S. Solution: a) Let J be an ideal of S and consider the set I = ϕ−1 (J) of R. Let a, b ∈ I. We show that I is a subgroup under addition using the one step subgroup criterion. ϕ(a) − ϕ(b) ∈ J ⇒ ϕ(a − b) ∈ J ⇒ a − b ∈ I. Then I is a subgroup under addition. We now show multiplication. Let r ∈ R. Then ϕ(r) · ϕ(a) ∈ J ⇒ ϕ(ra) ∈ J ⇒ ra ∈ I. So I is an ideal in R.

280

CHAPTER 5. RINGS

b) We show that this statement is true via example. Let ϕ : Z → Z[i] be a ring homomorphism with ϕ(a) = a. Then Z is obviously an ideal in Z but ϕ(Z) = Z is not an ideal in Z[i] since i · ϕ(1) = i ∈ / Z. c) Let ϕ be a surjective ring homomorphism from R to S and I be an ideal in R. Consider the set J = ϕ(I). Let a, b ∈ I so that ϕ(a), ϕ(b) ∈ J and a − b ∈ I. We show J is a subgroup under addition using the one step subgroup criterion. ϕ(a − b) ∈ J ⇒ ϕ(a) − ϕ(b) ∈ J. Now we show J is closed under multiplication by ring elements. Let s ∈ S. Since ϕ is surjective, there is some r ∈ R where ϕ(r) = s. Then ra ∈ I, ϕ(ra) ∈ J and s · ϕ(a) = ϕ(r) · ϕ(a) = ϕ(ra) ∈ J. So that J is closed under multiplication by any ring elements and J is an ideal. Exercise: 23 Section 5.5 Question: Suppose that R is a commutative ring with 1 6= 0. Prove that R is a field if and only if its only ideals are (0) and (1). Solution: (=⇒) Suppose that R is a field. Certainly (0) and (1) are distinct ideals. Now, any other ideal I distinct from those two must have some nonzero element a ∈ I. Since R is a field, we know that a is a unit. However this implies that I = (1) by Proposition 5.5.14. So (0) and (1) are the only ideals. (⇐=) Suppose that the only ideals of R are (0) and (1). Then consider the ideal (a) generated by any nonzero element a ∈ R. Now, (a) must be (0) or (1). It is clearly not (0) and so we must have (a) = (1). By Proposition 5.5.14, (a) contains some unit which looks like ra for some r ∈ R. So there is some s ∈ R so that s(ra) = (sr)a = 1. This shows that any nonzero element a is a unit which shows that R is a field.

Exercise: 24 Section 5.5 Question: Suppose that R is a commutative ring with 1 6= 0. Prove that a principal ideal (a) = R if and only if a is a unit. [Exercise 5.5.9 gives an example of a noncommutative ring where this result does not hold.] Solution: (=⇒) Suppose that (a) = R. Since R has an identity, since R is Pn we know that Pn(a) = RAR.PFurthermore, n commutative, then any element in (a) looks like i=1 ri (a)si = i=1 (ri si )a = ( i=1 ri si )a = qa for some q ∈ R. Then if (a) = R, by Proposition 5.5.14, we know that qa is a unit for some q ∈ R. Then there is some s ∈ R where s(qa) = (sq)a = 1. This shows that a is a unit. (⇐=) This is a direct application of Proposition 5.5.14.

Exercise: 25 Section 5.5 Question: Let I and J be ideals of a ring R. Prove that: a) I + J is an ideal of R and b) I ∩ J is an ideal of R. Solution: Let ik ∈ I and jk ∈ J be any elements in I and J respecitively where we will replace k with an index. a) Any element in I + J looks like i1 + j1 . We first show that this is a subgroup under addition using the one step subgroup criteria. i1 + j1 − (i2 + j2 ) = (i1 − i2 ) + (j1 − j2 ) ∈ I + J. Now we show that it is closed under multiplication on either side by any element r, s ∈ R. Then r(i1 + j1 )s = ri1 s + rj1 s = i2 + j2 ∈ I + J. This shows that I + J is an ideal. b) Let i ∈ I ∩ J. By Exercise 5.1.45 we know that this is a subring so we just need to show that it is closed under multiplication. For any r, s ∈ R we have r(i)s ∈ I since i ∈ I and r(i)s ∈ J since i ∈ J. So r(i)s ∈ I ∩ J. and I ∩ J is closed under multiplication by elements on either side. This shows that I ∩ J is an ideal.

5.5. IDEALS

281

Exercise: 26 Section 5.5 Question: Let C be an arbitrary (not necessarily finite) collection of ideals of a ring R. Prove that \

I

I∈C

is an ideal of R. T Solution: Let J = I∈C I. By Exercise 5.1.45 we know that this is a subring so we just need to show that it is closed under multiplication. Let i ∈ J. For any r, s ∈ R and any ideal I ∈ C. we know that r(i)s ∈ I since I is an ideal, so that r(i)s ∈ J. This shows that J is an ideal of R. Exercise: 27 Section 5.5 Question: Suppose that I and J are ideals in a ring R that are generated by certain finite sets of elements, say I = (a1 , a2 , . . . , am ) and J = (b1 , b2 , . . . , bn ). 1. Prove that I + J = (a1 , a2 , . . . , am , b1 , b2 , . . . , bn ). 2. prove that IJ is generated by the set of mn elements, {ai bj | i = 1, 2, . . . , m, j = 1, 2, . . . , n}. Solution: a) We will show each of the generators on the right hand side is in I + J. Without loss of generality, take a generator ai . ai = ai + 0 ∈ I + J. The same reasoning can be applied for each generator and (a1 , a2 , . . . , am , b1 , b2 , . . . , bn ) ⊆ I + J. Now, consider an arbitrary element i + j ∈ I + J. Each can be rewritten using the given generators and a set of weights from their own ring. So we rewrite as i + j = r1 a1 + r2 a2 + . . . + rm am + s1 b1 + s2 b2 + . . . + sn bn ∈ (a1 , a2 , . . . , am , b1 , b2 , . . . , bn ). This proves that I + J = (a1 , a2 , . . . , am , b1 , b2 , . . . , bn ). b) Consider an arbitrary element of IJ which looks like ij. We can rewrite this in terms of our generators, (r1 a1 + r2 a2 + . . . + rm am ) · (s1 b1 + s2 b2 + . . . + sn bn ) = Πi≤m,j≤n ri ai sj bj = Πi≤m,j≤n (ri ai sj )bj = Πj≤n (Πi≤m qi ai )bj

= Πi≤m,j≤n qi ai bj .

We note that (ri ai sj ) = Πi≤M qi ai for some set of coefficients since I is an ideal. The last step comes from rearranging and combining terms based on ai bj . It is easy to see that this element is contained in {ai bj | i = 1, 2, . . . , m, j = 1, 2, . . . , n}. Now we show the reverse inclusion by noticing that any generator ai bj is a member of the product IJ via (ai ) · (bj ). Then IJ = {ai bj | i = 1, 2, . . . , m, j = 1, 2, . . . , n}. Exercise: 28 Section 5.5 Question: Let I and J be ideals of a ring R. Prove that I ∪ J is not necessarily an ideal of R. Solution: We will provide a counter example. Consider the ideals 2Z and 3Z. Consider the union of the two ideals. Then we know that 2, 3 ∈ 2Z ∪ 3Z. If the union were an ideal, it would be a subgroup under addition and 3 − 2 = 1 should lie in 2Z ∪ 3Z. However, 1 6∈ 2Z and 1 6∈ 3Z. So the union is not a subgroup under addition and this shows that the union of ideals is not an ideal in general. Exercise: 29 Section 5.5 Question: Let I, J, K be ideals of a ring R. Prove that: 1. I(J + K) = IJ + IK; 2. I(JK) = (IJ)K. Solution: Let I, J, K be ideals of a ring R.

282

CHAPTER 5. RINGS

a) An element in I(J + K) has the form r = a1 (b1 + c1 ) + a2 (b2 + c2 ) + · · · + an (bn + cn ) for some positive integer n and ai ∈ I, bi ∈ J, and ci ∈ K. Then r = (a1 b1 + a2 b2 + · · · + an bn ) + (a1 c1 + a2 c2 + · · · + an cn ) which shows that r ∈ IJ + IK. Thus I(J + K) ⊆ IJ + IK. Now let s ∈ IJ + IK. Then s has the form a1 b1 + a2 b2 + · · · + am bm + a01 c1 + a02 c2 + · · · + a0n cn with ai , a0j ∈ I, bi ∈ J, and cj ∈ C. We can write this sum as s = a1 (b1 + 0) + · · · am (bm + 0) + a01 (0 + c1 ) + · · · + a0n (0 + cn ), which shows s as an element of I(J + K) since 0 ∈ J ∩ K. By mutual inclusion I(J + K) = IJ + IK. b) Let r ∈ I(JK). Then r has the form r = a1 (b11 c11 + · · · + b1n1 c1n1 ) + a2 (b21 c21 + · · · + b2n2 c2n2 ) + · · · + am (bm1 cm1 + · · · + bmnm cmnm ), for ai ∈ I, b ∈ J and c ∈ K. Upon distribution, r can be written as a sum of the form r = (a01 b01 )c01 + (a02 b02 )c02 + · · · + (a0n b0n )c0n , which shows that r ∈ (IJ)K. Thus, I(JK) ⊆ (IJ)K. It is a similar reasoning to show that (IJ)K ⊆ I(JK). Hence, the two ideals are equal. Exercise: 30 Section 5.5 Question: Let I1 ⊆ I2 ⊆ · · · ⊆ Ik ⊆ · · · be a chain (by in the partial order of inclusion) of ideals in a ring R. Prove that ∞ [ Ik k=1

is an ideal in R. Solution: Let J = ∪∞ k=1 Ik We first show that it is a subgroup under addition. Consider i1 , i2 ∈ J. Then for some l, m we have i1 ∈ Il and i2 ∈ Im . We will assume without losing generality that m ≥ l. Then Il ⊆ Im so that i1 ∈ Im . Then the difference, i1 − i2 ∈ Im ⇒ i1 − i2 ∈ J. By the one step subgroup criteria, J is a subgroup under addition. Now we show that it is closed under multiplication. Let i ∈ J, then for some Ik we know that i ∈ Ik . For any r ∈ R, we have ri, ir ∈ Ik ⇒ ri, ir ∈ J. This shows that J is closed under multiplication by any ring elements on the right and left. So J is an ideal of R. Exercise: 31 Section 5.5 Question: Let R be a commutative ring and let I be an ideal of R. Prove that the subset {an xn + · · · + a1 x + a0 ∈ R[x] | ak ∈ I k } is a subring of R[x] but not necessarily an ideal. Solution: We denote the subset with J. We first show that it is a subgroup under addition by using the one step subgroup criteria. We know that for any ak , bk ∈ I k that ak − bk ∈ I k . Then (an xn + · · · + a1 x + a0 ) − (bn xn + · · · + b1 x + b0 ) = (an − bn )xn + . . . + (a1 − b1 )x + (a0 − b0 ) ∈ J. So J is a subgroup under addition. Now we try an arbitrary product, (an xn + · · · + a1 x + a0 ) · (bn xn + · · · + b1 x + b0 ) =

2n X k X ( ai bk−i )xk k=0 i=0

Now we will show that ai bk−i ∈ I k for any k and i. Now, we know that ai = i1 i2 . . . ii ∈ I i and likewise bk−i = h1 h2 . . . hk−i ∈ I k−i . Then ak bk−i = i1 i2 . . . ii h1 h2 . . . hk−i ∈ I k−i+i = I k for any k and i. Then

5.5. IDEALS

283

Pk k the coefficient i=0 ai bk−i also lies in I . This implies that the product exists in J. So J is closed under multiplication. This shows that J is a subring. Now we note that if we multiply an arbitrary element on the right by x ∈ R[x] we get an xn+1 + . . . + a1 x2 + a0 x. We know that the powers of ideals satisfy I k+1 ⊆ I k so in order for this to hold we would need the opposite inclusion which would imply that I k+1 = I k for all k. This does not hold in general so J is not necessarily an ideal. Exercise: 32 Section 5.5 Question: Let Un (R) be the ring of upper triangular n × n matrices with coefficients in a ring R. Let I be an ideal in R. Prove that S = {(aij ) ∈ Un (R) | aij ∈ I j−i }. is a subring of Un (R) but not an ideal. Solution: We note that I 0 is defined as R. We first show that S is a subgroup under addition, consider two arbitrary members of S, (aij ) and (bij ). Then (aij ) − (bij ) = (aij − bij ). Since aij − bij ∈ I j−i for all j and i, we have that (aij − bij ) ∈ S. So S is a subgroup under addition by the one step subgroup criteria. Now we check that it is closed under product. By the definition of matrix multiplication, the (i, j)th entry of the product Pn Pj (aij ) · (bij ) is k=1 aik bkj . Since many of these entries are zero we can rewrite the sum as k=i aik bkj where we under stand that if i > j this summation is zero. Now if i ≤ j, then aik = a1 a2 . . . ak−i ∈ I k−i where ai ∈ I and bkj = b1 b2 . . . bj−k ∈ I j−k . Then the product aik bkj = a1 a2 . . . ak−i b1 b2 . . . bj−k ∈ I j−k+k−i = I j−i . Then the (i, j)th element is just a summation of elements in the ideal I j−i which implies that the (i, j)th element is in I j−i . So, (aij ) · (bij ) ∈ S and S is closed under products. This shows that S is a subring. To show this does not hold in general, assume that R contains an identity. Now we examine the product of an arbitrary element (aij ) ∈ S with a matrix (bij ) ∈ Un (R) where b1n = 1 and 0 elsewhere. It is easy to see that (aij ) · (bij ) = (cij ) is a matrix of all zeroes except c1n = a11 . In order for this to be a member of S, we need a11 ∈ I n−1 for all a11 ∈ I 0 = R, or R ⊆ I n−1 . Since this does not always hold for an arbitrary ideal I, S is not an ideal. Exercise: 33 Section 5.5 Question: Let R be a commutative ring. 1. Prove that if I and J are comaximal ideals, then IJ = I ∩ J. 2. Prove that if I1 , I2 , . . . , In is a finite collection of pairwise comaximal ideals, then I1 I2 · · · In = I1 ∩ I2 ∩ · · · ∩ In . Solution: Note that the notion of comaximal only applies to ideals in a commutative ring with an identity 1. a) By construction, we know that IJ ⊆ I ∩ J. If I and J are comaximal, then there exists a ∈ I and b ∈ J such that a + b = 1. Then consider and element c ∈ I ∩ J. We have c = c(a + b) = ca + cb = ac + cb. Since c ∈ J, then ac ∈ IJ and since c ∈ I, then bc ∈ IJ. Thus c ∈ IJ. This shows that I ∩ J ⊆ IJ. Thus IJ = I ∩ J. b) Let I1 , I2 , . . . , In be a finite collection of pairwise comaximal ideal. As above, we know that I1 I2 · · · In ⊆ I1 ∩ I2 ∩ · · · ∩ In by definition of the ideal operations. Now let c ∈ I1 ∩ I2 ∩ · · · ∩ In . Since all the ideals are pairwise comaximal, for any pair (i, j) with 1 ≤ i < j ≤ n, there exist aij and bij such that aij + bij = 1. Then   X c = c aij + bij  1≤i<j≤n

and upon expanding the product, we obtain a sum of products in which each product contains for Ik each with 1 ≤ k ≤ n. Hence, I1 ∩ I2 ∩ · · · ∩ In ⊆ I1 I2 · · · In . Thus, these two are equal.

284

CHAPTER 5. RINGS

Exercise: 34 Section 5.5 Question: Let R be a commutative ring and let I be an ideal in R. Prove that the radical defined in Definition 5.5.21 is an ideal. √ Solution: Let a, b ∈ I as defined in Definition 5.5.21. We first show that it is a subgroup under addition, consider a − b. Now, an and bm ∈ I for some positive integers n and m. Then consider (a − b)n+m =

n+m X i=0

(−1)n+m−i

n + m i n+m−i ab . i

When i ≤ n then n + m − i ≥ n + m − n = m so that bn+m−i = bk bm for some k and since bm ∈ I, we have that ((−1)n+m−i n+m ai bk )bm ∈ I. When i > n then ai = ak an for some k. Since an ∈ I we i n+m have ai bn+m−i = ((−1)n+m−i i ak )an (bn+m−i ) ∈ I. So (a − b)n+m is just a sum of elements of I, so √ √ we (a − b)n+m√∈ I ⇒ (a − b) ∈ I. So I is a subgroup under addition by the one step subgroup criteria. Now √ show that I is closed under products on either side by elements of R. Let r ∈ R be any element and a ∈ √ I n n n n where we know that ar = ra ∈ I. √ a ∈ I. Then (ra) = (ra)(ra) . . . (ra) = r a ∈ I. Since R is commutative, √ So I is closed under multiplication by elements of R. This shows that I is an ideal. Exercise: 35 Section 5.5 p p p Question: Let R = Z. Calculate the following radical ideals: a) (72); b) (105); c) (243). αk max (α1 ,α2 ,...αk ) 1 α2 Solution: Let n = pα = k·n 1 p2 . . . pk and consider any number q where pi |q for all i. Then q m for some k. Likewise if any number p q = k · n for some k, then q must have every prime present in its prime decomposition. This shows that (n) = (p1 p2 . . . pk ). p a) (72) = (6). p b) (105) = (105). p c) (243) = (3).

Exercise: 36 Section 5.5 √ √ Question: Let I be an ideal in a commutative ring R. Prove that the radical of I is again I. √ √ n m Solution: Let r be any element of the radical of I so that rn ∈ I. I for some m and √ √ Then (r ) ∈ √ m (rn )√ = rn rn√. . . rn = rnm ⇒ r is√an element of I. So the radical of I is a subset of I. Since for any √ r ∈ I, r1 ∈ I, this implies that I is a subset of the radical of I which proves equality. Exercise: 37 Section 5.5 Question: p Let R be a commutative ring. Show that the set N (R) of nilpotent elements is equal to the ideal (0). [For this reason, the subring of nilpotent elements in a commutative ring is often called the nilradical of R.] Solution: p p Let a ∈ N (R). This implies ak = 0 for some integer k. Then by definition, a ∈ (0). Suppose that a ∈ (0), then by definition of radical,pak ∈ (0) for some integer k. Then ak = 0 for some integer k and a ∈ N (R). This shows the equality N (R) = (0). Exercise: 38 Section 5.5 Question: In the ring Z, prove the following fraction ideal equalities. a) ((2) : (0)) = Z b) ((24) : (4)) = (6) c) ((17) : (15)) = (17). Solution: a) Since r · 0 = 0 ∈ (2) for all r ∈ Z, ((2) : (0)) = Z. b) Consider an element r ∈ R so that for all k ∈ Z we have r · (4k) = 24m for some m ∈ Z. Dividing both sides by 4 we get, r · k = 6 · m. Since this must work for every single k ∈ Z, using k = 1 show us that r must be a multiple of 6. Then ((24) : (4)) ⊆ (6).

5.6. QUOTIENT RINGS

285

Consider any r ∈ (6) which looks like 6k for some k ∈ Z. Then for any element 4m ∈ (4), we have 6k · 4m = 24(km) ∈ (24). This shows the other containment and so ((24) : (4)) = (6). c) Consider any element r ∈ ((17) : (15)). Then for any k ∈ Z, r · (15k) = 17m for some m ∈ Z. Considering k = 1, we see that r · 15 = 17m. Then we must have that r is a multiple of 17. Then ((17) : (15)) ⊆ (17). Now consider any element 17k ∈ (17). 17k · 15m = 17(15(km)) ∈ (17). This shows the equality ((17) : (15)) = (17).

5.6 – Quotient Rings Exercise: 1 Section 5.6 Question: Refer to Exercises 5.5.5. For subsets that were ideals, describe the elements and operations in the corresponding quotient ring. Solution: Only the subsets described in parts (a) and (c) are ideals. Consider the ideal I1 of polynomials whose coefficients are even. Two elements in the quotient ring an xn + · · · + a1 x + a0

and

bn xn + · · · + b1 x + b0

are equal if and only if the difference of the polynomials is in I1 so bi − ai is even for all i. Consequently, we can use a complete set of distinct representatives of elements in I1 polynomials of the form an xn + · · · + a1 x + a0 with ai ∈ {0, 1}. By considering the addition and multiplication of polynomials, we see that Z[x]/I2 ∼ = (Z/2Z)[x]. Now consider the ideal I3 of polynomials such that the 0th coefficient is even. Then two elements in the quotient ring an xn + · · · + a1 x + a0 and bn xn + · · · + b1 x + b0 are equal if and only if the difference of the polynomials is in I3 , so a0 − b0 is even. There are only two distinct elements in Z[x]/I3 , namely the set of polynomials with 0 as 0th coefficient, and the set of polynomials with 1 as the 0th coefficient. Furthermore, the addition and multiplication by polynomials shows that Z[x]/I3 ∼ = Z/2Z. Exercise: 2 Section 5.6 o n Question: Prove that the quotient ring Z[x]/(5x − 1) is isomorphic to the subring Z[ 51 ] = 5nk n ∈ Z, k ∈ N of Q. Solution: Consider the function ϕ : Z[x]/(5x − 1) → Q defined by ϕ(p(x)) = p(1/5), for all p(x) ∈ Z[x]. Since we defined this “function” on cosets based on a representative p(x) of a coset, we need to check that ϕ is well defined. Two elements a(x), b(x) ∈ Z[x] are in the same coset of (5x − 1) if and only if a(x) − b(x) ∈ (5x − 1). In other words, a(x) − b(x) = d(x)(5x − 1) for some d(x) ∈ Z[x]. So if a(x) and b(x) are two elements in the same coset of (5x − 1), then 1 a(1/5) = b(1/5) + d(1/5)(5 − 1) = b(1/5). 5 Hence ϕ is well-defined. The image of ϕ consists of all rational numbers of the form p(1/5), where p(x) ∈ Z[x], which precisely gives Z[ 51 ]. Now suppose that ϕ(p(x)) = 0. Then p(1/5) = 0. The only polynomials in Z[x] with p(1/5) = 0 are polynomials that are multiples of (5x − 1). Hence p(x) ∈ (5x − 1) and thus p(x) = 0 in the quotient ring. Thus ϕ is injective. We have shown that ϕ is an isomorphism onto Z[1/5]. Exercise: 3 Section 5.6 Question: Let [a, b] be an interval in R and consider the ring C 2 ([a, b], R) (real-valued functions over [a, b]

286

CHAPTER 5. RINGS

whose zeroeth, first, and second derivatives are continuous). The operations are addition and multiplication of functions. Let I = {f (x) ∈ C 2 ([a, b], R) | f (−1) = f 0 (−1) = f 00 (−1) = 0}. Show that I is an ideal and describe the quotient ring C 2 ([a, b], R)/I. Solution: Let f, g ∈ I. Then (f + g)(−1) = f (−1) + g(−1) = 0 (f + g)0 (−1) = f 0 (−1) + g 0 (−1) = 0 (f + g)00 (−1) = f 00 (−1) + g 00 (−1) = 0, so f + g ∈ I. Let f ∈ I and let h ∈ C 2 ([a, b], R). Then (f h)(−1) = f (−1)h(−1) = 0 (f h)0 (−1) = f 0 (−1)h(−1) + f (−1)h0 (−1) = 0 (f h)00 (−1) = f 00 (−1)h(−1) + f 0 (−1)h0 (−1) + f (−1)h00 (−1) = 0 Thus, I is close under multiplication by any element in the ring. We have shown that I is an ideal. Exercise: 4 Section 5.6 Question: Consider the ideal I = (x + 1, x2 + 1) in the ring R[x]. Prove that 2 ∈ (x + 1, x2 + 1). Deduce that I = R[x] so that R[x]/I ∼ = {0}, the trivial ring. Solution: ] (x2 + 1) − (x + 1)(x + 1) + 2(x + 1) = x2 + 1 − x2 − 2x − 1 + 2x + 2 = 2 ∈ I Since 2 is a unit in R[x], by Proposition 5.5.14 I = R[x]. It follows that R[x]/I is the trivial ring. Exercise: 5 Section 5.6 Question: Consider the ideal I = (x + 1, x2 + 1) in the ring Z[x]. Prove that 2 ∈ (x + 1, x2 + 1). Prove that I = (x + 1, 2). Prove that Z[x]/I ∼ = Z/2Z. Solution: (x2 + 1) − (x + 1)(x + 1) + 2(x + 1) = x2 + 1 − x2 − 2x − 1 + 2x + 2 = 2 ∈ I Since 2 ∈ I, it is clear that (x + 1, 2) ⊆ I. We will show that I = (x + 1, 2) by showing the second generator is contained in (x + 1, 2). Now (x + 1)(x + 1) − 2(x + 1) + 2 = x2 + 2x + 1 − 2x − 2 + 2 = x2 + 1 ∈ (x + 1, 2). This shows I ⊆ (x + 1, 2) so that (x + 1, 2) = I. Since x+1 ∈ I, in Z[x]/I we have the property that x̄ = −1̄. Then for any polynomial p(x), p(x) = p(−1) = n̄ for some integer n. Additionally, 2̄ = 0̄, which implies that every integer (and therefore every polynomial) is equivalent to 0̄ or 1̄. So the only elements in our quotient Z[x]/I are 1̄ and 0̄ and Z[x]/I ∼ = Z/2Z. Exercise: 6 Section 5.6 Question: Consider the quotient ring R[x]/(x3 + x − 2). a) In this quotient ring, calculate and simplify as much as possible the sum and product of x2 + 7x − 1 and 2x2 − x + 5. b) Prove that this quotient ring is not an integral domain. Solution: a) We have the relation that x̄3 = −x̄ + 2̄ which we can use to simplify polynomials with degree greater than three. So, 4

3

4

3

2

2

2

(x2 + 7x − 1) · (2x2 − x + 5) = 2x + 14x − 2x − x̄3 − 7x + x̄ + 5x + 35x − 5̄ ¯ − 5̄ = 2x + 13x + −4̄x̄2 + 36x̄ 2 ¯ ¯ − 5̄ = 2x(−x̄ + 2̄) + 13(−x̄ + 2̄) − 4x + 36x̄ 2

2

= −2x + 4x − 13x + 26 − 4x + 36x − 5̄ 2 ¯ = −6x + 53x + 21.

5.6. QUOTIENT RINGS

287

b) We can prove it is not an integral domain by finding two nonzero polynomials in our quotient that multiply to be 0̄. We can accomplish this by factoring the polynomial x3 + x − 2. We note that this polynomial has a root of 1, so, x3 + x − 2 = (x2 + x + 2)(x − 1). Since both polynomials on the right are nonzero members of our quotient, then (x2 + x + 2)(x − 1) = x3 + x − 2 = 0. Exercise: 7 Section 5.6 Question: Consider the quotient ring Z[x]/(2x3 + 5x − 1). a) In this quotient ring, calculate and simplify as much as possible the sum and product of 4x2 − 5x + 2 and 2x + 7. b) Show that x is a unit. Solution: Consider the quotient ring Z[x]/(2x3 + 5x − 1). Note that two elements a(x), b(x) ∈ Z[x] have a(x) = b(x) if and only if a(x) − b(x) ∈ (2x3 + 5x − 1), or in other words, a(x) − b(x) = d(x)(2x3 + 5x − 1) for some polynomial d(x). Thus, either a(x) − b(x) = 0 or deg(a(x) − b(x)) ≥ 3. a) We have 4x2 − 5x + 2 + 2x + 7 = 4x2 − 3x + 9. There is no simplification. For the product, we have 4x2 − 5x + 2 · 2x + 7 = 8x3 + 28x2 − 10x2 − 35x + 4x + 14. However, in this quotient ring, 2x3 + 5x − 1 = 0 so 2x3 = −5x + 1. Thus 4x2 − 5x + 2 · 2x + 7 = 4(−5x + 1) + 18x2 − 31x + 14 = 18x2 − 51x + 18. b) It is not hard to notice that (2x2 + 5) · x = 2x3 + 5x = 1. Hence x is a unit in the quotient ring. Exercise: 8 Section 5.6 Question: Let f (x) = x2 + 2 in F5 [x]. a) Prove that the quotient ring F5 [x]/(f (x)) has 25 elements. b) Prove that F5 [x]/(f (x)) contains no zero divisor. c) Deduce that F5 [x]/(f (x)) is a field. [Hint: See Exercise 5.1.23] Solution: a) In our quotient, we have the relation that x̄2 = 5̄. So any polynomial with degree greater than or equal to 2 can be reduced to a polynomial of the form ax + b̄ where a and b are members of F5 . So we have no more than 25 elements in our field. Now if ax + b = cx + d for two pairs of numbers not exactly the same, this implies that (a − c)x + (b − d) = 0̄ which can only happen for degree one polynomials if a = c and b = d. This shows that each of these 25 elements is unique so there are exactly 25 elements. b) Any element of the form 0x + a (with a nonzero) cannot be a zero divisor since it is an element of the field F5 . Now, any other element is a degree one polynomial. If we have a degree one polynomial and we multiply it with any of the 24 nonzero elements in our quotient, we will either get another degree one polynomial (which cannot be zero) or a degree two polynomial (which could be zero if it was a multiple of x2 + 2). So we need to show that x2 + 2 cannot be factored into degree one polynomials. This is equivalent to seeing if it has any roots, if it does not, then it is irreducible. So we check all numbers in F5 12 + 2 = 3 22 + 2 = 1 32 + 2 = 3 42 + 2 = 3 02 + 2 = 2. So the polynomial is irreducible and no degree one polynomial can multiply to be a multiple of x2 + 2 or, equivalently in our quotient, zero. This shows that there are no zero divisors. c) Our quotient has an identity (1) and no zero divisors making it an integral domain. Since our quotient is of finite size and an integral domain, we know from Exercise 5.1.23 that it must be a field.

288

CHAPTER 5. RINGS

Exercise: 9 Section 5.6 Question: Let R be a commutative ring with a 1 6= 0 and let a(x) ∈ R[x] with deg a(x) = n > 0 and an ∈ U (R). n−1 a) Prove that, in the quotient ring R[x]/(a(x)), the element x satisfies xn = −a−1 + · · · + a1 x + a0 ). n (an−1 x b) Prove that for every polynomial p(x), there exists a polynomial q(x) that is either 0 or has deg q(x) < n such that p(x) = q(x) in R[x]/(a(x)). [Hint: Use induction on the degree of p(x).] c) Prove that the polynomial q(x) described above is unique. Solution: Let R be a commutative ring with a 1 6= 0 and let a(x) ∈ R[x] with deg a(x) = n and an ∈ U (R). a) In the quotient ring R[x]/(a(x)), we have a(x) = 0 =⇒ an xn + · · · + a1 x + a0 = 0 =⇒ an xn = −(an−1 xn−1 + · · · + a1 x + a0 ) =⇒ an · xn = −(an−1 xn−1 + · · · + a1 x + a0 ) =⇒ an · xn = −(an−1 xn−1 + · · · + a1 x + a0 ) n−1 =⇒ xn = −a−1 + · · · + a1 x + a0 ). n (an−1 x

b) Suppose that 0 ≤ deg p(x) ≤ n−1. Then the statement is satisfies trivially since p(x) satisfies the conditions for q(x). Now suppose that the statement is true for all polynomials of degree m ≥ n − 1. Let p(x) have degree m + 1 with p(x) = pm+1 xm+1 + · · · + p1 x + p0 n−1 + · · · + a x + a ) + p xm + · · · + p . = pm+1 xm+1−n a−1 n (an−1 x 1 0 m 0

Setting the polynomial under the bar as r(x), we see that r(x) has degree m. Using the induction hypothesis, r(x) = q(x) for some polynomial of degree less that n. c) Suppose that p(x) = q1 (x) = q2 (x) where q1 (x) and q2 (x) both have degree less than n. Then q1 (x) − q2 (x) is a multiple of a(x), say with q1 (x) − q2 (x) = d(x)a(x). Since an ∈ U (R) it is not a zero divisor so the leading term of d(x)a(x) = d` an x`+n , where deg d(x) = `. Since an is not a zero divisor, then d` an 6= 0 and deg d(x)a(x) = deg d(x) + deg a(x). In particular, q1 (x) − q2 (x) = 0 or deg(q1 (x) − q2 (x)) ≥ n. Since both deg q1 (x), deg q2 (x) < n, then q1 (x) = q2 (x). Exercise: 10 Section 5.6 Question: Consider the ring Z[i] and the ideal I = (2 + i). We study the quotient ring Z[i]/(2 + i). 1. Prove that −1 + 2i, −2 − i, and 1 − 2i are in the ideal (2 + i). 2. Prove every element a + bi ∈ Z[i] is congruent modulo I to at least one element inside or on the boundary of the square with 0, 2 + i, 1 + 3i, and −1 + 2i. Prove also that the vertices of the square are congruent to each other modulo I. 3. Using division in C, show that none of the five elements in 0, i, 2i, 1 + i, 1 + 2i are congruent to each other and conclude that Z[i]/(2 + i) = {0, i, 2i, 1 + i, 1 + 2i}. 4. Write down the multiplication table in Z[i]/(2 + i) and deduce that this quotient ring is a field. 5. Find an explicit isomorphism between Z[i]/(2 + i) and F5 . Solution: Consider the ring Z[i] and the ideal I = (2 + i). We study the quotient ring Z[i]/(2 + i). a) −1 + 2i = i(2 + i) so −1 + 2i ∈ (2 + i). Also −2 − i = −1(2 + i) so −2 − i ∈ (2 + i). Finally and 1 − 2i = −i(2 + i) so 1 − 2i ∈ (2 + i). b) Two elements in Z[i]/(2 + i) are equivalent if they differ by a multiple a(2 + i), where a ∈ Z. However, viewed as elements in the complex plane, two elements different by a(2 + i) if their difference as vectors is an integer multiple of (2, 1). Similarly, two elements are equivalent in the quotient ring if they happen to differ by bi(2 + i) = b(−1 + 2i). This different corresponds to an integer multiple of (−1, 2). The integer multiples a(2, 1) + b(−1, 2) correspond to a square lattice on C as shown in the figure. So anything in the lattice Z[i] is equivalent to an element in the box as show or to a vertex of the square. Furthermore, the vertices of the square differ from each other by the elements mentioned in (a). Hence all the vertices are equivalent in Z[i]/(2 + i).

5.6. QUOTIENT RINGS

289

c) We want to verify that for no pair (z1 , z2 ) of the five complex numbers do we have z1 − z2 (z1 − z2 )(2 − i) = ∈ Z[i]. 2+i 5 This is obvious if z1 = 0. i

:

2i

:

(1 + i)

:

1 − 2i i − (1 + i) −2 + i i − (1 + 2i) −3 − i i − 2i = , = , = 2+i 5 2+i 5 2+i 5 2i − (1 + i) −1 + 3i 2i − (1 + 2i) −2 + i = , = 2+i 5 2+i 5 (1 + i) − (1 + 2i) −1 − 2i = . 2+i 5

This verifies that none of the five given elements are congruent to each other modulo the ideal (2 + i). Thus Z[i]/(2 + i) = {0, i, 2i, 1 + i, 1 + 2i}. d) The mutiplication table is × 0 i 2i 1 + i 1 + 2i 0 0 0 0 0 0 i 0 1+i i 1 + 2i 2i 2i 0 i 2i 1 + i 1 + 2i 1 + i 0 1 + 2i 1 + i 2i i 1 + 2i 0 2i 1 + 2i i 1+i From this multiplication table, we see that Z[i]/(2+i)−{0} is a group under the operation of multiplication. Since multiplication is commutative as has an identity as well, then Z[i]/(2 + i) is a field. e) We see that 2i serves as the multiplicative identity. Consider the function f : Z[i]/(2 + i) → F5 given by a f (a)

0 0

2i 1

1 + 2i 2

i 3

1+i 4

sets up an isomorphism between Z[i]/(2 + i) and F5 . Exercise: 11 Section 5.6 Question: Some rings with eight elements. a) Consider the (quotient) ring R1 = (Z/2Z)[x]/(x3 + 1). List all 8 elements of this ring and determine whether (and show how) they are units, zero-divisors or neither. b) Repeat the same question with the ring R2 = (Z/2Z)[x]/(x3 + x + 1). c) Consider also the ring R3 = Z/8Z of modular arithmetic modulo 8. The rings R1 , R2 and R3 all have 8 elements. Show that none of them are isomorphic to each other. Solution: a) We will drop the bars for convenience when listing the elements of the quotient. All eight elements are {0, 1, x, x + 1, x2 , x2 + 1, x2 + x, x2 + x + 1}. We first show pairs of units, x(x2 ) = x3 = −1 = 1. Now we show the rest are zero-divisors, (x + 1)(x2 + x + 1) = x3 + x2 + x2 + x + x + 1 = x3 + 1 = 0 (x2 + x)(x2 + x + 1) = x(x + 1)(x2 + x + 1) = x(0) = 0 (x2 + 1)(x2 + x + 1) = x2 (x + 1)(x2 + x + 1) = x2 (0) = 0. b) We again drop the bars. All eight elements are {0, 1, x, x + 1, x2 , x2 + 1, x2 + x, x2 + x + 1}. This time the relationship we have is that x3 = x + 1. We show the units now, x(x2 + 1) = x3 + x = 1 (x2 + x)(x + 1) = x3 + x = 1 x2 (x2 + x + 1) = x4 + x3 + x2 = x(x + 1) + x + 1 + x2 = x2 + x2 + x + x + 1 = 1. Which shows that every nonzero element is a unit.

290

CHAPTER 5. RINGS

c) Any isomorphism between the three rings must preserve the number of units and zero-divisors. R2 has no zero-divisors, but R3 and R1 each have 2 which means that R2 is not isomorphic to either of them. In R1 , we have the property that for any element r, r + r = 0 or, every element is its own additive inverse.. However, in R3 , 2 + 2 = 4 6= 0. This shows that additive inverses would not be preserved under any bijective function, so no isomorphism can exist. Exercise: 12 Section 5.6 Question: Show that (Z/7Z)[x]/(x2 − 3) is a field. Solution: The relationship of our quotient is x2 = 3. Any polynomial of degree greater than or equal to 2 can be reduced to a polynomial of the form ax + b. We have a total of 49 elements then. Now, all of the nonzero constants are units. Say two first degree polynomials in our quotient, ax + b and cx + d, multiplied to be zero. Then, ax + b · cx + d = acx2 + (ad + bc)x + bd. Every zero element in our quotient looks like p(x)(x2 − 3). Now, ac is not zero because neither a nor c is zero. So we must have that acx2 + (ad + bc)x + bd = e(x2 − 3) for some constant e. This could only occur if x2 − 3 were factorable in our original ring. Furthermore, x2 − 3 is factorable if and only if it has a root. So we show that it does not by trying each element of Z/7Z. 02 − 3 = 4 62 − 3 = 12 − 3 = 5 52 − 3 = 22 − 3 = 1 42 − 3 = 32 − 3 = 6. So x2 − 3 is not factorable and therefore two nonzero elements from our quotient could not multiply to zero. Since 1 functions as a unit, the quotient is a finite integral domain and by Exercise 5.1.23, it is also a field. Exercise: 13 Section 5.6 Question: Consider the ring Z[x] and define the ideals Ip = (px − 1), where p is a prime. 1. Prove that Z[x]/(I2 I3 I5 · · · Ip ) is isomorphic to

n ∈ Q | n ∈ Z and α ∈ N . i 2α2 3α3 · · · pαp

2. Prove that there exists no ideal I ∈ Z[x] such that Z[x]/I ∼ = Q. Solution: a) Note that Ip does not contain an constant polynomials as the 0 polynomials. In the quotient ring, Z[x]/Ip , for all m, n ∈ Z, the elements m = n are equal if and only m − n = 0, i.e., m = n. Also, the element x satisfies px = 1. It is natural then to consider the function ϕ : Z[x] → Q defined by ϕ(a(x)) = a p1 . This is similar to the evaluation homomorphism of Example 5.4.4 so this is a homomorphism. The kernel Ker ϕ consists of all polynomials that have p1 as a root. These are polynomials that have (px − 1) as a factor and h i h i thus Ker ϕ = Ip . Also, Im ϕ = Z 1 . Hence, by the First Isomorphism Theorem, Z[x]/Ip ∼ =Z 1 . p

p

For more than one prime, the exercise is incorrect. For more than one prime, Z[x]/(I2 I3 I5 · · · Ip ) will have zero divisors such as 2x − 1. On the other hand, n ∈ Q | n ∈ Z and αi ∈ N , R= 2α2 3α3 · · · pαp as a subring of the field Q, does not have zero divisors. A correction to the exercise is that using the reasoning in the previous paragraph, Z[x]/I2·3···p is isomorphic to R.

5.6. QUOTIENT RINGS

291

b) Assume that there exists an ideal I ⊆ Z[x] such that Z[x]/I ∼ = Q. Then by the First Isomorphism Theorem, there exists a surjective homomorphism ϕ : Z[x] → Q with I = Ker ϕ. Now, for a polynomial a(x) = an xn + · · · + a1 x + a0 , by homomorphism properties, ϕ(a(x)) = ϕ(an )ϕ(x)n + · · · + ϕ(a1 )ϕ(x) + ϕ(a0 ) = an · ϕ(1)ϕ(x)n + · · · + a1 · ϕ(1)ϕ(x) + a0 · ϕ(1). Let p1 be the largest prime occurring in the prime decomposition of the denominator of ϕ(1) (when written in reduced form) and let px be the largest prime occurring in the prime decomposition of the denominator of ϕ(x) (when written in reduced form). If p is any prime bigger than p1 and px , then by properties of arithmetic, p1 ∈ / Im ϕ. This contradicts the assumption. Hence, Q is not isomorphic to any quotient ring of Z[x]. Exercise: 14 Section 5.6 Question: Prove that every element in the quotient ring R[x, y]/(y −x2 ) can be written uniquely as a(y)+xb(y) where a(y) and b(y) are any polynomials in y and where x satisfies the relation x2 = y. Solution: In our quotient we have the relation x2 = ȳ. Then for powers of x we have, x2k = y k and x2k+1 = y k x. Then for any arbitrary polynomial p(x, y) we can split it up into two polynomials, p1 (x2 , y) + p2 (x, y), where p1 holds all terms with even powers of x and p2 holds all terms with odd powers fo x. p1 (x2 , y) can be rewritten entirely into a polynomial in y and p2 (x, y) = p3 (x2 , y)x for some p3 which can then be rewritten entirely into a polynomial in y. Then for some q1 (y) and q2 (y) we have p1 (x2 , y) + p2 (x, y) = q1 (y) + q2 (y)x. Exercise: 15 Section 5.6 Question: Let R be a commutative ring and let N (R) be the nilradical of R. Prove that N (R/N (R)) = 0. Solution: Consider the quotient Q = R/N (R). Suppose that some a ∈ Q, where an = 0. Then an ∈ N (R). Then for some k, (an )k = 0 ⇒ ank = 0. Which implies that a ∈ N (R) and a = 0. So the only element in N (R/N (R)) is 0. Exercise: 16 Section 5.6 Question: Let R be a ring and let I and J be ideals in R. 1. Prove that the function f : R/I → R/(I + J) defined by f (r + I) = r + (I + J) is a well-defined function. 2. Show that f is a homomorphism. 3. Show that Ker f ∼ = J/(I ∩ J). Solution: a) Suppose that both x and y represent the same coset in R/I. Then for some i ∈ I we have x + i = y. Then f (x+I) = x+(I +J). Now, x+(i+0) = y ∈ x+(I +J), so that f (x+I) = x+(I +J) = y+(I +J) = f (y+I). This shows the function is well-defined. b) This follows from addition of cosets. f ((x + I) + (y + I)) = f ((x + y) + I) = (x + y) + (I + J) = x + (I + J) + y + (I + J) = f (x + I) + f (y + I). c) Let x ∈ Ker f so that f (x + I) = x + (I + J) = 0 + (I + J). Then x is of the form i + j. So all elements in Ker f ∼ = (J + I)/I. By the Second Isomorphism Theorem, (J + I)/I ∼ = J/(I ∩ J). Exercise: 17 Section 5.6 Question: Let R be a ring and let I be an ideal of R. Prove that Mn (I) is an ideal of Mn (R) and show that Mn (R)/Mn (I)M ∼ =n (R/I).

292

CHAPTER 5. RINGS

Solution: Consider the function f : Mn (R) → Mn (R/I) defined by f (A) is the matrix whose ijth entry is aij in R/I. Since the congruence operation behaves well with respect to + and × in R, then this function f is a homomorphism. (With multiplication, the ij entry of f (AB) is ai1 b1j + ai2 b2j + · · · + ain bnj while the ijth entry of f (A)f (B) is āi1 b̄1j + āi2 b̄2j + · · · + āin b̄nj . These are the same. So we see that f is a homomorphism. Furthermore, ker f = Mn (I), all n × n matrices in which all the components are in the ideal I. Hence, by the First Isomorphism Theorem, Mn (I) is an ideal in Mn (R) and Mn (R)/Mn (I) ∼ = Mn (R/I). Exercise: 18 Section 5.6 Question: Let Un (R) be the set of upper triangular n × n matrices with coefficients in R. Prove that the subset I = {A ∈ Un (R) | aii = 0 for all i = 1, 2, . . . , n} is an ideal in Un (R) and determine Un (R)/I. Solution: We rewrite the definition of our ideal I. I = {A ∈ Mn (R) | aij = 0 when i ≥ j} Then the only requirement for a matrix M to be in I is that whenever i ≥ j, the (i, j)th entry is 0. We show that I is a subgroup using the one step subgroup criteria. Let A = (aij ) and B = (bij ) be two elements of I. Then their difference A − B = (aij − bij ) is in I since whenever i ≥ j, the (i, j)th entry of A − B is aij − bij = 0 − 0 = 0. Next we show that I is closed under multiplication on the left and right by elements of Un (R). Consider the arbitrary elements A = (aij ) of I and B = (bij ) of Un (R). Then we know that when i > j, bij = 0 and when i ≥ j, aij = 0. Consider the left product, B · A. Let i ≥ j. The (i, j)th entry of B · A is n X

bik akj =

i−1 X

bik akj +

bik akj =

i−1 X

0akj +

n X

bik 0 = 0.

k=i

k=1

k=i

k=1

k=1

n X

Then by our definition of I, B · A ∈ I. Now we examine A · B. Let i ≥ j again. Then the (i, j)th entry of A · B is n X k=1

aik bkj =

i X k=1

aik bkj +

n X k=i+1

aik bkj =

i X k=1

0bkj +

n X

aik 0 = 0.

k=i+1

So A · B ∈ I. This shows that I is closed under multiplication by ring elements. So I is an ideal of Un (R). We examine the quotient Un (R)/I. Consider the arbitrary elements of Un (R)/I, A = (aij ) and B = (bij ). Suppose that we have the coset equivalence, A + I = B + I. Then A − B = (aij − bij ) ∈ I, which implies that when i ≥ j, aij − bij = 0 or aij = bij . Then the cosets of Un (R)/I are exactly the elements which have the same elements along the diagonal. So Un (R)/I ∼ = {A ∈ Un (R) | aij = 0 when i 6= j}

Exercise: 19 Section 5.6 Question: Prove the Second Isomorphism Theorem for rings. (See Theorem 5.6.11) Solution: Let R be a ring, A a subring, and B an ideal. We first show that A + B is a subring. We use the one step subgroup criteria, (a1 + b1 ) − (a2 + b2 ) = (a1 − a2 ) + (b1 − b2 ) ∈ A + B. So that A + B is a subgroup under addition. Next we show that it’s closed under multiplication, remembering that since B is an ideal of R, ba, ab ∈ B for all a ∈ A. (a1 + b1 ) · (a2 + b2 ) = (a1 a2 + b1 a2 + a1 b2 + b1 b2 ) = (a1 a2 ) + (b1 a2 + a1 b2 + b1 b2 ) ∈ A + B

5.6. QUOTIENT RINGS

293

So A + B is closed under multiplication as well and is a subring. A ∩ B is an ideal of A from the results of Exercise 5.25.b using I = R = A and J = B. Now we set up the following map, ϕ : A + B → A/(A ∩ B), letting ϕ(a1 + b1 ) = a1 + A ∩ B. We show that this map is a ring homomorphism. First with addition, ϕ((a1 + b1 ) + (a2 + b2 )) = ϕ((a1 + a2 ) + (b1 + b2 )) = (a1 + a2 ) + A ∩ B = (a1 + A ∩ B) + (a2 + A ∩ B) = ϕ(a1 + b1 ) + ϕ(a2 + b2 ). Secondly with multiplication, ϕ((a1 + b1 ) · (a2 + b2 )) = ϕ((a1 a2 ) + (b1 a2 + a1 b2 + b1 b2 )) = a1 a2 + A ∩ B = (a1 + A ∩ B) · (a2 + A ∩ B) = ϕ(a1 + b1 ) · ϕ(a2 + b2 ). So ϕ is a ring homomorphism. Then a1 + b1 ∈ Ker ϕ if and only if a1 + b1 ∈ A ∩ B or a1 + b1 ∈ B. But then, a1 + b1 = 0 + b2 ∈ B. So Ker ϕ = B. Then by the First Isomorphism Theorem for rings, (A + B)/B ∼ = ϕ(A + B). Now we show that ϕ is surjective. For any a1 + A ∩ B ∈ A/(A ∩ B), a1 ∈ A, so a1 + 0 ∈ A + B and ϕ(a1 + 0) = a1 + A ∩ B. Then ϕ is surjective and (A + B)/B ∼ = A/(A ∩ B). Exercise: 20 Section 5.6 Question: Prove the Third Isomorphism Theorem for rings. (See Theorem 5.6.12) Solution: Let R be any ring and I ⊂ J be ideals of R. We first show that J/I is an ideal of R/I. Elements in J/I look like j1 + I. We show it is a subgroup under addition using the one step subgroup criteria, (j1 + I) − (j2 + I) = (j1 − j2 ) + I ∈ J/I. Lastly, we show it is closed under multiplication by elements of R/I. (r + I) · (j1 + I) = (r · j1 ) + I = j2 + I ∈ J/I (j1 + I) · (r + I) = (j1 · r) + I = j2 + I ∈ J/I. Now we set up the following map, ϕ : R/I → R/J by ϕ(r + I) = r + J. We show this is well defined. If r1 + I ≡ r2 + I, then r2 − r1 ∈ I. Since I ⊂ J, r2 − r1 ∈ J, r1 + J ≡ r2 + J, and ϕ(r2 + I) = ϕ(r1 + I). Next we show that ϕ is a ring homomorphism. ϕ((r1 + I) + (r2 + I)) = ϕ((r1 + r2 ) + I) = (r1 + r2 ) + J = (r1 + J) + (r2 + J) = ϕ(r1 + I) + ϕ(r2 + I).

ϕ((r1 + I) · (r2 + I)) = ϕ((r1 r2 ) + I) = (r1 r2 ) + J = (r1 + J) · (r2 + J) = ϕ(r1 + I) · ϕ(r2 + I). So ϕ splits over addition and multiplication and is a ring homomorphism. We examine the kernel. r + I ∈ Ker ϕ if and only if r ∈ J. This implies that r + I ∈ J/I. It is easy to see that any element of J/I must be in Ker ϕ as well. Then from the First Isomorphism Theorem for rings, (R/I)/(J/I) ∼ = ϕ(R/I). Now we will show that ϕ is surjective. Any element in R/J looks like r1 + J where r1 ∈ R. Then r1 + I ∈ R/I and ϕ(r1 + I) = r1 + J. So ϕ is surjective and (R/I)/(J/I) ∼ = J/I.

294

CHAPTER 5. RINGS

Exercise: 21 Section 5.6 Question: Prove the Fourth Isomorphism Theorem for rings. (See Theorem 5.6.14.) Solution: Let R be a ring and let I be an ideal. If A is a subring that contains the ideal I, then by the defining properties of ideals, I is also an ideal of A. Then A/I is a subset of the quotient ring R/I. However, A/I is a ring as well and its operations are inherited from R/I so A/I is a subring of R/I. Conversely, let S be a subring of R/I and consider the set π −1 (S), where π : R → R/I is the canonical projection. If x, y ∈ π −1 (S), then π(x − y) = π(x) − π(y) and this is in S since π(x), π(y) ∈ S. Furthermore, π(xy) = π(x)π(y). Thus xy ∈ π −1 (S). We have show that π −1 (S) is a subring of R and it must contain I since I = π −1 (0). By set theory properties S = π(π −1 (S)) = π −1 (S)/I. Furthermore, for all subrings A in R, we have π −1 (A/I) = A Thus, π and π −1 , applied to subrings of R and R/I, give a bijection between subrings of R containing I and subrings of R/I. Simply by set theory with U ⊆ V implies f (U ) ⊆ f (V ), we conclude that this correspondence between sets of subrings is inclusion preserving. Now let J be an ideal in R that contains the ideal I. Then J is a subring of R that is closed by left and right multiplication by any element in R. Then for all a ∈ J/I and all r ∈ R/I, we have r a = ra and a r = ar. But since J is an ideal, ar, ra ∈ J so ar and ra are in J/I. Finally, By Exercise 5.5.22, π −1 (J 0 ) is an ideal of R. This shows that under the correspondence between subrings of R containing I and the subrings of R/I, ideals correspond to ideals. Exercise: 22 Section 5.6 Question: Let R be a ring and let e be an idempotent element in the center C(R). Observe that the ideals (e) = Re and (1 − e) = R(1 − e) and prove that R ∼ = Re ⊕ R(1 − e). Solution: We make a few observations. First, e and (1 − e) are zero divisors, e · (1 − e) = e − e2 = e − e = 0. Second, (1 − e) is also idempotent, (1 − e)(1 − e) = 1 − e − e + e2 = 1 − e − e + e = 1 − e. We set up a function ϕ : Re ⊕ R(1 − e) → R with ϕ((r1 e, r2 (1 − e))) = r1 e + r2 (1 − e). We will show this is a ring homomorphism. First we prove it separates across addition, ϕ((r1 e, r2 (1 − e)) + (r3 e, r4 (1 − e))) = ϕ(((r1 + r3 )e, (r2 + r4 )(1 − e)) = (r1 + r3 )e + (r2 + r4 )(1 − e) = (r1 e + r2 (1 − e)) + (r3 e + r4 (1 − e)) = ϕ((r1 e, r2 (1 − e))) + ϕ((r3 e, r4 (1 − e))). Next, that it separates across multiplication, ϕ((r1 e, r2 (1 − e)) · (r3 e, r4 (1 − e))) = ϕ(((r1 r3 )e, (r2 r4 )(1 − e))) = r1 r3 e + r2 r4 (1 − e) = (r1 e + r2 (1 − e)) · (r3 e + r4 (1 − e)) = ϕ(r1 e, r2 (1 − e)) · ϕ(r3 e, r4 (1 − e)). The separation within R can be verified by doing the algebra and remembering which elements are zero-divisors and idempotents. So ϕ is a ring homomorphism. We next show that the kernel is zero. Suppose that for some arbitrary element, (r1 e, r2 (1 − e)) ∈ Re ⊕ R(1 − e), we have ϕ((r1 e, r2 (1 − e)) = 0. This implies that r1 e + r2 (1 − e) = 0.

5.6. QUOTIENT RINGS

295

We multiply on the right by e and find r1 e + r2 (0) = 0 ⇒ r1 e = 0. Likewise, we multiply on the right by (1 − e) and find r1 (0) + r2 (1 − e) = 0 ⇒ r2 (1 − e) = 0. Then our element is (0, 0) and Ker ϕ = 0. By the First Isomorphism Theorem for rings, Re ⊕ R(1 − e) ∼ = ϕ(Re ⊕ R(1 − e)). Finally, we show ϕ is surjective to complete the isomorphism. Given an arbitrary r ∈ R, the element (re, r(1 − e)) ∈ Re ⊕ R(1 − e) and ϕ((re, r(1 − e))) = re + r(1 − e) = re + r − re = r. So ϕ is surjective and Re ⊕ R(1 − e) ∼ = R. Exercise: 23 Section 5.6 Question: Consider the group ring Z[S3 ]. Show that the set I of elements α = a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2) satisfying ( a1 + a2 + a3 + a4 + a5 + a6 a1 − a2 − a3 − a4 + a5 + a6

=0 =0

is an ideal and show that Z[S3 ]/I ∼ = Z ⊕ Z. Prove that every element in Z[S3 ]/I can be written uniquely as a1 1 + a2 (1 2), with a1 , a2 ∈ Z. Solution: If we examine the two conditions, adding them shows us that a1 + a5 + a6 = 0 and likewise subtracting them shows us that a2 + a3 + a4 = 0. These are some extra helpful conditions to have. We will first show that it is closed under addition using the one step subgroup. Consider two elements in I, a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2) and b1 + b2 (1 2) + b3 (1 3) + b4 (2 3) + b5 (1 2 3) + b6 (1 3 2). (a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) − (b1 + b2 (1 2) + b3 (1 3) + b4 (2 3) + b5 (1 2 3) + b6 (1 3 2)) = (a1 − b1 ) + (a2 − b2 )(1 2) + (a3 − b3 )(1 3) + (a4 − b4 )(2 3) + (a5 − b5 )(1 2 3) + (a6 − b6 )(1 3 2) Then we show this element satisfies both conditions, (a1 − b1 ) + (a2 − b2 ) + (a3 − b3 ) + (a4 − b4 ) + (a5 − b5 ) + (a6 − b6 ) = (a1 + a2 + a3 + a4 + a5 + a6 ) − (b1 + b2 + b3 + b4 + b5 + b6 ) =0−0=0 and also (a1 − b1 ) − (a2 − b2 ) − (a3 − b3 ) − (a4 − b4 ) + (a5 − b5 ) + (a6 − b6 ) = (a1 − a2 − a3 − a4 + a5 + a6 ) − (b1 − b2 − b3 − b4 + b5 + b6 ) = 0 − 0 = 0. So the element satisfies both conditions and is a member of I. Then by the one step subgroup criteria, I is a subgroup under addition. Next we show that I is closed under multiplication by a member of Z[S3 ]. Every element of our ring looks like b1 + b2 (1 2) + b3 (1 3) + b4 (2 3) + b5 (1 2 3) + b6 (1 3 2) (where the coefficients no longer necessarily satisfy the conditions for being in I). However, because we can distribute, we just need to consider multiplication by an element that looks like b1 · σ where σ ∈ S3 and b1 ∈ Z. We can split this up as (b1 · 1)(1 · σ). Using the associative property, we can now just show that (b1 · 1) · α ∈ I and (1 · σ) · α ∈ I for any σ ∈ S3 , b1 ∈ Z, and α ∈ I. We first show for any b1 ∈ Z. b1 · α = b1 · (a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) = (b1 · a1 ) + (b1 · a2 )(1 2) + (b1 · a3 )(1 3) + (b1 · a4 )(2 3) + (b1 · a5 )(1 2 3) + (b1 · a6 )(1 3 2)

296

CHAPTER 5. RINGS

Now we show that the coefficients satisfy both conditions, (b1 · a1 ) + (b1 · a2 ) + (b1 · a3 ) + (b1 · a4 ) + (b1 · a5 ) + (b1 · a6 ) = b1 · (a1 + a2 + a3 + a4 + a5 + a6 ) = b1 · (0) = 0 (b1 · a1 ) − (b1 · a2 ) − (b1 · a3 ) − (b1 · a4 ) + (b1 · a5 ) + (b1 · a6 ) = b1 · (a1 − a2 − a3 − a4 + a5 + a6 ) = b1 · (0) = 0. So both conditions hold and I is closed under multiplication by an integer. Next we show that (1 · σ) · α ∈ I for any σ ∈ S3 . Now, σ = τ1 · τ2 . . . · τn for some transposition τ . Then given any element α ∈ I, if we can show that τ · α ∈ I for all transpositions τ , then σ · α = (τ1 · τ2 . . . · τn ) · α can be seen to be in I by using the associative property. Now, τ · α can be seen as a permutation of the coefficient a1 , a2 , . . . a6 . τ · α = τ · (a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) = aσ(1) + aσ(2) (1 2) + aσ(3) (1 3) + aσ(4) (2 3) + aσ(5) (1 2 3) + aσ(6) (1 3 2), where σ ∈ S6 . Then our first condition will always hold since addition is commutative and we can simply rearrange them. In our second condition, we have different weights for different coefficients, and so we need to determine if it will hold for each τ . (1 2) · α : a1 − a2 − a3 − a4 + a5 + a6 → a2 − a1 − a6 − a5 + a4 + a3 = (a2 + a4 + a3 ) − (a1 + a5 + a6 ) = 0 (1 3) · α : a1 − a2 − a3 − a4 + a5 + a6 → a3 − a5 − a1 − a6 + a2 + a4 = (a2 + a3 + a4 ) − (a1 + a5 + a6 ) = 0 (2 3) · α : a1 − a2 − a3 − a4 + a5 + a6 → a4 − a6 − a5 − a1 + a3 + a2 = (a2 + a3 + a4 ) − (a1 + a5 + a6 ) = 0. The last steps for each τ follows from our additional two conditions on the coefficients for each element of I. In order to show Z[S3 ]/I ∼ = Z ⊕ Z, we set up a map ϕ : Z[S3 ] → Z ⊕ Z where ϕ(a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) = (a1 + a5 + a6 , a2 + a3 + a4 ). This is based off of our extra two conditions. We note that if we add the two we get our first condition for I and if we subtract the first slot from the second, we get our second condition. Then it’s clear that the kernel of ϕ is exactly our set I. Then Z[S3 ]/I ∼ = ϕ(Z[S3 ]). Now we show ϕ is surjective. Given any element (m, n) ∈ Z⊕Z, we will construct an element in Z[S3 ]/I. Consider α = m+n(1 2)+0(1 3)+0(2 3)+0(1 2 3)+0(1 3 2), ϕ(α) = (m + 0 + 0, n + 0 + 0) = (m, n). Then ϕ is surjective and by the first isomorphism theorem of Rings, Z[S3 ]/I ∼ = Z ⊕ Z. Given an arbitrary element, a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2) ∈ Z[S3 ]/I, we show that this element is equal to an element of the form b1 + b2 (1 2). This is equivalent to showing that, under our previously defined map, ϕ : Z[S3 ] → Z ⊕ Z, we have ϕ(a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) = ϕ(b1 + b2 (1 2)) Then, suppose ϕ(a1 + a2 (1 2) + a3 (1 3) + a4 (2 3) + a5 (1 2 3) + a6 (1 3 2)) = (m, n). Then the element m + n(1 2) works. Exercise: 24 Section 5.6 Question: Let R be a PID and let I be an ideal in R. Prove that R/I is a PID. Solution: Consider an ideal J of R/I. By the Fourth Isomorphism Theorem for rings, there is some ideal K containing I where K/I ∼ = J. Then there is some ϕ : K → J, where ϕ(K) ∼ = J and Ker ϕ = I. Now, K ⊆ R, a PID, so K = (k) for some element k. We will claim that ϕ(k) generates all of J. Consider an arbitrary element of j ∈ J. There exists some kj ∈ K where ϕ(kj ) = j. Since K is principal, kj = r · k for some r ∈ R. Then ϕ(kj ) = ϕ(r · k) = ϕ(r) · ϕ(k) = rj · ϕ(k) = j This shows that any element j ∈ J is of the form rj · ϕ(k) for some rj ∈ R/I. Then J = (ϕ(k)) and is principal. Since every ideal is principal, R/I is a PID.

5.6. QUOTIENT RINGS

297

Exercise: 25 Section 5.6 Question: Consider the subset I = {(2m, 3n) | m, n ∈ Z}. Prove that I is an ideal in Z ⊕ Z and that (Z ⊕ Z)/I ∼ = Z/2Z ⊕ Z/3Z. Solution: We show that I is a subgroup under addition using the one step subgroup criteria. (2a, 3b) − (2c, 3d) = (2(a − c), 3(b − d)) ∈ I Now we note that multiplication in Z ⊕ Z is commutative and we show that I is closed under multiplication by a ring element. (m, n) · (2c, 3d) = (2(mc), 3(nd)) ∈ I So I is an ideal. Now we set up the map ϕ : Z⊕Z → Z/2Z⊕Z/3Z by ϕ(m, n) = (m (mod 2), n (mod 3)). It’s easy to see that the kernel of ϕ is I. We just need to show that ϕ is surjective. This follows from the fact that taking the integers mod 2 and mod 3 is surjective. Then from the first isomorphism theorem for rings, (Z ⊕ Z)/I ∼ = Z/2Z ⊕ Z/3Z. Exercise: 26 Section 5.6 Question: Let R and S be rings with identity 1 6= 0. Prove that every ideal of R ⊕ S is of the form I ⊕ J for some ideals I ⊆ R and J ⊆ S. Solution: Consider an arbitrary ideal I of R ⊕ S. Without loss of generality, consider the projection onto the ring S, ϕ : R ⊕ S → S by ϕ(r, s) = s. ϕ is easily shown to be a homomorphism with kernel (R, 0). We will show that ϕ(I) is an ideal of S. We use the one step subgroup criteria to show ϕ(I) is an additive subgroup of S. Given any two elements, (a, b), (c, d) ∈ I, ϕ((a, b)) − ϕ((c, d)) = (b − d) = ϕ((a − c, b − d)) ∈ ϕ(I) Now we show ϕ(I) is closed under right multiplication by elements of S, ϕ((a, b)) · s = b · s = bs = ϕ((a, bs)) = ϕ((a, b) · (1, s)) ∈ ϕ(I). As well as left multiplication, s · ϕ((a, b)) = s · b = sb = ϕ((a, sb)) = ϕ((1, s) · (a, b)) ∈ ϕ(I). So ϕ(I) is an ideal of S. Next we show that (0, ϕ(I)) ⊆ I. All elements in (0, ϕ(I)) look like (0, a). Since a ∈ ϕ(I), there must be some element (b, a) ∈ I. Since I is closed under multiplication (0, 1) · (b, a) = (0, a) ∈ I. Then (0, ϕ(I)) ⊆ I. If we let ψ be the projection onto R, we come to the fact that (ψ(I), 0) ⊆ I as well with ψ(I) being an ideal of R by the same reasoning as before. Then ψ(I) ⊕ ϕ(I) ⊆ I. It’s clear that this is an equivalence and ψ(I) ⊕ ϕ(I) = I. Then for any ideal I of R ⊕ S where R and S have identities, I = J ⊕ K for some ideals J ⊆ R and K ⊆ S. Exercise: 27 Section 5.6 Question: Solve the following system of congruences in Z: ( x ≡ 3 (mod 5) x ≡ 7 (mod 11). Solution: In order to solve the system of congruences, we look at the examples in the section for a model (or the proof of the Chinese Remainder Theorem). We have n1 = 5 and n2 = 11. In this situation, we also have n01 = 11 and n02 = 5. Now n01 ≡ 1 (mod 5) so it is its own inverse. On the other hand, the inverse to n02 = 5 modulo 11 is t2 = 9. Thus a solution for this system of congruences is x = 3 × 1 × 11 + 7 × 9 × 5 = 348.

298

CHAPTER 5. RINGS

The set of solutions to this system of equations is x ≡ 348 ≡ 18 (mod 55). Exercise: 28 Section 5.6 Question: Solve the following system of congruences in Z: ( x ≡ 4 (mod 10) x ≡ 17 (mod 21) Solution: The system of congruences satisfy all the conditions of the Chinese Remainder Theorem and we can use the congruences to find a unique solution modular 210. From the first congruence we know that x = 4 + 10k1 . We plug this into the second congruence, 4 + 10k1 ≡ 17

(mod 21)

10k1 ≡ 13

(mod 21)

We need an inverse for 10, note that 21 · 9 = 189 and 190 = 10 ∗ 19 so 19 will work. (19 · 10)k1 ≡ k1 ≡ 19 · 13 ≡ 16

(mod 21)

Then from this congruence we find that k1 = 16 + 21k2 . We plug this back into our first equation, x = 4 + 10(16 + 21k2 ) = 4 + 160 + 210k2 = 164 + 210k2 So, x ≡ 164 (mod 210). Exercise: 29 Section 5.6 Question: Solve the following system of congruences in Z:   x ≡ 2 (mod 3) x ≡ 3 (mod 5)   x ≡ 9 (mod 13) Solution: The system of congruences satisfies all the conditions of the Chinese Remainder Theorem and we can use the congruences to find a unique solution modular 195. From the first congruence we know that x = 2 + 3k1 . We plug this into the second congruence, 2 + 3k1 ≡ 3

(mod 5)

3k1 ≡ 1

(mod 5)

We use 2, the inverse for 3. (2 · 3)k1 ≡ k1 ≡ 2

(mod 5)

Then from this congruence we find that k1 = 2 + 5k2 . We plug this back into our first equation, x = 2 + 3(2 + 5k2 ) = 2 + 6 + 15k2 = 8 + 15k2 .

5.7. MAXIMAL IDEALS AND PRIME IDEALS

299

We plug this into our last equation, 8 + 15k2 ≡ 9

(mod 13)

2k2 ≡ 1

(mod 13).

We use 7 as the inverse for 2, (7 · 2)k2 ≡ k2 ≡ 7

(mod 13).

From this congruence we have, k2 = 7 + 13k3 . We plug this back into our most recent equation for x, x = 8 + 15(7 + 13k3 ) = 8 + 105 + 195k3 = 113 + 195k3 . So x ≡ 113 (mod 195).

5.7 – Maximal Ideals and Prime Ideals Exercise: 1 Section 5.7 Question: Show that the ideal {(2m, 3n) | m, n ∈ Z} is not a prime ideal. Solution: Denote the ideal by I and consider the element (2, 3) ∈ I. (2, 3) = (2, 1) · (1, 3). Note that neither (2, 1) nor (1, 3) exist in I so that I is not prime. Exercise: 2 Section 5.7 Question: Consider the ideal (x) in the polynomial ring R[x]. a) Prove that (x) is a prime ideal if and only if R is an integral domain. b) Prove that (x) is a maximal ideal if and only if R is a field. Solution: a) Consider the mapping ϕ : R[x] → R where for any polynomial p(x) = pn xn + . . . + p1 x + p0 we have ϕ(p(x)) = p0 . Any element in the kernel has a 0th coefficient of 0 and so can be written as q(x) · x ∈ (x) for some q(x) ∈ R[x]. Likewise, any element in (x) does has a 0th coefficient of 0 and so is in Ker ϕ. Then Ker ϕ = (x). It is clear that ϕ is surjective and R[x]/(x) ∼ = R. From Proposition 5.7.7, we have that R is an integral domain if and only if (x) is a prime ideal. b) By the same mapping as in part a, we have that R[x]/(x) ∼ = R. Then by Proposition 5.7.3, (x) is maximal if and only if R[x]/(x) and R are fields. Exercise: 3 Section 5.7 Question: Let R be a ring with an identity 1 6= 0. Prove that if the set of non-units is an ideal M , then M is the unique maximal ideal in R. Conversely, prove that if R is a commutative ring that contains a unique maximal ideal M , then R − M is the set of all units. Solution: Suppose that the set of non-units is an ideal M . By Krull’s Theorem, there exists a maximal ideal I containing M . Assume that M ( I. Then I contains an element u that is not an element of M . Hence u is a unit. Let v = u−1 . Then for all r ∈ R, the element rvu = r is in the ideal I. Hence I is all of R. This is a contradiction since I is a maximal ideal. We conclude that I = M so M is a maximal ideal. Furthermore, knowing that M is a maximal ideal, assume that there exists another maximal ideal J 6= M . Then J − M is non-empty so there exists a unit u ∈ J. By the same reasoning as before, we deduce that J = R, which contradicts the assumption that I si a maximal ideal. We conclude that M is the unique maximal ideal of R. Now suppose that R is a commutative ring (with an identity 1 6= 0) that contains a unique maximal ideal M . Let a be any non-unit in R. Since R is commutative, (a) = Ra so (a) = R if and only if 1 ∈ (a) if and only if a

300

CHAPTER 5. RINGS

is a unit of R. However, since a is not a unit, then (a) ( R, so by Krull’s Theorem, there exists a maximal ideal M 0 containing the ideal (a). Since there exists only one unique maximal ideal M , then a ∈ M for all non-units a. In order to be a proper ideal of R, M does not contain any units of R. Thus M consists precisely of all the non-units of R. Exercise: 4 Section 5.7 Question: Consider the subset R ⊂ Q of fractions ab for which, in reduced form, 19 does not divide b. a) Show that R is a subring of Q. b) Determine the set of units in R. c) Use Exercise 5.7.3 to conclude that the set of non-units in R is the unique maximal ideal in R. Solution: a) We show R is a subgroup under addition. Let ab and dc be two elements of R in reduced form. Then a c ad − bc − = . b d bd b and d not being divisible by 19 implies that bd is also not divisible by 19 since 19 is prime. Then the difference is in R and R is a subgroup under addition. We now check multiplication, a c ac · = ∈ R. b d bd This shows that R is a subring of Q. b) We need to just make sure that if ab ∈ R, ab ∈ R as well. This happens if a is also not divisible by 19. Then the set of units is a 19 6 | a and 19 6 | b}. { b c) We need to show that the set of non units is an ideal in R. The set of non units is {

a ∈ R a = 19 · k for some k ∈ Z}. b

We denote this set by I and show it is closed as a subgroup under addition. 19kd − 19mb 19(kd − mb) 19k 19m − = = ∈ I. b d bd bd We show that I is closed under multiplication by ring elements. a 19k 19(ak) · = ∈ I. b d bd Then I is an ideal and by the results of Exercise 5.7.3, the set of non units is the unique maximal ideal of R. Exercise: 5 Section 5.7 Question: Consider the ring Un (R) of upper triangular n × n matrices with real coefficients. Fix an integer k with 1 ≤ k ≤ n. Prove that the set Ik = {A ∈ Un (R) | akk = 0} is a maximal ideal in Un (R). Solution: Let A, B ∈ Un (R). It is obvious that if C = A + B, then the (k, k) entry is akk + bkk . Also, if C = AB, then the (k, k) entry is n X ckk = aki bik . i=1

But aki = 0 if k < i and bik = 0 if i < k. Hence, the only nonzero term in the summation is ckk = akk bkk . We have shown that the function ϕk : Un (R) → R defined by ϕk (A) = akk , i.e., that picks off the kth diagonal element, is a ring homomorphism.

5.7. MAXIMAL IDEALS AND PRIME IDEALS

301

Note that Ik = ker ϕk . By the First Isomorphism Theorem of rings, we know that Ik is an ideal and that Un (R)/Ik ∼ = R, which is a field. We cannot directly use Proposition 5.7.3 because Un (R) is not commutative. Now every field has exactly two ideals. Thus, since Un (R)/Ik is a field, by the Fourth Isomorphism Theorem, the only ideals that contain Ik are Ik and Un (R). Thus, Ik is a maximal ideal. [The point here is that R/M is a field implies that M is a maximal ideal, regardless of whether R is commutative or not. However, if we know that R is commutative, then R/M is a field if and only if M is maximal.] Exercise: 5 Section 5.7 Question: Consider the ring Un (Z) of upper triangular n × n matrices with real coefficients. Fix an integer k with 1 ≤ k ≤ n. Prove that the set Ik = {A ∈ Un (Z) | akk = 0} is a prime ideal in Un (Z) that is not maximal. Solution: Let A, B ∈ Un (Z). It is obvious that if C = A + B, then the kk entry is akk + bkk . Also, if C = AB, then the kk entry is n X ckk = aki bik . i=1

But aki = 0 if k < i and bik = 0 if i < k. Hence, the only nonzero term in the summation is ckk = akk bkk . We have shown that the function ϕk : Un (Z) → Z defined by ϕk (A) = akk , i.e., that picks off the kth diagonal element, is a ring homomorphism. Note that Ik = ker ϕk . By the First Isomorphism Theorem of rings, Un (Z)/Ik ∼ = Z. Since Z is an integral domain, then, by a proposition in this section, Ik is a prime ideal. On the other hand, since Un (Z)/Ik is not a field, then Ik is not a maximal ideal. Exercise: 7 Section 5.7 Question: Let R = C 0 (R, R) and consider the set of functions of compact support, I = {f ∈ C 0 (R, R) | Supp(f ) is closed and bounded }. [Recall that a subset S of R is bounded if there exists some c such that S ⊆ [−c, c].] 1. Prove that I is an ideal. 2. Prove that any maximal ideal that contains I is not equal to any of the ideals Ma described in Example 5.7.12. Solution: Let R = C 0 (R, R) and call I the subset of functions of compact support, which means functions, the closures of which are compact functions. As subsets of R, a function of compact support is just one whose support is a bounded subset of R. a) I is nonempty since it contains the 0 function. Let f, g ∈ I and suppose that f is 0 outside a set [−c1 , c1 ] and that g is 0 outside a set [−c2 , c2 ] with ci ≥ 0. If c = max{c1 , c2 }, then f (x) = g(x) = 0 and so f (x) − g(x) = 0 for all x outside [−c, c]. Thus f − g has compact support, so is in I. Let f ∈ I and let f ∈ C 0 (R, R). For all x ∈ R, if f (x) = 0, then f (x)h(x) = 0. Hence, Supp(f h) ⊆ Supp(f ) so Supp(f h) is bounded so f h ∈ I. This proves that I is an ideal of R. b) The ideals Ma consist of subsets of R of continuous functions such that f (a) = 0. Fix any a ∈ R. Then Consider the function  0 if x ≤ a − 1    1 + (x − a) if a − 1 ≤ x ≤ a f (x) =  1 − (x − a) if a ≤ x ≤ a + 1    0 if a + 1 ≤ x. This is a tent function centered at a. Then f (x) is continuous has Supp(f ) = (a − 1, a + 1). Thus the closure of Supp(f ) is the compact interval [a − 1, a + 1]. So f has compact support and f (a) 6= 0. Hence, there is no real number a such that f (a) = 0 for all f ∈ I. Thus, the ideal I is not a subset of Ma for any a ∈ R.

302

CHAPTER 5. RINGS

Exercise: 8 Section 5.7 Question: This exercise asks the reader to prove the following modification of Proposition 5.7.3. Let R be any ring. An ideal M is maximal if and only if the quotient R/M is a simple ring. [A simple ring is a ring that contains no ideals except the 0 ideal and the whole ring.] Solution: Let M be a maximal ideal of a ring R. Consider R/M . Utilizing the Fourth Isomorphism Theorem for rings, we see that there is a bijection between the ideals of R containing M and the ideals of R/M . There are only two ideals of R containing M , these are R and M . Then there can only be two ideals in R/M , which must necessarily be R/M and (0). This shows that R/M is a simple ring. Let M be an ideal of R and suppose that R/M is a simple ring. Then the only ideals in R/M are R/M and (0). Using the bijection from the Fourth Isomorphism Theorem for rings, this implies that the only ideals containing M in R are M and R. Then M is a maximal ideal. Exercise: 9 Section 5.7 Question: In noncommutative rings, we call a left ideal a maximal left ideal (and similarly for right ideals) if it is maximal in the poset of proper left ideals ordered by inclusion. Prove that maximal left ideals (and maximal right ideals) exist under the same conditions as for Krull’s Theorem. Solution: Krull’s Theorem is stated for two-sided ideals. To generalize Krull’s Theorem to the situation described by the exercise, we merely need to make a minor adjustment to the proof of the exercise. We do the proof for left ideals since the proof for right ideals is identical. Let R be a ring with an identity and let I be any proper left ideal. Let S be the set of all proper left ideals that contain I. Then S is a nonempty set (since it contains I) and this is still partially ordered by inclusion. We consider a chain C in S. We claim that the chain C again has the upper bound [ J= A ∈ CA. The point is to show that J is a proper left ideal of R. Let a, b ∈ J. Then a ∈ A1 and b ∈ A2 where A1 , A2 ∈ C. But since C is a chain, either A1 ⊆ A2 or A2 ⊆ A1 . Let’s assume the former without loss of generality. Then both a and b are in the ideal A2 . Thus, b − a ∈ A2 ⊂ J so J is an additive subgroup of R by the one-step subgroup test. Now let r ∈ R and let a ∈ J. Then a is an element of some A ∈ C. Since A is a left ideal we have ra ∈ A ⊂ J. Thus, we have shown that J is an ideal. Since R has an identity, then an ideal is proper if and only if it does not contain the identity. Since all ideals in S do not contain the identity, then 1 ∈ / J. Thus J 6= R, so J is proper. By Zorn’s Lemma, since every chain in S has an upper bound, the set S has a maximal element. Thus, there exists a maximal left ideal containing I. Exercise: 10 Section 5.7 Question: Prove that the set of matrices {A ∈ M2 (R) | a11 = a21 = 0} is a maximal left ideal of M2 (R). (See Exercise 5.7.9) Find a maximal right ideal in M2 (R). Solution: We denote the set by I and we show that it is a left ideal. Consider two matrices in I, A = (aij ) and B = (bij ). A − B = (aij ) − (bij ) = (aij − bij ). Then a11 − b11 = 0 and a21 − b21 = 0 and A − B ∈ I. By the one step subgroup criterion, I is a subgroup under addition. We now show I is closed under left multiplication by ring elements. Let C = (c12 ) ∈ M2 (R). 2 X C ·A=( cil alj ). l=1

Then we check to see it satisfies our conditions, 2 X

c1l al1 = c11 a11 + c12 a21 = c11 0 + c12 0 = 0

l=1 2 X l=1

c2l al1 = c21 a11 + c22 a21 = c21 0 + c22 0 = 0.

5.7. MAXIMAL IDEALS AND PRIME IDEALS

303

So both our conditions hold and C · A ∈ I. So I is a left ideal. Suppose that there was a bigger ideal J that contained I. Then J has an element of the form b11 b12 b21 b22 Where at least one of b11 and b21 is not zero. Without loss of generality, suppose b21 is nonzero. If this is a unit, then J is all of M2 (R). Assume that this element is not a unit and its determinant, b11 b22 − b12 b21 , is 0. then we show that J must contain a unit, and therefore must be all of M2 (R). Consider the element of I and J, 0 1 . 0 0 Since J is closed under addition,

b11 b21

b12 + 1 b22

∈ J.

Then we examine this matrices determinant, b11 b22 − (b12 + 1)b21 = b11 b22 − b12 b21 − b21 = −b21 6= 0. So we have found a unit and J = M2 (R). If the case was that b11 was nonzero and the determinent was zero, we would add the matrix, 0 0 . 0 1 We have shown that any left ideal that strictly contains I is M2 (R). So I is maximal. For a maximal right ideal, consider the set {A ∈ M2 (R) | a11 = a12 = 0}. Exercise: 11 Section 5.7 Question: Find all the prime ideals in Z ⊕ Z. Solution: By Exercise 5.6.26, we know that the ideals of Z ⊕ Z are of the form I ⊕ J where I and J are ideals of Z. Let I be an arbitrary prime ideal in Z ⊕ Z. Since I is an ideal, (0, 0) ∈ I. Now, Z ⊕ Z is not an integral domain. We have (0, 0) = (1, 0) · (0, 1). So that either (1, 0) ∈ I or (0, 1) ∈ I. This implies that either Z ⊕ 0 ⊆ I or 0 ⊕ Z ⊆ I. Then our only options are ideals of the form I ⊕ Z and Z ⊕ J. Without loss of generality, let’s examine the former case. Let ϕ : Z ⊕ Z → Z/I by ϕ(a, b) = a + I. The kernel of this surjective ring homomorphism is I ⊕ Z. By Proposition 5.7.7, I ⊕ Z is an prime ideal if and only if Z/I is an integral domain. This only happens when I = pZ for some prime p, I = Z, or I = 0. Then all of our prime ideals are contained in the following 5 categories, Z ⊕ Z, 0 ⊕ Z, Z ⊕ 0, Z ⊕ pZ, pZ ⊕ Z. Exercise: 12 Section 5.7 Question: Prove that (y − x2 ) is a prime ideal in R[x, y]. Prove also that it is not maximal. Solution: From the propositions in this section, to answer these questions, we need to determine R[x, y]/(y−x2 ). There are a few ways to determine this quotient ring. First, we note that in the quotient ring, y − x2 = 0 so y = x2 . Consequently, every polynomial p(x, y) ∈ R[x, y], in the quotient ring we have p(x, y) = p(x, x2 ). Polynomials of this form make up all polynomials in x. Hence, R[x, y]/(y − x2 ) = R[x]. Alternatively, consider the function ϕ : R[x, y] → R[x] defined by ϕ(p(x, y)) = p(x, x2 ).

304

CHAPTER 5. RINGS

It is easy to see that this is a homomorphism. Furthermore, ker ϕ = {p(x, y) | p(x, x2 ) = 0}. Every polynomial in this ideal satisfies p(x, y) = p(x, y) − 0 = p(x, y) − p(x, x2 ). By an appropriate factoring, we see that p(x, y) must be a multiple q(x, y)(y − x2 ). Hence ker ϕ = (y − x2 ). By the First Isomorphism Theorem, since ϕ is surjective, R[x, y]/(y − x2 ) ∼ = R[x]. By the propositions in this section, since R[x] is an integral domain, then (y − x2 ) is a prime ideal. Since R[x] is not a field, then the ideal (y − x2 ) is not maximal. Exercise: 13 Section 5.7 Question: Prove that (y, x2 ) is not prime in R[x, y]. Solution: Every element of (y, x2 ) looks like p(x, y) · y + q(x, y) · x2 . Now consider the element xy + 1x2 = x · (y + x). If (y, x2 ) is prime, then either x or y + x exist in (y, x2 ). However, clearly neither of them is contained in (y, x2 ) so (y, x2 ) is not a prime ideal. Exercise: 14 Section 5.7 Question: Prove that the principal ideal (x2 + y 2 ) is prime in R[x, y] but not in C[x, y]. Solution: We prove that R[x, y]/(x2 + y 2 ) is an integral domain. It contains 1 and is commutative so we must show that it has no zero divisors. We can view the ring R[x, y] as R[y][x] and then we view the polynomial x2 + y 2 as a quadratic polynomial with variable x (and constant y). Then in R[y][x]/(x2 + y 2 ) every element can be expressed uniquely as a(y) + b(y)x, for polynomials a(y), b(y) ∈ R[y], where x satisfies the property that x2 = −y 2 . Then 0 = (a(y) + b(y)x)(c(y) + d(y)x) = (a(y)c(y) − y 2 b(y)d(y)) + (a(y)d(y) + b(y)c(y))x. Case 1: a(y) 6= 0. Then d(y) = −b(y)c(y)/a(y) from the component of x and then plugging this into the constant component we must have a(y)c(y) + y 2

b(y)2 c(y) c(y) = 0 =⇒ a(y)2 + y 2 b(y)2 = 0, a(y) a(y)

in R[y]. Hence, since a(y) 6= 0, we must have c(y) = 0. But then from d(y) = −b(y)c(y)/a(y) we see that d(y) = 0 as well. Case 2: a(y) = 0. Then b(y)d(y) = b(y)c(y) = 0. So b(y) = 0 or c(y) = d(y) = 0. In both cases, if (a(y) + b(y)x)(c(y) + d(y)x) = 0, then one of the factors is 0. Hence, R[y][x]/(x2 + y 2 ) has no zero divisors and is an integral domain. By Proposition 5.7.7, (x2 + y 2 ) is a prime ideal. In the quotient ring C[x, y]/(x2 + y 2 ), there are zero divisors, namely x + iy and x − iy, since multiplied together they give 0. Thus, C[x, y]/(x2 + y 2 ) is not an integral domain and so (x2 + y 2 ) is not a prime ideal. Exercise: 15 Section 5.7 Question: Show that in C 1 (R, R) the ideal I = {f | f (2) = f 0 (2) = 0} is not a prime ideal. Solution: Let f1 (x) and f2 (x) be two functions with fi (2) = 0 but fi0 (2) 6= 0. Neither of these functions is an element of I. However, f1 (2)f2 (2) = 0 and d (f1 (x)f2 (x))] = f10 (2)f2 (2) + f1 (2)f20 (2) = f10 (2) · 0 + 0 · f20 (2) = 0. dx x=2 Thus, f1 (x)f2 (x) ∈ I. This shows that I is not a prime ideal. Exercise: 16 Section 5.7 Question:

5.7. MAXIMAL IDEALS AND PRIME IDEALS

305

Show by example that the intersection of two prime ideals is in general not another prime ideal. Solution: Consider the ring R = Z/2Z ⊕ Z/2Z. Within this ring, there are two nontrivial ideals I = (0, 1) and J = (1, 0) . Then R/I ∼ = Z/2Z which is a field. Then both I and J are maximal and prime. Now = R/J ∼ I ∩ J = (0, 0) . This only contains the zero element. However, (1, 0) · (0, 1) = (0, 0) but neither (1, 0) nor (0, 1) exist in I ∩ J. Then I ∩ J is not a prime ideal and this serves as a counter example. Exercise: 17 Section 5.7 Question: Let R be any ring. 1. Show that the intersection of two prime ideals P1 and P2 is prime if and only if P1 ⊆ P2 or P2 ⊆ P1 . 2. Conclude that the intersection of two distinct maximal ideals is never a prime ideal. Solution: Let R be any ring. 1. Suppose that P1 and P2 are prime and that P1 ∩ P2 is also a prime ideal. We know that P1 P2 ⊆ P1 ∩ P2 by properties of ideal operations. Hence, since P1 ∩P2 is prime, then by definition P1 ⊆ P1 ∩P2 or P2 ⊆ P1 ∩P2 . But for any two sets U and V , we have U ⊆ U ∩ V if and only if U ⊆ V . Hence, P1 ⊆ P2 or P2 ⊆ P1 . Conversely, suppose that P1 ⊆ P2 or P2 ⊆ P1 . Without loss of generality, suppose that P1 ⊆ P2 . Then P1 = P1 ∩ P2 so P1 ∩ P2 is prime. 2. Let M1 and M2 be maximal ideals. If M1 ∩ M2 is prime, then since M1 M2 ⊆ M1 ∩ M2 , we know that M1 ⊆ M1 ∩ M2 or M2 ⊆ M1 ∩ M2 . Using the same result of set theory, we deduce that M1 ⊆ M2 or vice versa. Hence, by definition of maximality, M1 = M2 . In particular, the intersection of two distinct maximal ideals cannot be prime.

Exercise: 18 Section 5.7 Question: Let R be a commutative ring. Prove that the nilradical N (R) is a subset of every prime ideal. k

k−1

k−1

k

k−1

Solution: Let I be a prime ideal. First we note that if a2 ∈ I, then a2 · a2 = a2 ∈ I and a2 ∈ I. 2k k Then by induction, if a ∈ I, a ∈ I. Now consider any element a ∈ N (R) where a = 0 and any prime ideal I. Let m be the smallest power of 2 greater than or equal to k so that am = 0. Then am = 0 ∈ I and by our previous reasoning, a ∈ I. This shows that N (R) ⊆ I for any prime ideal I. Exercise: 19 Section 5.7 Question: Show that a commutative ring with an identity 1 6= 0 is an integral domain if and only if {0} is a prime ideal. Solution: We denote the ring by R. Suppose that R is a integral domain. Now consider the ideal (0). If a · b = 0, then since R is an integral domain, either a or b is 0. Then a or b is contained in (0). This show that {0} is a prime ideal. Now suppose that we know {0} is a prime ideal. Then if a · b = 0 for any a and b then at least one of them must be 0. Then there are no nontrivial zero-divisors and R is an integral domain. Exercise: 20 Section 5.7 Question: Let ϕ : R → S be a ring homomorphism and let Q be a prime ideal in S. Prove that ϕ−1 (Q) is a prime ideal in R. Solution: Let A and B be ideals in R such that AB ⊆ ϕ−1 (Q). Then ϕ(AB) = ϕ(A)ϕ(B) ⊆ Q. Now ϕ(A) and ϕ(B) are not necessarily ideals in S. Let us denote by (ϕ(A)) the ideal generated by ϕ(A) and by (ϕ(B)) the ideal in S generated by ϕ(B). Since ϕ(A)ϕ(B) ⊆ Q, then (ϕ(A))(ϕ(B)) ⊆ Q. Since Q is a prime ideal in S, then (ϕ(A)) ⊆ Q or (ϕ(B)) ⊆ Q. Now, in general, ϕ−1 (ϕ(A)) 6= A but nonetheless A ⊆ ϕ−1 (ϕ(A))), precisely because ϕ(A) ⊆ ϕ(A). Hence A ⊆ ϕ−1 ((ϕ(A))) and similarly for B. Since (ϕ(A)) ⊆ Q or (ϕ(B)) ⊆ Q, we deduce that A ⊆ ϕ−1 ((ϕ(A))) ⊆ ϕ−1 (Q)

or

B ⊆ ϕ−1 ((ϕ(B))) ⊆ ϕ−1 (Q).

306

CHAPTER 5. RINGS

We deduce that ϕ−1 (Q) is a prime ideal. Note that if R is a commutative ring, we could prove the result mores imply as follows. Let r, s ∈ R such that rs ∈ ϕ−1 (Q). Then ϕ(rs) ∈ Q but since ϕ is a homomorphism, then ϕ(r)ϕ(s) ∈ Q. Since Q is a prime ideal, then ϕ(r) ∈ Q or ϕ(s) ∈ Q. Hence r ∈ ϕ−1 (Q) or s ∈ ϕ−1 (Q). This shows that ϕ−1 (Q) is a prime ideal.

Exercise: 21 Section 5.7 Question: Let ϕ : R → S be a ring homomorphism and let P be a prime ideal in R. Prove that the ideal generated by ϕ(P ) is not necessarily a prime ideal in S. Solution: There is an easy case to examine why this is not true. Let R be any ring and let P be any prime ideal in the ring. Consider any ring S which contains nontrivial zero divisors. Let ϕ : R → S be the trivial ring homomorphism where ϕ(r) = 0 for all R. Then ϕ(P ) = {0} but {0} is not a prime ideal in S by Exercise 5.7.19. Exercise: 22 Section 5.7 Question: Let R be a ring and let S be a subset of R. Prove that there exists an ideal that is maximal (by inclusion) with respect to the property that it “avoids” (does not contain) the set S. Solution: Let Z be the set of ideals in R that do not contain the set S. In order for the exercise to be true, it is necessary that at least one ideal avoids the set S, i.e., that Z is nonempty. The collection Z is a partial order with the relation ⊆. Let C be any chain in Z. Consider the set [ J= I. I∈C

Using reasoning identical to Krull’s Theorem, J is an ideal and it does not contain the set S. First note that J is nonempty since 0 ∈ J. Also, if f, g ∈ J and r ∈ R, then there exists I0 ∈ C such that f, g ∈ I0 . Then f − g ∈ I0 ⊆ J. Finally, since I0 is an ideal, rf ∈ I0 ⊆ J and f r ∈ I0 ⊆ J. Thus J is an ideal. The ideal J does not contain S since no ideal I ∈ C contains S. Hence, J ∈ Z. By Zorn’s Lemma, since every chain in Z has an upper bound, Z contains maximal elements.

Exercise: 23 Section 5.7 Question: Prove that the nilradical of a commutative ring is equal to the intersection of all the prime ideals of that ring. Solution: By Exercise 5.7.18, N (R) is contained in every prime ideal so, \ N (R) ⊆ P. P prime ideal

The reverse inclusion is the difficult part. Let r be an element that is not nilpotent. We claim that there exists some prime ideal that does not contain r. Let Z be the collection of ideal that do not contain any element of the form rn for all n ≥ 1. The collection Z is not empty since (0) ∈ Z. By Exercise 5.7.22, Z contains a maximal element P . We prove that P is a prime ideal. Assume otherwise. Then there exist x, y ∈ / P with xy ∈ P . The set of elements z for which xz ∈ P is an ideal (in particular the fraction ideal (P : (x))) that properly contains P since y ∈ (P : (x)) − P . By the maximality condition of P , the ideal (P : (x)) must contain some power rm . Furthermore, the set, I = {z ∈ R |, rm z ∈ P } is also an ideal that properly contains P . Consequently, again by the maximality of P , we deduce that I contains some power rn of r. Hence, rm+n ∈ P . This gives a contradictions so the assumption that P is not prime is false. Thus P is a prime ideal. We conclude that for every nonnilpotent element r ∈ R, there exists some prime ideal P that does not contain r. Consequently, the intersection of all prime ideals cannot contain a nonnilpotent element. In other words, \ P ⊆ N (R). P prime ideal

5.7. MAXIMAL IDEALS AND PRIME IDEALS

307

This gives the desired reverse inclusion that establishes that \ N (R) =

P.

P prime ideal

Exercise: 24 Section 5.7 √ Question: Let R be a commutative ring and let I be an ideal. Prove that the radical ideal I is equal to the intersection of all prime ideals that contain I. Solution: Let R be a commutative ring and I and ideal. The nilradical N (R/I) consists of √ elements r + I Under the such that rn + I = I, that is rn ∈ I, for some positive integer n. In other words, N (R/I) = I/I. √ bijective correspondence of the Fourth Isomorphism Theorem, the nilradical N (R/I) corresponds to I. By Exercise 5.7.23, N (R/I) is the intersection of all the prime ideals in R/I. Let C be the prime ideals in R/I. Referring to the projection homomorphism π : R → R/I, by Exercise 5.7.20, π −1 (Q) is a prime ideal in R for all Q ∈ C. Again by the Fourth Isomorphism Theorem, I ⊆ π −1 (Q) for all Q ∈ C. It is not hard to show that for any function f : A → B between sets, if {Vi }i∈I is a collection of subsets in B, then ! \ \ −1 f Vi = f −1 (Vi ). i∈I

i∈I

Thus, √

 I = π −1 (N (R/I)) = π −1 

 \

Q =

Q∈C

\

π −1 (Q).

Q∈C

Now {π −1 (Q) | Q ∈ C} might not be all the prime ideals in R that contain I. So we have show that \ √ P ⊆ I. P prime: I⊆P

√ Conversely, let r ∈ I. Then rk ∈ I for some positive integer k. Let P be any prime ideal √ that contains I. Then rk ∈ P . Since P is commutative, by Proposition 5.7.6, r ∈ P . Since every element of I is in every prime √ ideal that contain P , then I is a subset of the intersection of all prime ideals that contain P . With the above paragraph, this proves that \ √ I= P. P prime: I⊆P

Exercise: 25 Section 5.7 Question: Consider the polynomial ring R[x1 , x2 , x3 , . . .] with real coefficients but a countable number of variables. Define I1 = (x1 ), I2 = (x1 , x2 ), and Ik = (x1 , x2 , . . . , xk ) for all integers k. Prove that Ik is a prime ideal for all positive integers k and prove that I1 ( I2 ( · · · ( Ik · · ·

Solution: We first claim that R = R[x1 , x2 , x3 , . . .] is an integral domain. It is commutative and it contains an identity 1. Furthermore, given two nonzero polynomials p and q in R, they can involve only a finite number of terms,. each of which involves a finite number of variables. Thus p and q are in a subring that involves only a finite number of variables. Polynomial rings in a finite number of variables with coefficients in R contain no zero divisors. Hence, pq 6= 0 in R. Thus R is an integral domain. It is obvious that I1 ( I2 ( · · · ( Ik · · · and that this chain never terminates. We now just need to prove that each Ik is a prime ideal. We consider the quotient ring R/Ik = R[xk+1 , xk+2 , . . .] ∼ = R. Hence, for all k ≥ 1, the quotient ring R/Ik is an integral domain. By Proposition 5.7.7, Ik is a prime ideal.

6 | Divisibility in Commutative Rings 6.1 – Divisibility in Commutative Rings Exercise: 1 Section 6.1 Question: List all the divisors of (6, 14) in Z ⊕ Z. Solution: The divisors of (6, 14) are (1, 1), (1, 2), (1, 7), (1, 14), (2, 1), (2, 2), (2, 7), (2, 14), (3, 1), (3, 2), (3, 7), (3, 14), (6, 1), (6, 2), (6, 7), (6, 14) along with any sign changes on either of the components. (There are a total of 4 × 16 = 64 divisors.) Exercise: 2 Section 6.1 Question: Prove that if a|b and a|c in a commutative ring, then a|(b + c). Solution: Suppose that a|b and a|c in a commutative ring R. Then there exist r, s ∈ R such that ar = b and as = c. By distributivity, b + c = ar + as = a(r + s). Thus, a | (b + c) with r + s as the quotient. Exercise: 3 Section 6.1 Question: Let R be a noncommutative ring. Prove that the relation on R of left-divisible is a transitive relation. Solution: We say that a is a left divisor of b if there exists r ∈ R such that b = ra. Suppose that a is a left divisor of b and that b is a left divisor of c. Then there exist r, s ∈ R such that b = ra and c = sb. Then, c = s(ra) = (sr)a by associativity in R. Thus a is a left divisor of c. The relation of left-divisibility is a transitive relation. Exercise: 4 Section 6.1 Question: Prove that a subring of an integral domain that contains the identity is again an integral domain. Solution: Let R be an integral domain and let S be a subring that contains the identity 1. For all a, b ∈ S, they are in R so ab = ba. Hence, the multiplication is commutative in S. Also, suppose that a, b ∈ S with ab = 0. Then ab = 0 ∈ R so a = 0 or b = 0. Thus S does not contain any zero divisors. We conclude that S is also an integral domain. Exercise: 5 Section 6.1 Question: Let R be a commutative ring with a 1 6= 0. Prove or disprove that if R/I is an integral domain, then R is an integral domain (and I is prime). Solution: The hypothesis that if R/I is an integral domain then R must be is false. Consider the following counterexample. Let R = Z[x]/(x2 − 1). This is not an integral domain since x − 1 and x + 1 are zero divisors in R. However, (x) is an ideal in R and R/(x) ∼ = Z, which is an integral domain. Exercise: 6 Section 6.1 Question: Let R be an integral domain. Prove that the relation a ' b defined by a is an associate to b is an equivalence relation. Solution: Let ' be the relation of associate on an integral domain R. Reflexivity: For all a ∈ R, we have a × 1 = a, where 1 is a unit. Thus a ' a. Symmetry: Let a, b ∈ R with a ' b. Then b = au for some unit u. Let v ∈ R be the element such that uv = 1. Then bv = auv = a. Hence, b ' a. Transitivity: Let a, b, c ∈ R such that a ' b and b ' c. Then there exist u1 , u2 ∈ U (R) such that b = au1 and c = bu2 . But then c = au1 u2 . If v1 and v2 are the inverses of u1 and u2 respectively, then (u1 u2 )(v1 v2 ) = 1 so u1 u2 is a unit. Hence, a ' c. 309

310

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

These prove that ' is an equivalence relation on R. Exercise: 7 Section 6.1 √ √ Question: Let D be a square-free integer and consider Z[ D] along with the norm N (a + b D) = a2 − Db2 . √ Prove that if N (α) is a prime number, then α is irreducible in Z[ D]. √ Solution: Suppose that α can be written α = βγ with β, γ ∈ Z[ D]. Since this norm is multiplicative, we have N (α) = N (β)N (γ). Furthermore, since N (α) is a prime integer either N (β) = 1 or N (γ) = 1. Thus, by Proposition 6.1.7, β or γ is a unit. Thus, α is irreducible. Exercise: 8 Section 6.1 Question: Use Exercise 6.1.7 to find 5 irreducible elements in Z[i] that are not associates to each other. Solution: We simply need to find 5 non-associate irreducible elements in Z[i] and it is sufficient to check that N (a + bi) = a2 + b2 is prime to decide that a + bi is irreducible. 2+i

N (2 + i) = 5

3 + 2i

N (3 + 2i) = 13

4+i

N (4 + i) = 17

5 + 2i

N (5 + 2i) = 29

5 + 4i

N (5 + 4i) = 41

Exercise: 9 Section 6.1 √ Question: Use Exercise 6.1.7 to find 5 irreducible elements in Z[ 3] that are not associates to each other. √ √ Solution: The quadratic norm function on √ Z[ 3] satisfies N (a + b 3) = |a2 − 3b2 |. Nonassociate elements will have different norm values. The element 2 + 3 is a unit and the following remaining elements all have different norms √ √ √ √ 3 + 3, 4 + 3, 1 + 2 3, 1 + 3.

Exercise: 10 Section 6.1 √ Question: Find all the divisors of 21 in Z[ −3]. √ √ Solution: √ All the divisors of 21 in Z are ±1, ±3, ±7, and ±21. We note that 3 = − −3 −3 and (2 + √ −3)(2 − −3) = 7. Hence, √ 21 will have a number more divisors. Also, if αβ = 21 in Z[ −3], then N (α)N (β) =√441 = 32 × 72 . With theorems at our disposal right now, √ we can solve this√exercise by determining the α ∈ Z[ −3] with N (α) | 441 and from these see if α|21 in Z[ −3]. Now N (a + b −3) = a2 + 3b2 and for a given nonnegative integer n, there are a finite number of pairs (a, b) such that a2 + 3b2 = n. So we can list all relevant pairs by enumeration. If we find the corresponding divisors √ β, the we only need to check α with N (α) ≤ 441 = 21, since the β will have 21 ≤ N (β) ≤ 441. • N (α) = 1 implies α = ±1 and the corresponding divisors β are β = ±21. √ √ • N (α) = 3 implies α = ± −3 and the corresponding divisors β are ∓7 −3. √ √ √ • N (α) = 7 implies α = ±(2 + −3) and ±(2 − −3) and the corresponding divisors β are β = ±(6 − 3 −3) √ and ±(6 + 3 −3). • N (α) = 9 implies α = ±3 and the corresponding divisors β are β = ±7. √ √ • N (α) = 21 implies α = ±(3 + 2 −3) and α = ±(3 − 2 −3), with each other serving as corresponding divisors β. √ This list incorporates as the divisors of 21 in Z[ −3]. Exercise: 11 Section 6.1 Question: Consider the ring Z[i].

6.1. DIVISIBILITY IN COMMUTATIVE RINGS

311

a) Prove that if a + bi and c + di are such that (a + bi)(c + di) ∈ Z, then c + di is ±(a − bi). b) Prove that for all pairs (a, b) ∈ Z2 , the expression a2 + b2 is never congruent to 3 modulo 4. c) Prove that 2 is not irreducible. d) Deduce that for a prime number p, the element p = p + 0i is irreducible in Z[i] if p ≡ 3 (mod 4). Solution: Consider the ring Z[i]. a) Suppose that a + bi and c + di are such that (a + bi)(c + di) ∈ Z. Since (a + bi)(c + di) = (ac − bd) + (ad + bc)i, we must have ad + bc = 0. Note that if a = 0, then c must be 0 but d can be anything. On the other hand, if a 6= 0, then d = − bc a so c bc c + di = c − i = (a − bi). a a Consequently, c + di is some rational multiple of a − bi. This rational multiple ac must satisfy ac (a2 + b2 ) ∈ Z 2 2 so, if m is the denominator of a +b when expressed in reduced form, then c can be any multiple of m. a b) (This is Exercise 2.2.8.) c) The number 2 is not irreducible because 2 = (1 + i)(1 − i) and since N (1 ± i) = 2, then neither 1 + i nor 1 − i is a unit. d) Assume that there exists α, β ∈ Z[i] such that αβ = p + 0i, for a prime number p ∼ = 3 (mod 4) in Z. Then N (αβ) = p2 = N (α)N (β). If neither α nor β is a unit, then we must have N (α) = p = N (β). However, both N (α) and N (β) are the sum of two integer squares. By part (c), such a sum cannot be congruent to 3 modulo 4. This leads to a contradiction. Hence, p is irreducible in Z[i]. Exercise: 12 Section 6.1 √ √ Question: Prove that Z[ 3 2] is an integral domain. Prove that if a, b ∈ Z such that a3 − 2b3 = ±1 then a + b 3 2 is a unit. √ √ Solution: The ring Z[ 3 2] is a subring of the field R. Furthermore, it contains 1 so by Exercise 6.1.4, Z[ 3 2] is an integral domain. Suppose that a3 − 2b3 = ±1. Then √ √ √ 2 3 3 3 (a + b 2)(a2 − ab 2 + b2 2 ) = a3 − 2b3 = ±1 √ √ √ 2 so a + b 3 2 is a unit with inverse ±(a2 − ab 3 2 + b2 3 2 ). Exercise: 13 Section 6.1 √ √ Question: Consider the ideal I = (3, 2 + −5) in the ring Z[ −5]. a) Prove that 1 ∈ / I so you can conclude that I is a proper ideal of the ring. √ √ b) Use the quadratic norm to show that the only elements α ∈ Z[ −5] such α | 3 and α | (2 + −5) are units. c) Deduce that I is not a principal ideal. √ √ Solution: Consider the ideal I = (3, 2 + −5) in the ring Z[ −5]. √ √ √ a) Suppose that 1 ∈ I. Then there exist a + b −5, c + d −5 ∈ Z[ −5] such that √ √ √ √ 1 = 3(a + b −5) + (2 + −5)(c + d −5) = (3a + 2c − 5d) + (3b + c + 2d) −5. Thus, there mist exist integers a, b, c, d such that ( ( 3a + 2c − 5d = 1 3a + 2c − 5d = 1 =⇒ 3b + c + 2d = 0 c = −3b − 2d

( =⇒

3a − 6b − 9d = 1 c = −3b − 2d.

This has a contradiction because 3a − 6b − 9d is a multiple of 3, while 1 is not. Hence 1 ∈ / I. This implies that I ( R, so I is a proper ideal. √ b) The quadratic norm N (a + b −5) = a2 + 5b2 is √ multiplicative. Furthermore, it is easy to check by a proof by cases that there does not exist an α ∈ Z[ −5] such that N (α) = 3. Suppose that √ √ √ α divides 3 and 2 + −5 with αz = 3 and αw = 2 + −5. Then N (α) divides N (3) = 9 and N (2 + −5) = 9, though N (α) 6= 3. Thus N (α) = 1 or 9. Assume that N (α) = 9. Then N (z) and √ N (w) are both equal to√1. Thus z and w are units so α is an associate of 3 and also an associate of 2 + −5. However, 9 and 2 + −5 are not multiples of each other so α cannot be associate to both √ at the same time. Thus N (α) = 1 and we deduce that the only elements that divide both 3 and 2 + −5 are units.

312

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

c) Consequently, I is a strict ideal of R and I * (α) for any α. Thus, in particular I ( (α) for all non-unit α. Exercise: 14 Section 6.1 Question: Let ϕ : R → S be an injective homomorphism between integral domains. a) Prove that if s ∈ Im ϕ is an irreducible element in S, then r = ϕ−1 (s) is an irreducible element in R. b) Prove by a counterexample that if r is an irreducible element in R, then ϕ(r) is not necessarily irreducible in S. Solution: Let ϕ : R → S be an injective homomorphism between integral domains. a) Let s ∈ Im ϕ and since ϕ is injective there exists a unique r ∈ R such that ϕ(r) = s. Since s is irreducible, whenever s = ab in S, then a or b is a unit. In particular, whenever s factors into s = ab in Im ϕ, then a or b is a unit in Im ϕ. By the First Isomorphism Theorem, because ϕ is injective R ∼ = Im ϕ so ϕ−1 (a) or −1 −1 ϕ (b) is a unit in R. Thus ϕ (s) is irreducible. √ b) An easy example to consider is the ϕ : Z → Z[ 2]. The element 2 is irreducible √ √ √ inclusion homomorphism √ in Z but ϕ(2) = 2 = 2 2, and 2 is not a unit in Z[ 2]. Exercise: 15 Section 6.1 √ √ Question: Prove that there are no√elements α ∈ Z[ 10] with N (α) = 3. Conclude that the elements 7 + 2 10 and 3 are irreducible elements in Z[ 10]. √ √ Solution: The quadratic norm on Z[ 10] has N (a + b 10) = |a2 − 10b2 |. Thus a2 − 10b2 = ±3. Considering this expression modulo 10, we see that a2 ≡ 3 or 7 (mod 10). However, modulo 10, the squares are 0, 1, 4, 5, 6, and 9. Consequently, √ there is no pair (a, b) of integers such that |a2 − 10b2 | = 3. Note that N (7 + 2 10) = |49 − 40| = 9 and N (3) = 9. In either case, if N (αβ) = 9, then since there does√not exist an element of norm 3, we conclude that one of N (α) and N (β) is 1 while the other is 9. Hence, 7 + 2 10 √ and 3 are irreducible elements in Z[ 10]. Exercise: 16 Section 6.1 Question: Let R be an integral domain. Prove that, if it exists, a least common multiple of a, b ∈ R is a generator for the (unique) largest principle ideal containing (a) ∩ (b). Conclude that in a PID, if (a) ∩ (b) = (m), then m is a least common multiple of a and b. Solution: Since R is commutative, an element x is in the ideal (a) if and only if x = ar for some r ∈ R. Hence, an element x in the ideal (a) ∩ (b) is such that x = ar for some r ∈ R and x = bs for some s ∈ R. Thus, x is a common multiple of a and b. Suppose that a least common multiple m exists to a and b. Then m | x for all x ∈ (a) ∩ (b) so x ∈ (m). Hence, (a) ∩ (b) ⊆ (m). If there were a larger principal ideal containing (a) ∩ (b), then (m) ⊆ (m0 ). But this would mean that m0 | m, which contradicts the least condition for m. Thus (m) is the largest principle ideal containing (a) ∩ (b). If R is a PID, then every ideal is principal so (a) ∩ (b) = (c) for some (c). Now if m0 is a common multiple to a and b, then m0 ∈ (a) ∩ (b). Thus, (m0 ) ∈ (a) ∩ (b). Consequently, in a PID, if m is a least common multiple of a and b, then (m) ⊆ (a) ∩ (b) and (a) ∩ (b) ⊆ (m). The result follows. Exercise: 17 Section 6.1 Question: Prove that in a PID, every irreducible element is prime. Solution: Let R be a PID. We first prove that an ideal (r) is maximal if and only if r is irreducible. Suppose that (m) is a maximal ideal. Suppose that m = ab. Then m ∈ (a) so (m) ∈ (a). Thus, since (m) is maximal (a) = (m) or (a) = R, which implies that a ' m or a is a unit. The converse also holds. Next, we claim that an ideal (p) is prime if and only if the element p is prime. Suppose that (p) is prime. Then p | ab if and only if ab ∈ (p), if and only if a ∈ (p) or b ∈ (p), if and only if p | a or p | b. (This uses Proposition 5.7.6.) Let R be a PID and suppose that p ∈ R is a nonzero prime element. By Proposition 5.7.10, the prime ideal (p) is a maximal ideal, so by what we proved above, p is an irreducible element. Exercise: 18 Section 6.1 Question: Let R be an integral domain and let a, b ∈ R. Prove that a and b have a greatest common divisor d with a = dk and b = d`, then k and ` are relatively prime.

6.1. DIVISIBILITY IN COMMUTATIVE RINGS

313

Solution: Let R be an integral domain and let a, b ∈ R. Suppose that a and b have a greatest common divisor d with a = dk and b = d`. Assume that k and ` are not relatively prime. Then k = ck 0 and ` = c`0 where c is not a unit. Then a = dck 0 and b = dc`0 . Thus, d | dc. Since d is a greatest common divisor of a and b, then dc | d. Hence, dcr = d for some r ∈ R. But then d(cr − 1) = 0 and since R is an integral domain and d 6= 0, we have cr = 1 so c is a unit, leading to a contradiction. Hence, k and ` are relatively prime. Exercise: 19 Section 6.1 √ Question: Prove that Proposition 2.1.14 does not hold when we replace Z with Z[ 10]. Solution: We use an example inspired by Example 6.1.13. Consider the product √ √ 2 · 3 = 6 = (4 − 10)(4 + 10). √ In Exercise 6.1.15, we saw that no element in Z[10] has a norm of 2, 3, 7, or 8. The norms of 2, 3, 4 − 10, and √ 4 + √10 are respectively 4, 9, 6, and 6. These norm values can only nontrivially factor into 2 and 3. Hence, 2, √ √ 3, 4 − 10, and 4 + 10 are irreducible. Now, as irreducible elements, the only common divisors of 2 and 4 − 10 are units. Hence, they are relatively prime. Furthermore, √ √ 2 | (4 − 10)(4 + 10) √ √ but again by √ norm considerations we see that 2 - 4 − 10 and 2 - 4 + 10. Hence, Proposition 2.1.14 does not hold for Z[ 10]. Exercise: 20 Section 6.1 Question: Let p1 , p2 , q1 , q2 be irreducible elements in an integral domain R such that none are associates to any of the others and p1 p2 = q1 q2 . Prove that p1 q1 q2 and p1 p2 q1 do not have a greatest common divisor. Solution: Let r = p1 q1 q2 and s = p1 p2 q1 . Notice that p1 q1 divides r and s. Furthermore, since q1 q2 = p1 p2 , then r = p21 p2 . Thus p1 p2 also divides s. So p1 q1 and p1 p2 are common divisors of r and s. Since p2 and q1 are not associates of each other, then p1 p2 and p1 q1 are not associates of each other. Consequently, one does not divide the other and in particular this implies that they are not greatest common divisors of r and s. Assume that r and s have a greatest common divisor d. Then by definition, p1 p2 | d and p1 q1 | d. Thus p1 p2 x = d and p1 q1 y = d for some x, y ∈ R. Furthermore, neither x nor y is a unit since we already know that p1 p2 and p1 q1 are not greatest common divisors of r and s. Since d divides s, we have p1 p2 xu = r = p1 p2 q1 for some u ∈ R. By the cancellation law, xu = q1 . Similarly since d | r, we have p1 q1 yv = p1 p2 q1 so we conclude that yv = q2 . Since q1 and q2 are irreducible, then u and v are units and x and y are associates of q1 and of q2 respectively. We conclude that d = u0 p1 p2 q1 = v 0 p1 q1 q2 , where u0 and v 0 are units. Thus, since q1 q2 = p1 p2 , we have u0 p1 p2 q1 − v 0 p21 p2 = 0 =⇒ p1 p2 (u0 q1 − v 0 p1 ) = 0 =⇒ q1 = v 0 p1 /u0 . But this is a contradiction since we assumed that p1 and q1 are not associates of each other. This contradicts the assumption that r and s have a greatest common divisor. Exercise: 22 Section 6.1 Question: Prove Proposition 6.1.16. Solution: Let R be an integral domain. Suppose that a and b possess a greatest common divisor d. Since d divides a and divides b, then d | ab. Also, we can write a = kd and b = ld, for some k, l ∈ R. Let m be the element such that md = ab. From md = dkb, we get m = bk and similarly m = al and hence m is a common multiple of a and b. Let m0 be another common multiple of a and b that satisfies m0 | ab. Obviously, such a common multiple exists since ab itself is an example of one. Set m0 = pa and m0 = qb for some p and q in R. Let us assume that p and q are relatively prime because if not we could divide both expressions m0 = pa and m0 = qb by a common divisor of p and q. Then m0 = pdk = qdl −→ pk = ql. Now we use the result that if a and b have a greatest common divisor d, then for any c ∈ R, a greatest common divisor to ac and bc exists is an associate to dc. This is proved in Exercise 6.1.25. With exercises in this order, the students would need to prove this as a lemma.

314

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Since k and l are relatively prime, then 1 is a greatest common divisor. Thus pk and pl have a greatest common divisor that is an associate to p. Since pk = ql, we deduce that ql and pl have a greatest common divisor of p. However, since q and p are relatively prime, a greatest common divisor to ql and pl is l. Hence, p ' l. Similarly q ' k. Then m0 = pa ' la ' m. If p and q were not relatively prime but had a common divisor of D, then we would find that m0 ' mD. In other words, m | m0 and we deduce that m is a least common multiple. By construction, dm = ab. The reverse implication is similar in proof. Exercise: 23 Section 6.1 Question: Lt R be a gcd-domain. Prove that gcd(a, b) lcm(a, b) ' ab for all nonzero a, b ∈ R. Solution: Let d be a greatest common divisor of a and b. Write a = dk and b = d`. Then M = ab/d = a` = bk is a least common multiple; see Exercise 22 of this section. Thus for any gcd(a, b) and any lcm(a, b), we have gcd(a, b) lcm(a, b) ' ab. Exercise: 24 Section 6.1 Question: Let R be a gcd-domain. Prove that gcd(a, gcd(b, c)) ' gcd(gcd(a, b), c) and also that lcm(a, lcm(b, c)) ' lcm(lcm(a, b), c) for all nonzero a, b, c ∈ R. Solution: Let d1 be a greatest common divisor to b and c and let d2 be a greatest common divisor to a and d1 . Also let d3 be a greatest common divisor to a and b, and let d4 be a greatest common divisor to d3 and c. By transitivity of divisibility, d2 divides both b and c. Since d2 divides a and b, it divides d3 , and since it divides both d3 and c, then d2 divides d4 . Hence d2 | d4 . By an entirely parallel argument, we show that d4 | d2 . Thus d2 ' d4 . This establishes gcd(a, gcd(b, c)) ' gcd(gcd(a, b), c). The proof for least common multiples is similar. Let m1 be a least common multiple to b and c and let m2 be a least common multiple to a and m1 . Also let m3 be a least common multiple to a and b, and let m4 be a least common multiple to m3 and c. By transitivity of divisibility, m2 is a multiple to both b and c. Since m2 is a multiple to a and b, it is a multiple of m3 , and since it is a multiple to both m3 and c, then m2 is a multiple of m4 . Hence m4 | m2 . By an entirely parallel argument, we show that m2 | m4 . Thus m2 ' m4 . This establishes lcm(a, lcm(b, c)) ' lcm(lcm(a, b), c).

Exercise: 25 Section 6.1 Question: Prove that gcd(ac, bc) ' gcd(a, b)c for all nonzero a, b, c ∈ R. Prove also that lcm(ac, bc) ' lcm(a, b)c. Solution: Let d1 be a greatest common divisor to a and b and let d2 be a greatest common divisor to ac and bc. Suppose that a = d1 k and b = d1 `. Then ac = cdk and bc = cd`. Hence, d1 c is a common divisor to both ac and bc and thus cd1 | d2 . Furthermore, c divides both ac and bc so c divides d2 with cd0 = d2 for some d0 in R. But then cd0 | ac and cd0 | bc so d0 | a and d0 | b. Thus d0 | d1 and therefore d2 = cd0 | cd1 . This shows that cd1 ' d2 . This proves that gcd(ac, bc) ' gcd(a, b)c. The proof for least common multiple is similar. Let m1 be a least common multiple to a and b and let m2 be a least common multiple to ac and bc. Suppose that m1 = ap and m1 = bq for some p, q ∈ R. Then m1 c = apc = bqc. Hence, m1 c is a common multiple to ac and bc. Thus, m1 c | m2 . Furthermore, c divides both ac and bc so c divides any common multiple, and in particular, c | m2 with cm0 = m2 for some m0 ∈ R. But then ac | cm0 and bc | cm0 so a | m0 and b | m0 . Thu, m1 | m0 since m0 is a common multiple of a and b. Hence, m2 = cm0 | cm1 . Together with cm1 | m2 , we deduce that cm1 ' m2 . This proves that lcm(ac, bc) ' lcm(a, b)c.

Exercise: 26 Section 6.1 Question: Prove that gcd(a, b) ' 1 and gcd(a, c) ' 1 if and only if gcd(a, bc) ' 1. Solution: Suppose that gcd(a, b) ' 1 and gcd(a, c) ' 1. Using the previous exercises, we have 1 ' gcd(a, b) ' gcd(a, gcd(ab, bc)) ' gcd(gcd(a, ab), bc) ' gcd(a, bc),

6.2. RINGS OF FRACTIONS

315

where it is obvious that gcd(1, b) ' 1 so gcd(a, ab) ' a. Conversely, suppose that gcd(a, bc) ' 1. Then 1 ' gcd(a, bc) ' gcd(gcd(a, ab), bc) ' gcd(a, gcd(ab, bc)) ' gcd(a, b gcd(a, c)). But then if gcd(a, b) is not a unit, then gcd(a, bx) is not a unit for any x ∈ R. Hence, we deduce that gcd(a, b) ' 1. By an identical reasoning, we also deduce that gcd(a, c) ' 1. Exercise: 27 Section 6.1 Question: Prove that if gcd(a, b) ' 1 and a|bc, then a|c. Solution: Note that in a gcd-domain, r | s if and only if gcd(r, s) ' r. Suppose that gcd(a, b) ' 1 and a | bc. By Exercise 6.1.25, c ' gcd(ac, bc). Then, using previous exercises, gcd(a, c) ' gcd(a, gcd(ac, bc)) ' gcd(gcd(a, ac), bc) ' gcd(a, bc) ' a. We conclude that a | c. Exercise: 28 Section 6.1 Question: Prove that if gcd(a, b) ' 1, a|c, and b|c, then ab|c. Solution: Suppose that gcd(a, b) ' 1, a | c and b | c. (See the remark at the beginning of the previous exercise as well as all the exercises on gcd-domains.) Since gcd(a, b) ' 1, then gcd(ac, bc) ' c. Then gcd(ab, c) ' gcd(ab, gcd(ac, bc)) ' gcd(gcd(ab, ac), bc) ' gcd(a gcd(b, c), bc) ' gcd(ab, bc)

since b | c

' b gcd(a, c) ' ba

since a | c.

Thus ab | c. Exercise: 29 Section 6.1 Question: A Bézout domain is an integral domain in which the sum of any two principal ideals is again a principal ideal. Prove that a Bézout domain is a gcd-domain. Solution: Let R be a Bézout domain. Let a, b ∈ R. By the definition of a Bézout domain, (a) + (b) = (d) for some d ∈ R. This means that for all r, s ∈ R, there exists a q such that ar + bs = dq. In particular, with (r, s) = (1, 0) we see that d | a and with (r, s) = (0, 1) we see that d | b. Thus, d is a common divisor of a and b. Note that since d ∈ (a) + (b), there exist r, s ∈ R such that d = ar + bs. Now suppose that d0 is any common divisor of a and b. Then d0 c1 = a and d0 c2 = b. Hence, d = d0 (c1 r + c2 s) ans so d0 | d. This shows that d is a greatest common divisors of a and b. Since a and b were taken as arbitrary, then every pair of nonzero elements in R has a greatest common divisor. Therefore every Bézout domain is an integral domain.

6.2 – Rings of Fractions Exercise: 1 Section 6.2 Question: Prove that × is distributive over + in any ring of fractions D−1 R. Solution: Let dr11 , dr22 , dr33 ∈ D−1 R. Then r r1 = × d d1 and

r2 r3 + d2 d3

=

r1 (r2 d3 + r3 d2 ) r1 r2 d3 + r1 r3 d2 = d1 d2 d3 d1 d2 d3

r0 r1 r2 r1 r3 r1 r2 r1 r3 r1 r2 d1 d3 + r1 r3 d1 d2 = × + × = + = . 0 d d1 d2 d1 d3 d1 d2 d1 d3 d21 d2 d3

316

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

We notice that d0 r = d21 d2 d3 (r1 r2 d3 + r1 r3 d2 ) = d1 d2 d3 (r1 r2 d1 d3 + r1 r3 d1 d2 ) = dr0 . Thus, the two fractions are equal, which means that × is distributive over +. Exercise: 2 Section 6.2 Question: Let D = {2a 3b | a, b ∈ N} as a subset of Z. Prove that D−1 Z is isomorphic to Z 16 even though D 6= {1, 6, 62 , . . .}. Solution: The ring D−1 Z consists of fractions of the form n 2a 3b such that n ∈ Z and a, b ∈ N. In the chance that a = b, we see that these fractions include those whose denominator is a power of 6. Hence, Z 61 ⊆ D−1 R. On the other hand, given any fraction n/(2a 3b ), we have n2b 3a n2b 3a n = = . 2a 3b 2a 3b 2b 3a 6a+b This shows that D−1 R ⊆ Z

1 6 . By mutual inclusion, these two sets are equal.

Exercise: 3 Section 6.2 1 . Question: Let D = {2a 3b 5c | a, b, c ∈ N} as a subset of Z. Prove that D−1 Z is isomorphic to Z 30 −1 Solution: The ring D Z consists of fractions of the form n 2a 3b 5c such that n ∈ Z and a, b, c ∈ N. In the chance a = b = c, we see that these fractions include those whose that 1 denominator is a power of 30. Hence, Z 30 ⊆ D−1 R. On the other hand, given any fraction n/(2a 3b 5c ), let m be any integer greater than or equal to max{a, b, c}. Then n 2a 3b 5c This shows that D−1 R ⊆ Z

1 30

=

n2m−a 3m−b 5m−c n2m−a 3m−b 5m−c = . m m m 2 3 5 30m

. By mutual inclusion, these two sets are equal.

Exercise: 4 Section 6.2 Question: Let D be the subset in Z of all positive integers that are products of powers of primes of the form 4k + 1. Prove that D is multiplicatively closed. Prove also that D−1 Z is neither isomorphic to Z n1 for any integer n nor isomorphic to Z(p) for a prime number p. Solution: Let m and n be integers of the form 4k + 1. Then m ≡ 1 (mod 4) and n ≡ 1 (mod 4). Thus mn ≡ 1 (mod 4) so mn has the same form 4k + 1. Consequently, any products of powers of numbers congruent to 1 modulo 4 remains congruent to 1 modulo 4. The described set D is multiplicatively closed. In the ring of fractions Z n1 as a subring of Q, there can be only a finite number of primes appearing in the denominator of a fraction when expressed in reduced form. However, there are an infinite number of primes of the form 4k + 1 so in the denominators of fractions in D−1 Z, even in reduced form, will have an infinite number of primes in their factorization. On the other hand, the ring Z(p) consists of fractions of the form nd such that p | d. Such denominators, even when expressed in reduced form, can include primes of the form 4k + 3. Hence, D−1 Z cannot be equal to Z(p) for any prime p. Exercise: 5 Section 6.2 Question: Let R = Z/12Z and let D = {3, 9}. Determine the number of elements in the ring of fractions D−1 R and exhibit a unique representative for all the fractions in D−1 R. Solution: At the outset, every element of D−1 R is an equivalence class of the form a3 or 9b , where a, b ∈ Z/12Z. However, we note that b 3b 3b = = . 9 27 3

6.2. RINGS OF FRACTIONS

317

Hence every element in D−1 R can be written as a3 with a ∈ Z/12Z. Two elements a3 and 3b in D−1 R are equal if and only if (3a − 3b)u = 0 for some u ∈ {3, 9}. But 3u(a − b) = 0 if and only if a − b is a multiple of 4 (in Z/12Z). Thus every fraction in D−1 R is equivalent to one of the four fractions 1 2 3 0 , , , . 3 3 3 3

Exercise: 6 Section 6.2 a Question: Let R = Z/100Z and let D = {2 | a ≥ 1}. Determine the number of elements in the ring of −1 fractions D R and describe a unique representative for all the fractions in D−1 R. a Solution: In D−1 R, a fraction of the form 25/2 has 25 100 a = a+2 = 0. 2 2 a

Therefore, for any fraction of the from n/2 , if n = 25q + r with 0 ≤ r < 25, then n 25q r r a = a + a = a. 2 2 2 2 Now in Z/100Z, the powers of 2 never reach 1 but instead are eventually periodic. The successive powers of 2 are (with bars removed) 2, 4, 8, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 4, 8, 16, . . . a

So every fraction n/2 , by multiplying top and bottom by an appropriate power of 2, we can rewrite the fraction as n/4 with 0 ≤ n < 25. This expression for the fraction is unique because a b = ⇐⇒ (4a − 4b)2k = 0, 4 4 which is equivalent to a − b is a multiple of 25. With the supposition that 0 ≤ a, b < 25, then −25 < a − b < 25, so a − b = 0 and thus a = b. In particular, D−1 R contains 25 elements. 2 [In fact, it is easiest to see that D−1 R ∼ = Z/25Z by noting first the 76 = 76 in Z/100Z. Thus, if we write every fraction in D−1 R as n/76 with 0 ≤ n < 25, then a b a+b + = 76 76 76

and

a b ab ab · = 2 = . 76 76 76 76

] Exercise: 7 Section 6.2 Question: Prove that if D contains only units, then D−1 R is isomorphic to R. Solution: If a ring has units then it must have a 1. Consider the homomorphism ϕ : R → D−1 R defined in Lemma 6.2.5. Units cannot be zero divisors so ϕ is injective. Now consider any element ur ∈ D−1 R. We can rvu r view the function ϕ as ϕ(x) = xu u . Then if v is the inverse to u, we have ϕ(rv) = u = u . This shows that ϕ is surjective. Consequently, R is isomorphic to D−1 R. Exercise: 8 Section 6.2 Question: Let R be an integral domain and let F be its field of fractions. Prove that as rings of rational expressions F (x) = R(x). Solution: It is obvious that R(x) is a subring of F (x) since R ⊆ F so every rational expression with coefficients in R can be viewed as a rational expression with coefficients in F . Conversely, let p(x) q(x) be a rational expression with p(x), q(x) ∈ F [x]. Suppose that we write p(x) =

am m a1 a0 x + ··· + x + bm b1 b0

and

q(x) =

cn m c1 c0 x + ··· + x + dn d1 d0

318

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

with ai , bi , cj , dj ∈ R. Then since R ⊆ R[x] as constant polynomials, then in F (x) we have p(x) (bn · · · b1 b0 )(dn · · · d1 d0 )p(x) = . q(x) (bn · · · b1 b0 )(dn · · · d1 d0 )q(x) But this product on the numerator and denominator clears all the fractions in the polynomials’ coefficients so (bn · · · b1 b0 )(dn · · · d1 d0 )p(x) ∈ R[x] and (bn · · · b1 b0 )(dn · · · d1 d0 )q(x) ∈ R[x]. Thus, F (x) ⊆ R(x) and therefore these two sets are equal. Exercise: 9 Section 6.2 Question: Let R be a commutative ring and let a ∈ R − {0}. Set D1 = {a, a2 , a3 , . . .}

D2 = {ak , ak+1 , ak+2 , . . .},

and

where k is any positive integer. Prove that D1−1 R is isomorphic to D2−1 R. Solution: Consider the function ϕ : D1−1 R → D2−1 R defined by ϕ

r am

=

rak−1 . am+k−1

We note that since m ≥ 1 for fractions in D1−1 R, then m + k − 1 ≥ k so the images of ϕ have the correct form of an element in D2−1 R. We check that ϕ is a well-defined function. (Recall that we need to do this whenever discussing functions on equivalence classes.) If arm = asn , then (ran − sam )a` = 0 for some ` ≥ 1. Then, by multiplying this expression by a2k−2 , we have (ran+2k−2 − sam+2k−2 )a` = 0 so

rak−1 sak−1 = . am+k−1 an+k−1

Thus, ϕ is a function. To check that ϕ is a ring homomorphism, note that n r r s s rak−1 sak−1 ra + sam (ran + sam )ak−1 ϕ m + n =ϕ = + = ϕ + ϕ = a a am+n am+n+k−1 am+k−1 an+k−1 am an and ϕ

rs r s s rsak−1 rsa2k−2 rak−1 sak−1 × = ϕ = = = × = ϕ × ϕ . am an am+n m+n+k−1 am+n+2k−2 am+k−1 an+k−1 am an

r

k−1

k−1

sa For injectivity, suppose that ϕ(r/am ) = ϕ(s/an ). Then ara m+k−1 = an+k−1 and

(ran+2k−2 − sam+2k−2 )a` = 0 for some ` ≥ k. Then a`+2k−2 (ran − sam ) = 0, which shows that arm = asn . This shows that ϕ is injective. To see that ϕ is surjective, consider asm with m ≥ k. Notice that ϕ(s/am ) =

sak−1 s = m am+k−1 a

because sam+k−1 − (sak−1 )am = 0. Hence, we have established that ϕ is a bijective homomorphism. Exercise: 10 Section 6.2 Question: Let R = Z[x]/(x2h− (mi+ n)x + mn) for some integers m 6= n. Let D = {(x − n)k | k ∈ N}. Prove that D−1 R is isomorphic to Z

1 m−n

.

Solution: In the quotient ring R = Z[x]/(x2 − (m + n)x + mn), we notice that (x − m)(x − n) = 0 so (without the bars over top) the element x − n is a zero divisor. Elements in D−1 R are fractions of the form r/(x − n)k for some r ∈ R or in other words, p(x) . (x − n)k

6.2. RINGS OF FRACTIONS

319

Then we have m (x − m)(x − n) m m x x−m+m x−m = = + = + = . k k k k k+1 k (x − n) (x − n) (x − n) (x − n) (x − n) (x − n) (x − n)k h i 1 This inspires the function f : R → Z m−n defined by f

p(x) (x − n)k

=

p(m) . (m − n)k

Since taking various polynomials p we can make p(m) be any integer, we see that f is surjective. Furthermore, since we just saw that every element in D−1 R can be written as a/(x − n)k for some a ∈ Z and some k ∈ Z, we −1 see that the functions if is injective. It is easy to tell that f is a homomorphism so we conclude that D R is h 1 . isomorphic to Z m−n Exercise: 11 Section 6.2 Question: Let D be a multiplicatively closed subset of a commutative ring R, that does not contain the 0. Suppose that a is a zero divisor in D with ab = 0 for some element b ∈ R. If ϕ : R → D−1 R is the standard mapping of a ring into the ring of fractions D−1 R, show that ϕ(a) is a unit and ϕ(b) = 0. Solution: Note that if D does not contain 0 but does contain a zero divisor a, then a is a zero divisor but not a nilpotent element. 2 0 The element ϕ(a) must be a unit since ϕ(a) = aa with inverse aa2 . However, ϕ(b) = ba a = a so this is the 0 −1 in D R. Exercise: 12 Section 6.2 Question: Prove that the ideals of D−1 R are in bijection with the ideals of R that do not intersect D. Solution: (Recall that rings of fractions are only in the context of commutative rings. Also, recall that some of the proof below is required since we do not assume that the ring has an identity.) Recall the function ϕ : R → D−1 R defined in Lemma 6.2.5. To an ideal J in D−1 R associate the set J 0 ∈ R defined as J 0 = {r ∈ R | ∃d ∈ D,

r ∈ J}. d

Let a, b ∈ J 0 . Then there exists d1 , d2 ∈ D with da1 and db2 in J. Then a d2 d1 ∈J · d1 d1 d22

and

b d2 d1 ∈ J. d2 d21 d2

Taking the difference of the above two elements, we see that a−b ad1 d2 − bd1 d2 = ∈ J. d21 d22 d1 d2 ar Thus a − b ∈ J 0 . Furthermore, for all r ∈ R, the element da1 dr = dd ∈ J so ar ∈ J 0 . Thus, we have shown that 1 J 0 is an ideal. On the other hand, let I be an ideal in R and consider the set D−1 I. Let da1 , db2 ∈ D−1 I with a, b ∈ I. Then

a b d2 a − d1 b − = d1 d2 d1 d2 but d2 a − d1 b ∈ I by properties of ideals so this difference is in D−1 I. Also, for all dr ∈ D−1 R, r a ra = d d1 dd1 but ra ∈ I by properties of ideals. Thus, D−1 I is an ideal in D−1 R. Now if I is an ideal in R such that I ∩ D 6= ∅, then there exists some d ∈ I ∩ D. Then D−1 I would contain the element dd which is the identity so we would have D−1 I = D−1 R.

320

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS Furthermore, for any ideal I in R we have (D−1 I)0 = {r ∈ R | ∃d ∈ D,

r ∈ D−1 I} = I, d

while for any ideal J in D−1 R, we have D−1 (J 0 ) = J. Thus, the described processes are inverses to each other. Since any ideal I that intersects D nontrivially has D−1 I = D−1 R, we see that there is a bijection between proper ideals of D−1 R and ideals of R that do not intersect D and R ↔ D−1 R. Exercise: 13 Section 6.2 Question: Let ϕ : R → D−1 R be the standard homomorphism of a commutative ring into a ring of fractions. Prove that if I is a principal ideal in D−1 R, then ϕ−1 (I) is also a principal ideal. Solution: Let I be a principal ideal in D−1 R with I = ad , where we assume that a is not divisible by any element of d (except possibly a unit). Note that since d2 /d is a unit in D−1 R, the ideal I could just as easily be a d2 generated by ad d = d · d or any unit multiple. Hence, I = (ϕ(a)). 0 0 By definition, ϕ−1 (I) = {r ∈ R | rd d0 ∈ I}, where d is some element of D. 0 ad −1 Note that d0 ∈ I so a ∈ ϕ (I) and therefore (a) ⊆ ϕ−1 (I). Conversely, let r ∈ ϕ−1 (I). Then for any d0 ∈ D ad r0 rd0 = 0 d d d00 0

r for some r0 ∈ R and some d00 ∈ D. Now not all elements of the form ad d d00 are in Im ϕ. This will only happen if 0 00 00 00 00 r = r d for some r ∈ R and some d ∈ D. Then we deduce that r = ar00 so that ϕ−1 (I) ⊆ (a). This shows that ϕ−1 (I) is a principal ideal. Setting

Exercise: 14 Section 6.2 Question: Let R be a commutative ring and let P be a prime ideal. a) Prove that that the set of non-units in RP is the ideal PP . b) Deduce that RP has a unique maximal ideal. Solution: Let R be a commutative ring and let P be a prime ideal. a) The set of units in RP is { rs | r, s ∈ / P }. Hence the set of non-units in RP is I = { rs | s ∈ / P, r ∈ P }. Thus, r a this is an ideal in RP because if s , b ∈ I then r a rb − sa − = s b sb and rb − sa ∈ P because P is an ideal. Also, if rs ∈ RP and ab ∈ I, then rs × ab = ra sb ∈ I because a ∈ P and P is an ideal. Thus I is an ideal and we denote it by PP . b) By a previous exercise, whenever the set of non-units is a ideal, then it is the unique maximal ideal of the ring. Hence RP has a unique maximal ideal. Exercise: 15 Section 6.2 Question: Consider the ring R = R[x, y] and the prime ideal P = (x, y). Prove that the elements in the localization RP are rational expressions of the form r(x, y) 1 + xp(x, y) + yq(x, y)

for p(x, y), q(x, y), r(x, y) ∈ R[x, y].

Solution: The localization RP is the ring of fractions in which the denominator consists of the elements R − P . Elements in R − P are polynomials in x and y that have a nonzero constant term, i.e., have the form c + xp(x, y) + yq(x, y), where c 6= 0. Thus, elements in RP are rational expressions of the form r(x, y) r(x, y)/c = , c + xp(x, y) + yq(x, y) 1 + xp(x, y)/c + yq(x, y)/c which proves the exercise.

6.3. EUCLIDEAN DOMAINS

321

Exercise: 16 Section 6.2 Question: Let F [[x]] be the ring of formal power series with coefficients in a field F . We denote the field of fractions of F [[x]] by F ((x)). Prove that the elements of F ((x)) can be written as X ak xk for some integer N . k≥−N

[Such series are called formal Laurent series and F ((x)) is called the field of formal Laurent series over F . See also Exercise 5.4.29.] Solution: An element in F ((x)) has the form P∞ an xn Pn=0 ∞ n n=0 bn x for an , bn ∈ F for all n ∈ N. Let n = N be the least nonnegative integer such that bn 6= 0. Then setting b0n = bn+N , we have P∞ n n=0 an x P ∞ xN ( n=0 b0n xn ) ∞ X with b00 6= 0. Since F is a field and b00 is nonzero, then it is a unit in F . By Exercise 5.2.19(c), b0n xn is a unit n=0

in F [[x]], so that P∞ n 1 n=0 an x P = N ∞ N 0 n x x ( n=0 bn x )

! X n=0

∞cn xn

=

∞ X

c0n xn ,

n=−N

where c0n = cn+N .

6.3 – Euclidean Domains Exercise: 1 Section 6.3 Question: Perform the Euclidean division for β = 32 + 8i divided by α = 3 + 8i. Solution: This division occurs in the ring of Gaussian integers. We have in C β (32 + 8i)(3 − 8i) 160 232 16 13 = = − i = (2 − 3i) + − i . α 73 73 73 73 73 Since

16 73

1 ≤ 12 and − 13 73 ≤ 2 , the quotient in Z[i] is 2 − 3i and the remainder is

32 + 8i − (3 + 8i)(2 − 3i) = 2 + i. The integer division is therefore 32 + 8i = (3 + 8i)(2 − 3i) + (2 + i). It is clear that 5 = |2 + i|2 < |3 + 8i|2 = 73.

Exercise: 2 Section 6.3 Question: Perform the Euclidean division for β = 719 − 423i divided by α = 24 − 38i. Solution: This division occurs in the ring of Gaussian integers. We have in C β (719 − 423i)(24 + 38i) 33 17 = = + i. α 2020 2 2 Using the methodology spelled out in Example 6.3.2, there are four choices for p and q that are acceptable for the Euclidean division. We take p = 16 and q = 8, so that “the” quotient is 16 + 8i and the remainder is 719 − 423i − (24 − 38i)(16 + 8i) = 31 − 7i. The integer division is therefore 719 − 423i = (24 − 38i)(16 + 8i) + (31 − 7i). It is clear that 1010 = |31 − 7i|2 < |24 − 38i|2 = 2020. Exercise: 3 Section 6.3 Question: Perform the Euclidean Algorithm on β = 24 + 17i and α = 13 − 16i. Deduce the generator d of the ideal I = (α, β). Solution: Let β = 24 + 17i and α = 13 − 16i.

322

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS 8 • β/α = 85 + 121 85 i so (according to the method described in this section) the Gaussian integer quotient is i and the remainder is 8 + 4i: 24 + 17i = (13 − 16i)i + 8 + 4i

• (13 − 16i)/(8 + 4i) = 12 − 49 i. In terms of choosing closest integers, we have a choice in relation to 21 . We take as a Gaussian integer quotient 1 − 2i and the remainder is −3 − 4i: 13 − 16i = (8 + 4i)(1 − 2i) + (−3 − 4i). • (8 + 4i)/(−3 − 4i) = − 85 + 45 i, so the quotient will be −2 + i and the remainder is given by: 8 + 4i = (−3 − 4i)(−2 + i) + (−2 − i). • (−3 − 4i)/(−2 − i) = 2 + i so this is the quotient and there is no remainder. −3 − 4i = (−2 − i)(2 + i). According to the Euclidean Algorithm, the last nonzero remainder in this algorithm, −2 − i is a greatest common divisor of α and β. Hence I = (24 + 17i, 13 − 16i) = (−2 − i) = (2 + i). Exercise: 4 Section 6.3 Question: Perform the Euclidean Algorithm on β = 14 + 23i and α = 42 + 3i. Deduce the generator d of the ideal I = (α, β). Solution: divisions.

We perform the Euclidean Algorithm without describing all the steps in each of the Euclidean

14 + 23i = (42 + 3i)i + (17 − 19i) 42 + 3i = (17 − 19i)(1 + i) + (6 + 5i) 17 − 19i = (6 + 5i)(−3i) + (2 − i) 6 + 5i = (2 − i)(1 + 3i) + 1 2 − i = 1(2 − i) + 0 In particular, the ideal I = (α, β) is all of Z[i] and can be generatd by 1. Exercise: 5 Section 6.3 Question: Perform the Extended Euclidean Algorithm associated to Exercise 6.3.3 to find a linear combination of α and β that give a greatest common divisor of α and β. Solution: To run the Extended Euclidean Algorithm on β and α, we start at the next to last row and proceed as (−2 − i) = (8 + 4i) − (−2 + i)(−3 − 4i) = (8 + 4i) − (−2 + i)[(13 − 16i) − (8 + 4i)(1 − 2i)] = −(−2 + i)(13 − 16i) + (1 + 5i)(8 + 4i) = −(−2 + i)(13 − 16i) + (1 + 5i)[(24 + 17i) − (13 − 16i)i] = (1 + 5i)(24 + 17i) + (7 − 2i)(13 − 16i).

Exercise: 6 Section 6.3 Question: Perform the Extended Euclidean Algorithm associated to Exercise 6.3.4 to find a linear combination of α and β that give a greatest common divisor of α and β.

6.3. EUCLIDEAN DOMAINS Solution: follows:

323

To perform the Extended Euclidean Algorithm, we begin at the penultimate row and proceed as 1 = (6 + 5i) − (1 + 3i)(2 − i) = (6 + 5i) − (1 + 3i)[(17 − 19i) − (−3i)(6 + 5i)] = (−1 − 3i)(17 − 19i) + (10 − 3i)(6 + 5i) = (−1 − 3i)(17 − 19i) + (10 − 3i)[42 + 3i − (1 + i)(17 − 19i)] = (10 − 3i)(42 + 3i) + (−14 − 10i)(17 − 19i) = (10 − 3i)(42 + 3i) + (−14 − 10i)[14 + 23i − i(42 + 3i)] = (−14 − 10i)(14 + 23i) + 11i(42 + 3i).

Exercise: 7 Section 6.3 Question: Perform polynomial division of 4x4 + 3x3 + 2x2 + x + 1 by 2x2 + x + 1 in Q[x]. Solution: The desired polynomial division is + 21 x − 14

2x2 2x2 + x + 1

4x4 +3x3 +2x2 +x −(4x4 +2x3 +2x2 ) x3 −(x3 + 12 x2

+1

+x +1 + 12 x)

.

− 12 x2 + 12 x +1 −(− 21 x2 − 14 x − 14 ) 3 4x

+ 45

Exercise: 8 Section 6.3 Question: Perform polynomial division of 4x4 + 3x3 + 2x2 + x + 1 by 2x2 + x + 1 in F7 [x]. Solution: The desired polynomial division is 2x2 2

2x + x + 1

4

3

4x −(4x4

+3x +2x3 x3 −(x3

+4x

+5

+2x2 +x +2x2 )

+1

+4x

2

3x2 −(3x2

+x +1 +4x) +4x +5x

+1 +5)

6x

+3

Exercise: 9 Section 6.3 Question: Perform polynomial division of 4x4 + 3x3 + 2x2 + x + 1 by 2x2 + x + 1 in F13 [x]. Solution: The desired polynomial division is (with bars over top of elements in F13 omitted) 2x2 2

2x + x + 1

4

3

+7x +3 2

4x +3x +2x +x +1 −(4x4 +2x3 +2x2 ) x3 +x +1 −(x3 +7x2 +7x) 6x2 +7x +1 −(6x2 +3x +3) 4x

+11

.

324

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Exercise: 10 Section 6.3 Question: Use the Euclidean Algorithm in Q[x] to find a generator of the principal ideal I = (2x4 + 7x3 + 4x2 + 13x − 10, 3x4 + 5x3 − 16x2 + 14x − 4).

Solution: To say ourselves some trouble with fractions, we point out that the ideal can be alternately generated by I = (3x4 + 5x3 − 16x2 + 14x − 4, x4 − 2x3 − 20x2 + x + 6) where the second new generator is the difference of the first two generators. We perform the Euclidean Algorithm on this pair of polynomials. We have 3x4 + 5x3 − 16x2 + 14x − 4 = (x4 − 2x3 − 20x2 + x + 6)3 + (11x3 + 44x2 + 11x − 22) 6 1 4 3 2 3 2 x− + (3x2 + 9x − 6) x − 2x − 20x + x + 6 = (11x + 44x + 11x − 22) 11 11 11 11 3 2 2 x+ + 0. 11x + 44x + 11x − 22 = (3x + 9x − 6) 3 3 Thus, the ideal I = (3x2 + 9x − 6) = (x2 + 3x − 2). Exercise: 11 Section 6.3 √ √ Question: Prove that Z[ 2] is a Euclidean domain with the norm N (a + b 2) = |a2 − 2b2 | as the Euclidean function. √ Solution: the strategy in Example 6.3.2 for Gaussian integers to the ring Z[ 2] with the norm √ We compare √ N (a + b 2) = |a2 − 2b2 |√as the Euclidean function. If α, β ∈ Z[ 2], then we consider the division β/α in R and write the result as r√+ s 2 with r, s ∈ Q. We take p to be the closest integer √ to r and q the closest integer to s. The element (p + q 2) will √ be the quotient. Then we take ρ = β − (p + q 2)α, which will be the remainder. √ β If we call θ = α − (p + q 2) = (p − r) + (q − s) 2, then both the coefficients p − r and q − s are between 12 and 12 . Then also θα = ρ so N (ρ) = N (α)N (θ). But N (θ) = |(p − r)2 − 2(q − s)2 |. The largest this expression can get is when p − r = 0 and q − s = 21 . Thus N (θ) ≤ 12 . Consequently, N (ρ) ≤ 12 N (α), so N (ρ) < N (α). √ We also check that since N (γ) ≥ 1 for all γ ∈ Z[ 2], then N (γδ) ≥ N (δ). We have shown that N satisfies the desired conditions for a Euclidean function. Exercise: 12 Section 6.3 √ √ Question: Perform the Euclidean division of β = 10 + 13 2 by α = 2 + 3 2. Solution: In R we have √ √ β (10 + 13 2)(2 − 3 2) 29 2 √ = = + 2. α −14 7 7 We choose p = 4√and q = 0. So the √ quotient of the Euclidean division is 4 and the remainder is 2 + that 2 = N (2 + 2) < 14 = N (2 + 3 2).

√

2. Note

Exercise: 13 Section 6.3 √ √ Question: Perform the Euclidean division of β = 25 − 3 2 by α = −1 + 13 2. Solution: In R we have √ √ β (25 − 3 2)(−1 − 13 2) 53 322 √ = =− + 2. α −337 337 337 √ √ We choose p = 0 and q = the Euclidean division is 2 and the remainder is −1 + 2 2. √ 1. So the quotient of √ Note that 7 = N (−1 + 2 2) < 337 = N (−1 + 13 2). Exercise: 14 Section 6.3 √ √ Question: Prove that Z[ −2] is a Euclidean domain with the norm N (a + b −2) = a2 + 2b2 as the Euclidean function.

6.3. EUCLIDEAN DOMAINS

325

√ Solution: We model the Euclidean division after √ what is done with the Gaussian integers. Let α = a + b −2 √ and let β = c + d −2 6= 0 be two elements in Z[ −2]. Consider the division √ √ √ √ a (a + b −2)(c − d −2 (ac + 2bd) + (bc − ad −2) a + b −2 √ = = . = β c2 + 2d2 c2 + 2d2 c + d −2 √ 2 This is an element of Q[ −2]. Let p the closest + 2d2 ) and let q be the closest integer √ √ integer to (ac + 2bd)/(c 2 2 to (bc − ad)/(c + 2d ) and consider γ = p + q −2. Since γ ∈ Z[ −2], then √ ρ = α − βγ ∈ Z[ −2]. This leads us to an expression of the form α = βγ + ρ in which N (ρ) ≥ 0 and in fact N (ρ) is a positive norm. We need to show that N (ρ) < N (β). √ − γ β. Since this norm also is multiplicative on Q[ −2], then By construction ρ = α − βγ = α β N (ρ) = N

α − γ N (β). β

√ However, γ = p + q −2 was chosen in such a way that the components of α β − γ are at most fractions between − 12 and 21 . Hence 2 2 α 1 1 3 N −γ ≤ +2 = . β 2 2 4 this shows that N (ρ) ≤ 34 N (β) and in particular, N (ρ) < N (β). √ Finally, we know that for all α, β ∈ Z[ −2], N (αβ) = N (α)N (β). Furthermore, since all nonzero elements √ in Z[ −2] have √ a norm greater or equal to 1, we have N (β) ≥ N (αβ). This establishes that N is a Euclidean function on Z[ −2] and proves the exercise. Exercise: 15 Section 6.3 Question: Let R be an integral domain. Prove that R[x] is an Euclidean domain if and only if R is a field. Solution: Let R be an integral domain. Corollary 6.3.6 already established one direction of the desired if and only if statement. Now suppose that R[x] is a Euclidean domain. In particular, R[x] is a PID. However, if R contains a nonzero element a that is not a unit, then (a, x) is not a principal ideal. Thus, in order for R[x] to be a PID, R must be a field. Exercise: 16 Section 6.3 Question: In Q[x], determine the monic greatest common divisor d(x) of a(x) = x4 − 2x + 1 and b(x) = x2 − 3x + 2. Using the Extended Euclidean Algorithm, write d(x) as a Q[x]-linear combination of a(x) and b(x). Solution: The Euclidean Algorithm on a(x) = x4 − 2x + 1 and b(x) = x2 − 3x + 2 gives: x4 − 2x + 1 = (x2 − 3x + 2)(x2 + 3x + 7) + (13x − 13) 2 1 x2 − 3x + 2 = (13x − 13) x− +0 13 13 The monic greatest common divisors of a(x) and b(x) is d(x) = (x − 1). We find a Q[x]-linear combination of a(x) and b(x) for d(x) from just the first line of the algorithm: x−1=

1 4 1 1 1 (x − 2x + 1) − (x2 + 3x + 7)(x2 − 3x + 2) = a(x) − (x2 + 3x + 7)b(x). 13 13 13 13

Exercise: 17 Section 6.3 Question: In F2 [x], determine the greatest common divisor d(x) of a(x) = x4 + x3 + x + 1 and b(x) = x2 + 1. Using the Extended Euclidean Algorithm, write d(x) as a F2 [x]-linear combination of a(x) and b(x).

326

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Solution: The Euclidean Algorithm on a(x) = x4 + x3 + x + 1 and b(x) = x2 + 1 gives: x4 + x3 + x + 1 = (x2 + 1)(x2 + x + 1) + 0 The greatest common divisor is x2 + 1. It is obviously in linear combination of a(x) and b(x) via 0a(x) + 1b(x).

Exercise: 18 Section 6.3 Question: Prove Proposition 6.3.7. Conclude that for all nonzero polynomials p(x) ∈ F [x], the quotient ring F [x]/(p(x)) is either a field or is not an integral domain. Solution: In a PID, an element is irreducible if and only if it is prime. (See Exercise 6.1.17.) Furthermore, with principal ideals in integral domains, an ideal (r) is maximal if and only if r is irreducible and (r) is prime if and only if r is prime. Suppose that p(x) is irreducible. Then (p(x)) is a maximal ideal and thus F [x]/(p(x)) is a field. Conversely, if F [x]/(p(x)) is a field, then (p(x)) is maximal and therefore p(x) is irreducible. Suppose that F [x]/(p(x)) is an integral domain. Then (p(x)) is a prime ideal. Since F [x] is a PID, (p(x)) is therefore a maximal ideal. Hence, the polynomial p(x) is irreducible. Consequently, F [x]/(p(x)) is a field. The result follows. Exercise: 19 Section 6.3 Question: Let R be a Euclidean domain with Euclidean function d. Let n be the least element in the set S = {d(r) | r ∈ R − {0}}. By the well-ordering of Z, S has a least element n. Show that all elements s ∈ R such that d(s) = n are units. Solution: Let a ∈ R − {0}. Then by the property that d(b) ≤ d(ab) for all a, b ∈ R, we see that d(1) ≤ d(1a) = d(a). Hence, d(1) is less than d(a) for all a ∈ R − {0}. Hence, d(1) = n. If u is a unit in R with v ∈ R such that uv = 1, then d(u) ≤ d(uv) = d(1) but since d(1) = n is the least value of d, then d(u) = n. Hence, all units in R satisfy d(u) = n. Conversely, let u ∈ S. Then doing the Euclidean division of 1 by u, we get 1 = uq + r where r = 0 or d(r) < d(u). But since there does not exist r ∈ R such that d(r) < d(u), then we see that r = 0, so that 1 = uq for some q ∈ R. This proves that u is a unit. Exercise: 20 Section 6.3 Question: Let R be a Euclidean domain with Euclidean function d. 1. Prove that in R, any two nonzero elements a and b have a least common multiple. 2. Prove that least common multiples of a and b have the form

ab where d is a greatest common divisor. d

Solution: Let R be a Euclidean domain with Euclidean function d. 1. By the Euclidean algorithm on a Euclidean domain, we know that any two nonzero elements have a greatest common divisor. Furthermore, by Proposition 6.1.16, those two elements must also have a least common multiple. 2. This result is built into proofs of various exercises in Section 6.1.

6.4 – Unique Factorization Domains Exercise: 1 Section 6.4 Question: In the ring Z[i], write a factorization of 11 + 16i. Solution: The quadratic norm of 11 + 16i is N (11 + 16i) = 121 + 256 = 377 = 13 × 29. If 11 + 16i has any nontrivial factors then they would have to have norm 13 or 29. Note that ±2 ± 3i and ±3 ± 2i have norm 13. Now in C, 11 + 16i 70 1 = − i 2 + 3i 13 13

6.4. UNIQUE FACTORIZATION DOMAINS

327

so 2 + 3i is not a factor but

11 + 16i = 5 + 2i. 3 + 2i This last element has norm 29 so it is irreducible. Thus 11 + 16i = (3 + 2i)(5 + 2i) is a factorization into irreducibles. Exercise: 2 Section 6.4 Question: In the ring Z[i], write a factorization of 9 − 19i. Solution: The norm N (a + bi) = a + bi of 9 − 19i is N (9 − 19i) = 442 = 2 × 13 × 17. Consequently, if 9 − 19i factors into α1 α2 · · · αk , then 2 × 13 × 17 = N (α1 )N (α2 ) · · · N (αk ). We see that 9 − 19i can have at most 3 prime (irreducible) factors in Z[i]. We first try to factor out 1 + i since N (1 + i) = 2. We find that 1 + i divides 9 − 19i and 9 − 19i = (1 + i)(−5 − 14i). There are two nonassociate elements in Z[i] of norm 13: 3 + 2i and 3 − 2i. We find that 3 + 2i does not divide −5 − 14i in Z[i] but 3 − 2i does and −5 − 14i = (3 − 2i)(1 − 4i). Note that N (1 − 4i) = 17, which is prime so 1 − 4i is irreducible. Hence a prime factorization is 9 − 19i = (1 + i)(3 − 2i)(1 − 4i).

Exercise: 3 Section 6.4 √ √ Question: Write a factorization of 12 + 3 2 in the UFD Z[ 2]. √ Solution: We had seen in an Hence, it is a PID and therefore √ earlier section√that Z[ 2] is a Euclidean domain. √ a UFD. Then norm of 12 + 3 2 is N (12 + 3 2) = √ |144 − 18| = 126 = 2 × 32 × 7. Irreducible factors of 12 + 3 2 must have a norm that divides 126. We note that 2 has norm 2 which leads us to observe that √ √ √ √ √ 12 + 3 2 = 2(3 + 6 2) = 2 · 3 · (1 + 2 2). √ √ Now N (1 + 2 2) = 7 and since this is prime then 1 + 2 2 is irreducible. On the other hand N (3) = 9 and √ we could possibly look for elements in Z[ 2] of norm 3. We claim that there do not exist elements of norm 3. Consider the expression a2 − 2b2 modulo 8. The squares modulo 8 can only take on three values: 0, 1, and 4. By trying all nine cases for values of a2 and b2 , we see that a2 − 2b2 can take the values 0, 1, 2, 4, 6, 7 but not 3 or 5. We note |a2 − 2b2 | is never 3. In particular, 3 is an irreducible √ the −3 ≡ 5 (mod 8) so we conclude that √ element in Z[ 2]. Thus, the above factorization of 12 + 3 2 is a factorization into irreducibles. Exercise: 4 Section 6.4 √ √ Question: Write a factorization of −10 + 13 2 in the UFD Z[ 2]. √ Solution: Note that N (−10 √+ 13 2) = 238, which has a prime factorization in Z of 238 = 2 × 7 × 17. We first note that we can factor out 2, which has norm 2 √ √ √ −10 + 13 2 = 2(13 − 5 2). √ The elements ±1 ± 2 2 have norm 7. We find that √ 13 − 5 2 33 31 √ √ = − 2, 7 7 1+2 2 √ so 1 + 2 2 is not a factor. However, √ √ 13 − 5 2 √ = −1 + 3 2, 1−2 2 √ √ √ √ which leads to the factorization into irreducibles: −10 + 13 2 = 2(1 − 2)(−1 + 3 2). Exercise: 5 Section 6.4 √ Question: Consider factorizations in Z[ −5].

328

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

√ a) Prove that there is no element α ∈ Z[ −5] such that N (α) = 7, N (α) = 11 or N (α) = 13. √ √ b) Deduce that 7, 11, 3 + −5, 1 + 2 −5 are irreducible elements. √ c) Prove that 9 + 4 −5 is an irreducible element. √ Solution: We work in Z[ −5]. √ √ a) The quadratic norm function N : Z[ −5] → N is defined as N (a + b −5) = a2 + 5b2 . We note first that the norm is unchanged if either a or b changes sign, so we only need to consider √ situations with a, b ≥ 0. We can proceed by cases. We note that if |b| ≥√2 then b2 ≥ 20 and N (a + b −5) ≥ 20, regardless of the value of a. If a ≥ 5, then a2 ≥ 25 so N (a + b −5) ≥ 25. Hence, as we look for elements with various norms, to check if there are any of norm less than 20, we only need to consider elements with 0 ≤ b ≤ 1 and 0 ≤ a ≤ 4. We get √ √ N (0 + √−5) = 5 N (0 + 0√−5) = 0 N (1 + 0√−5) = 1 N (1 + √−5) = 6 N (2 +√ −5) = 9 N (2 + 0√−5) = 4 N (3 + 0√ −5) = 9 N (3 + √−5) = 12 N (4 + 0 −5) = 16 N (4 + −5) = 21 None of these are 7, 11, or 13. Another way to solve this problem consists of looking at the expression a2 + 5b2 = k, with k = 7, 11, or 13, in modular arithmetic for various bases. If k = 7 consider the equation modulo 5: a2 ≡ 2 (mod 5) but this has no solutions. If k = 11, consider the equation modulo 4: a2 + b2 ≡ 3 (mod 4). It is easy to check that this has no solutions (since every square modulo 4 is either 0 or 1). Finally, if k = 13, then consider the equation modulo 5 again: a2 ≡ 3 (mod 5) but this has no solutions. b) We notice that N (7) = 49 but if we could write 7 = αβ, then N (7) = 49 = N (α)N (β). However, there are no elements of norm 7 so N (α) = 1 or N (β) = 1, i.e., α is a unit or β is a unit. This shows that 7 is irreducible. We notice that N (11) = 121 but if we could write 11 = αβ, then N (11) = 121 = N (α)N (β). However, there are no elements of norm 11 so N (α) = 1 or N (β) = 1, i.e., α is a unit or β is a unit. This shows that 11 is irreducible. √ √ Notice √ that N (3 + −5) = 14 = 2 × 7. However, there√are no elements in Z[ −5] of norm 2 or 7, so if 3 + −5 = αβ, then either α or β is a unit. Hence 3 + −5 is irreducible. √ Notice √ finally that N (1 + 2 −5) = 21 and again since there are no elements of order 7, we cannot have 1 + 2 −5 = αβ unless either α or β is a unit. √ c) N (9 + 4 −5) = 161 = 23, there cannot exist one √ 7 × 23. Whether or not there exists an element of norm √ of order 7, so if 9 + 4 −5 = αβ, then either α or β is a unit. Hence 9 + 4 −5 is irreducible. Exercise: 6 Section 6.4 Question: Let R and S be two factorization domains and let f : R−{0} → S −{0} be a surjective multiplicative function (f (ab) = f (a)f (b) for all a, b ∈ R) such that f (1) = 1. 1. Prove that for all a ∈ R, a is a unit in R if and only if f (a) is a unit in S. 2. Prove that if f (a) is irreducible in S, then a is irreducible in R. 3. Find an example in which a is irreducible but f (a) is not. 4. Explain how a prime factorization of f (a) may help in determining a prime factorization of a. Solution: Let R and S be two factorization domains and let f : R−{0} → S −{0} be a surjective multiplicative function (f (ab) = f (a)f (b) for all a, b ∈ R) such that f (1) = 1. 1. Let a be a unit in R. Then there exists b ∈ R with ab = 1. Then 1 = f (1) = f (ab) = f (a)f (b). Thus, f (a) is a unit. Conversely, suppose that f (a) is a unit. Then there exists some s ∈ S with f (a)s = 1. Since f is surjective, there exists some b ∈ R such that f (b) = s. Then 1 = f (a)f (b) = f (ab). Caveat: Just because f (a) is a unit does not imply that a is. For the rest of the exercise, we will generalize and consider multiplicative functions that are not necessarily surjective but satisfy the condition in part (a).

6.4. UNIQUE FACTORIZATION DOMAINS

329

2. Suppose that a is not irreducible in R, with a = xy such that neither x nor y is a unit. Then f (a) = f (x)f (y) with neither f (x) nor f (y) a unit. Thus f (a) is not irreducible. Taking the contrapositive of what we just prove establishes that if f (a) is irreducible, then a is irreducible. 3. We can change the exercise slightly for a more general result and assume that if f is not surjective but that a is a unit if√and only if f (a) is a unit. In this situation, Exercise 6.4.3 gives an example because 3 is irreducible in Z[ 2] but N (3) = 9 is not irreducible in Z. 4. Suppose that a = r1α1 r2α2 · · · r`α` . Then f (a) = f (r1α1 r2α2 · · · r`α` ) = f (r1 )α1 f (r2 )α2 · · · f (r` )α` . Though f (ri ) need not be irreducible, f (ri ) must appear in the prime factorization of f (a) in S. So if we know the prime factorization of f (a), we look for elements ri ∈ R such that f (ri ) is a prime factor or perhaps a product of a few of the prime factors in f (a) and then check if ri divides a. Consequently, if there is a limited number of ri that equal a given element in S, instead of looking through all possible elements of R for factors of a, we only need to look for those that are combinations of prime factors of f (a).

Exercise: 7 Section 6.4 Question: Use the factorizations in (6.9) to show that max(α1 ,β1 ) max(α2 ,β2 ) max(α` ,β` ) r2 · · · r`

m = r1 is a least common multiple of a and b. Solution: We note that

max(α1 ,β1 )−α1 max(α2 ,β2 )−α2 max(α` ,β` )−α` r2 · · · r` a

m = u−1 r1 and

max(α` ,β` )−β` max(α1 ,β1 )−β1 max(α2 ,β2 )−β2 · · · r` b r2

m = v −1 r1

Thus, m is a common multiple of a and b, since all the powers are nonnegative. On the other hand, let m0 be a common multiple to a and b with m0 = pa and m0 = qb. By considering the prime factorizations of m0 = pa and m0 = qb, we see that for all 1 ≤ i ≤ `, the irreducible ri (or an associate thereof) must appear in the prime factorization of m0 . In particular, riαi | m0 and riβi | m0 . Thus, the prime factorization of m0 must contain riγi with γi ≥ αi and γi ≥ β)i, which is equivalent to γi ≥ max(αi , βi ) for all i. Then γ −max(α1 ,β1 ) γ2 −max(α2 ,β2 ) γ −max(α` ,β` ) r2 · · · r` ` m,

m0 = wr1 1

for some unit w. Hence, m | m0 , which shows that m is a least common divisor. Exercise: 8 Section 6.4 Question: Let p be a prime number in Z. Recall that the localization of Z by the prime ideal (p), denoted by Z(p) , consists of all fractions whose denominator is not divisible by p. Discuss what variations can exist in the factorizations of elements in Z(p) . Solution: In Z(p) , the elements are ab where p - b. Therefore, every fraction of the form m n where p does not divide either m or n is a unit. Using the unique factorization of a and b as elements in Z, we see that every fraction ab ∈ Z(p) has the form upk , where u is a unit in Z(p) and k ∈ N. Hence, Z(p) is a UFD but has only one associate of irreducibles, namely p. Exercise: 9 Section 6.4 Question: Consider the quotient ring R = R[x, y]/(x3 − y 5 ). Prove that x̄ and ȳ are irreducible elements in R. Deduce that R is not a UFD. Solution: Since x3 = y 5 in the ring R, then every element p(x, y) ∈ R can be expressed as a polynomial in x and y but only using powers of xk with k ≤ 2. Furthermore, this expression is unique in the following sense.

330

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Suppose that p1 (x, y) and p2 (x, y) are two polynomials with p1 (x, y) = p2 (x, y) and both p1 (x, y) and p2 (x, y) have no terms with a degree higher than 2 on the variable x. Since p1 (x, y) = p2 (x, y), then p1 (x, y) − p2 (x, y) = a(x, y)(x3 − y) for some polynomial a(x, y) ∈ R[x, y]. However, if a(x, y) 6= 0, then considering the term of lowest degree in x in a(x, y), the product a(x, y)(x3 − y 5 ) has a power of x that is greater than 2. Hence, we must have a(x, y) = 0 so p1 (x, y) = p2 (x, y). Now suppose that p(x, y) q(x, y) = x, where p(x, y) and q(x, y) have no terms in xk with k ≥ 3. If we write p(x, y) = p0 (y) + p1 (y)x + p2 (y)x2 and q(x, y) = q0 (y) + q1 (y)x + q2 (y)x2 , where pi (y), qi (y) are polynomials in y. Note, we are viewing R[x, y] as R[y][x]. Then, with all these operations occurring in the quotient ring, and because x3 = y 5 , in R, x = (p0 (y)q0 (y) + y 5 p0 (y)q1 (y) + y 5 p1 (y)q2 (y)) + (p1 (y)q0 (y) + p0 (y)q1 (y) + y 5 p2 (y)q2 (y))x (p2 (y)q0 (y) + p2 (y)q1 (y) + p0 (y)q2 (y))x2 . We note from the term that is constant with respect to x that y 5 | p0 (y)q0 (y). Furthermore, by considering the component in front of the x variable, p0 (y) and q0 (y) cannot both be divisible by y or else, the coefficient polynomial in front of x would be a multiple of y. Hence, either p0 (y) or q0 (y) must have a constant term, while the other is then divisible by y 5 . Without loss of generality, suppose that q0 (y) has a constant term (and that y 5 divides p0 (y)). Since elements in R can be represented in a unique manner by polynomials whose degree in x is less than or equal to 2, the above equation leads to a system  5 5  p0 (y)q0 (y) + y p0 (y)q1 (y) + y p1 (y)q2 (y) = 0 p1 (y)q0 (y) + p0 (y)q1 (y) + y 5 p2 (y)q2 (y) = 1   p2 (y)q0 (y) + p2 (y)q1 (y) + p0 (y)q2 (y) = 0. For fixed q0 (y), q1 (y), q2 (y), using linear algebra we find that (as rational expressions in y) (p0 , p1 , p2 ) =

(y 5 (q12 − q0 q2 ), q02 − y 5 q1 q2 , y 5 q22 − q0 q1 ) . q23 y 10 + q13 y 5 + q03 − 3q0 q1 q2 y 5

In order for these to be polynomials, we need the denominator to be a constant. We note that the constant comes only from the q0 (y)3 term in the denominator since the remaining terms are multiples of y 5 . Assume that q0 (y), q1 (y), and q2 (y) are not all 0. Set deg q2 (y) = a, deg q1 (y) = b, and deg q0 (y) = c. The degrees (in y) of the four terms in q23 y 10 + q13 y 5 + q03 − 3q0 q1 q2 y 5 , are respectively 3a + 10, 3b + 5, 3c, and a + b + c + 5. Note that 3a + 10, 3b + 5 and 3c all have distinct remainders when divides by 3, so they canno be equal. Since this denominator polynomial must have degree 0, in particular the terms of top degree must cancel. Since 3a + 10, 3b + 5 and 3c are all distinct, the two top degree terms must come from −3q0 q1 q2 y 5 and one of the other three terms. This leads to three cases. The first case is when there are two leading terms, one from q23 y 10 and one from −3q0 q1 q2 y 5 . Then 3a + 10 = a + b + c + 5, so 2a = b + c − 5. Furthermore, 3b + 5 ≤ 3a + 10 so b ≤ a + 35 and hence b ≤ a + 1, since a, b, c ∈ N. Also 3c ≤ 3a + 10, which implies that c ≤ a + 3. From these last two inequalities, we deduce that b + c − 4 ≤ 2a, which contradicts 2a = b + c − 5. The three other cases all lead to contradictions and what this contradicts is that q0 (y), q1 (y), q2 (y) are all nonzero. Thus, q1 (y) = 0 or q2 (y) = 0 (since q0 (y) has a nonzero constant term). Then −3q0 q1 q2 y 5 = 0. If q1 (y) = 0, then for the top degrees of the remaining terms in the denominator to cancel, we would need the degrees 3a + 10 = 3c, a contradiction unless q2 (y) = 0. Similarly, if q2 (y) = 0, then q1 (y) = 0. Hence, the denominator reduces to is q0 (y)3 , which then must be a constant. Thus, q(x, y) = c, where c ∈ R∗ , a unit in the polynomial ring. Thus, x is irreducible in R. By a similar reasoning, y is an irreducible element in R. In the quotient ring R, the identity x3 = y 5 gives two non-equivalent factorizations of the same element into irreducible elements. We can tell the factorizations are not equivalent particularly because this involve a different number of irreducible factors. Exercise: 10 Section 6.4 Question: Consider the quotient ring R = R[x, y, z]/(x2 − yz). Prove that x̄, ȳ, and z̄ are irreducible elements in R. Deduce that R is not a UFD.

6.4. UNIQUE FACTORIZATION DOMAINS

331

Solution: Consider the quotient ring R = R[x, y, z]/(x2 − yz). Elements in R have the form a(x, y, z) with a(x, y, z) ∈ R[x, y, z] where x2 = y · z. Because of this relation, the class a(x, y, z) may have many different representatives. However, we notice that the total degree of each term (the sum of all the powers on x, y, and z in any term) remains the same under any change of representative. Then by degree considerations x is the unique polynomial of degree with terms of total degree less than 2 that represents x (and similarly for y and z). Furthermore, x cannot arise as the product of a(x, y, z)) b(x, y, z), where a and b do not have terms of total degree 1 or less. Now assume that x = a1 x + b1 y + c1 z + d1 · a2 x + b2 y + c2 z + d2 . Among other conditions, we have x = (d1 a2 + d2 a1 )x + (d1 b2 + d2 b1 )y + (d1 c2 + d2 c1 )z + d1 d2 which implies that d1 d2 = 0. Assume without loss of generality that d2 = 0. Then x = d1 a2 x + d1 b2 y + d1 c2 z. From this, we cannot have d1 or a2 equal to 0, so then we must have b2 = c2 = 0. Thus, one of the factors of x must be a2 x with a2 6= 0. This is an associate of x so x is irreducible. A similar argument works for y and z. Furthermore, the units in R over c with c a nonzero real number. Thus, since x2 = y z, this expression gives two nonequivalent factorizations of the same element in R. Exercise: 11 Section 6.4 Question: Let F be a field. Show that the subring F [x4 , x2 y, y 2 ] of the polynomial ring F [x, y] is not a UFD. Solution: In the ring F [x4 , x2 y, y 2 ], we have (x2 y)2 = (x4 )(y 2 ). The ring F [x, y] is a UFD so we know the primes factorization of x4 , x2 y, and y 2 as x · x · x · x, x · x · y and y · y. However, the elements x, x2 , x3 , and y are not in F [x4 , x2 y, y 2 ] so anytime x4 = ab, then a or b is a unit and similarly for x2 y and y 2 . Hence, x4 , x2 y and y 2 are irreducible. Thus, (x2 y)2 = (x4 )(y 2 ) gives two inequivalent factorizations of some ring element into irreducible elements and so F [x4 , x2 y, y 2 ] is not a UFD. Exercise: 12 Section 6.4 Question: Let R be a UFD and let p(x) ∈ R[x]. Prove that if p(ci ) = 0 for n distinct constants c1 , c2 , . . . , cn ∈ R, then p(x) = 0 or deg p(x) ≥ n. Conclude that if p(x) is a polynomial that is 0 or of degree less than n that has n distinct roots, then p(x) is the 0 polynomial. Solution: Caveat: This exercise belongs with section 6.5. In fact, this result is proved in Corollary 6.5.10. Exercise: 13 Section 6.4 Question: Let R be a UFD and let D be a multiplicatively closed set that does not contain 0. Prove that the ring of fractions D−1 R is a UFD. Solution: Let R be a UFD and let D be a multiplicatively closed set that does not contain 0. Every element 0 in D−1 R is written as dr with r ∈ R and d ∈ D. An element dr is a unit if and only if there exists some dr 0 such that d1 r r0 =1= d d0 d1 for any d1 ∈ D. Since R is an integral domain, this is equivalent to d1 rr0 = d1 dd0 , which is equivalent to rr0 = dd0 . Hence, r divides an element in D. Conversely, if r divides an element in D, say with rc = d0 , for some c ∈ R and d0 ∈ D, then for any d, d1 ∈ D we have rcdd1 = d0 dd1 =⇒

rcd d1 r cd = =⇒ = 1. dd0 d1 d d0

Thus dr is a unit. We have show that dr is a unit if and only if r divides some element of D. We claim that an element dr is irreducible if and only if r is irreducible and does not divide an element of D. Suppose that r is irreducible and does not divide an element of D. Consider the fraction dr and suppose that r1 r2 r d = d1 d2 . This is equivalent to rd1 d2 = dr1 r2 . Consider the factorization into irreducibles of both sides of this expression. Since the factorization is unique, the irreducible r be an associate to one of the irreducibles in dr1 r2 and hence to one of the irreducible factors of d, r1 , or r2 . Suppose that r | r1 . All the other irreducibles in the

332

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

factorization must divide an element of d. Hence, all the irreducibles in the factorization of r2 divide elements of D. Thus, dr22 is a unit in D−1 R, which shows that dr is an irreducible. Conversely suppose that r is not irreducible or r divides some element of D. If r divides some element of D, then any fraction dr is a unit in D−1 R so is not an irreducible element in D−1 R. Now suppose that r does not divides an element in D but that r = ab with neither a nor b a unit in R, then for any denominator d ∈ D, we have abd a bd r = = . d dd d d Since r does not divide an element in d, then neither do a or b. Thus ad is not a unit and bd d is not a unit either. Thus, dr is not irreducible. This establishes the if and only if claim. Since R is a UFD, then r = ur1 r2 · · · rn , where u is a unit in R and r1 , r2 , . . . , rn are irreducible elements in R. Then dr = ud r1 r2 · · · rn . Without loss of generality, suppose that ri with 1 ≤ i ≤ ` are irreducible factors in r that divide to some element in D. Then r ur1 r2 · · · r` r`+1 d r`+2 d rn d = ··· , d d d d d is a factorization of dr into a unit and irreducible elements in D−1 R. To prove that the factorization into irreducibles is unique, suppose that r0 rm r0 r0 r1 r2 ··· = 10 20 · · · n0 d1 d2 dm d1 d2 dn be two factorizations into irreducible elements in D−1 R. Note that r1 , r2 , . . . , rm , r10 , r20 , . . . , rn0 are irreducible elements in R that do not divide any element of D. Then r1 r2 · · · rm d01 d02 · · · d0n = r10 r20 · · · rn0 d1 d2 · · · dm . We consider the (unique) factorization into irreducible elements of the above expression in R. Whether on the right or the left, the factorizations have the same number of irreducible factors that divide some element of D. These are precisely r1 , r2 , . . . , rm and r10 , r20 , . . . , rn0 since any irreducible factor of d01 d02 · · · d0n or of d1 d2 · · · dm divides some element of D. In particular n = m. Furthermore, there is some permutation π ∈ Sn of the ri and ri0 such that ri0 is an associate of rπ(i) say with ri0 = urπ(i) . But then udπ(i) rπ(i) ri0 = 0 di d0i dπ(i) r0

so the same permutation π maps irreducible elements drii to associates of corresponding di0 . Thus, the factorization i is unique. Exercise: 14 Section 6.4 Question: Let R be an integral domain and let r ∈ R be an irreducible element. Prove that ordr (ab) = ordr (a) + ordr (b) for all a, b ∈ R − {0}. Solution: Let a, b ∈ R, with R an integral domain. Suppose that ordr (a) = α and ordr (b) = β. Then rα | a and rβ | b. Thus by properties of divisibility, rα+β divides ab. This shows that ordr (ab) ≥ ordr (a) + ordr (b). Suppose that a = rα a0 with r - r0 and b = rβ b0 with r - b0 . Then ab = rα+β a0 b0 . In a UFD, since an irreducible is prime, then r - a0 and r - b0 implies that r - a0 b0 . Thus, in a UFD, rα+β+1 - ab, so ordr (ab) = ordr (a) + ordr (b).

Exercise: 15 Section 6.4 Question: Let R and r be as in Exercise 6.4.14. Let (R − {0})/ ' be the set of equivalence classes of associate nonzero elements in R. def

1. Prove that ordr : (R − {0})/ ' → N given by ordr ([a]) = ordr (a) for all a ∈ R is a well-defined function. 2. Prove that ordr : (R − {0})/ ' → N is a monotonic function between the posets ((R − {0})/ ', |) and (N, ≤). Solution: In this exercise, we do not need to assume that R is a UFD. Let r be a fixed irreducible element in R.

6.4. UNIQUE FACTORIZATION DOMAINS

333

a) Suppose that a, a0 ∈ R − {0} are associates to each other with a0 = ua, with u ∈ U (R). Suppose that rα | a, and a0 = ua, then rα | a0 . Conversely, since a = u−1 a0 , if rα | a0 , then by transitivity of divisibility, rα | a. In other words, rα | a ⇐⇒ rα | a0 . Hence, if ordr (a) = α, then rα | a and rα+1 - a, which implies that rα | a0 and rα+1 - a0 . Hence, def

ordr (a0 ) = ordr (a). This shows that ordr : (R − {0})/ ' → N given by ordr ([a]) = ordr (a) is a welldefined function. b) Suppose that a | b so that b = ac. Then, by Exercise 6.4.14, ordr (b) ≥ ordr (a) + ordr (c) ≥ ordr (a). This shows that ordr : (R − {0})/ ' → N is a monotonic function between the posets ((R − {0})/ ', |) and (N, ≤). Exercise: 16 Section 6.4 Question: Let R be a UFD and let a ∈ R − {0}. Let Sa be a set of irreducible elements that divide a and such that no two elements in Sa are associates. Prove that Sa is a finite set and that Y rordr (a) . a' r∈Sa

Solution: Let R be a UFD and let a ∈ R − {0}. We can suppose also that a is not a unit. Let a = r1 , r2 , . . . , rn be a factorization of a into irreducible elements. Consider the equivalence relation ' on the set {r1 , r2 , . . . , rn }. It 0 be a complete list of distinct representatives of the equivalence will possess equivalence classes. Let r10 , r20 , . . . , rm classes occurring in {r1 , r2 , . . . , rn }/ '. By the definition of a UFD, every element in Sa must be an associate to 0 }. Hence, |Sa | = m ≤ n. Thus Sa is a finite set. a unique element in {r10 , r20 , . . . , rm Since in a UFD every pair of nonzero elements has a greatest common divisor, we can use the results of Exercises 6.1.23 to 6.1.28. For all r ∈ Sa , we have rordr (a) | a but rordr (a)+1 - a. Since gcd(r1 , r2 ) ' 1 for all r1 , r2 ∈ Sa , and therefore gcd(r1α1 , r2α2 ) ' 1 for any α1 , α2 ∈ N, then using Exercise 6.1.28, we deduce that Y p= rordr (a) r∈Sa

divides a. Suppose that a = pc. Since all irreducible divisors of a have an associate in Sa , and all powers of such an irreducible r that divide a also divide p, then c has no irreducible divisors. Hence c is a unit and the exercise follows. Exercise: 17 Section 6.4 Question: Let R be an integral domain. Suppose that for all a ∈ R − {0}, any set S of irreducible elements that divide a in which no two elements are associates is finite. Prove that R is a unique factorization domain if and only if for all a ∈ R − {0} and all such sets S, Y a' rordr (a) . r∈S

Solution: Exercise 6.4.16 proved one direction of this exercise. We now prove the converse: Suppose that R is an integral domain such that for all a ∈ R − {0} and subsets Sa as described in the previous exercise, we have Y a' rordr (a) . r∈S

We claim that R must be a UFD. Assume that R is not a unique factorization. A first way R might not be a UFD is if there exists an element a such that a cannot be written as the product of irreducible elements. In this case, it is impossible to have Y a' rordr (a) . r∈S

A second way in which R might not be a UFD is if for some a ∈ R, the element a has non-unique factorization. Suppose that a has a factorization a = ur1α1 r2α2 · · · rnαn where u is a unit and where r1 , r2 , . . . , rn are irreducible

334

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

elements that are not associate to each other. If factorizations into irreducible elements are not unique for a, then either riαi +1 | a or there is another irreducible element s ∈ / {r1 , r2 , . . . , rn } such that s | a. But then either ri or s divide ! 1 Y ordr (a) . r a r∈S

Thus a is not an associate of

Y

rordr (a) .

r∈S

6.5 – Factorization of Polynomials Exercise: 1 Section 6.5 Question: Prove that the following polynomials in Z[x] are irreducible. a) 9x2 − 11x + 1 b) 2x2 + 5x + 7 c) x3 + 4x + 3 Solution: The following polynomials are in Z[x]. If a polynomial p(x) ∈ Z[x] has a root pq in Q, then it has a linear factor of x − pq and by Gauss’ Lemma it has a linear factor of (qx − p). If a polynomial is a quadratic or a cubic in Z[x] then it is irreducible if and only if it doe not have a root in Q. a) If p(x) = 9x2 − 11x + 1, then by the Rational Root Theorem, the possible roots are ±1, ± 31 , ± 19 . We have p(1) = −1,

p(−1) = 21,

p

1 5 =− , 3 3

1 17 p − = , 3 3

p

1 1 =− , 9 9

1 7 p − = . 9 3

There are no rational roots, so p(x) has no linear factors, so since it is a quadratic, it is irreducible. b) If p(x) = 2x2 + 5x + 7, then we √ can also see if p(x)√has rational roots using the quadratic formula. The roots in C of p(x) are 41 (−5 ± 25 − 56) = 14 (−5 ± −31). These are not rational so p(x) has no rational roots, so it has no linear factors. Since it is quadratic, it is irreducible. c) If p(x) = x3 + 4x + 3, then by the Rational Root Theorem, the possible roots are ±1, ±3. We have p(1) = 8,

p(−1) = −2,

p(3) = 42,

p(−3) = −36.

Since p(x) has no rational roots, it has linear factor in Z[x]. Since p(x) is a cubic polynomial, this means that it is irreducible. Exercise: 2 Section 6.5 Question: Prove that the following polynomials in Z[x] are irreducible: a) x3 + 2x2 + 3x + 4 b) x4 + x3 + 1 c) x4 + 3x2 − 6 Solution: Checking reducibility. a) Since p(x) = x3 + 2x2 + 3x + 4 is a cubic polynomial, it is reducible if and only if it has a linear factor, if and only if it has a root in Q. By the Rational Root Theorem, the only possible rational roots of p(x) are ±1, ±2, ±4. It is obvious that none of the positive elements 1, 2, or 4 are roots of p(x). On the other hand, p(−1) = 2, p(−2) = −2, and p(−3) = −14. Hence p(x) has no rational roots and hence is irreducible (since it is a cubic with no roots). b) It is easy to see from the Rational Root Theorem that q(x) = x4 + x3 + 1 has no rational roots. Hence it does not have a linear factor in Z[x]. Suppose that q(x) factors into two quadratics x4 + x3 + 1 = (x2 + ax + b)(x2 + cx + d)

6.5. FACTORIZATION OF POLYNOMIALS

335

then we must have  a+c=1    d + b + ac = 0  bc + ad = 0    bd = 1. The equation only gives two possibilities for the pair (b, d), namely (1, 1) or (−1, −1). We already have c = 1 − a. If (b, d) = (1, 1), then we have 2 + a − a2 = 0 and a + c = 0. It is impossible for a + c = 0 and a + c = 1 so we’ve reached a contradiction. If (b, d) = (−1, −1), then the third equation becomes −a − c = 0 which again contradicts the equation a + c = 1. Hence, q(x) does not factor into two quadratics so is irreducible. c) The polynomial x4 + 3x2 − 6 satisfies the hypotheses of Eisenstein’s Criterion with the prime ideal of (n) = (3) so it is irreducible. Exercise: 3 Section 6.5 Question: For each of the following polynomials in F5 [x], decide if it is irreducible and if it is reducible give a complete factorization. a) x2 + 3x + 4 b) x3 + x2 + 2 c) x4 + 3x3 + x2 + 3 Solution: For this exercise, we do not directly use the rational root theorem to look for roots but we only need to test all the elements in the field since it is finite. a) If p(x) = x2 + 3x + 4, then p(0) = 4,

p(1) = 8 = 3,

p(2) = 14 = 4,

p(3) = 22 = 2,

p(4) = 32 = 2.

Hence, p(x) has no roots in F5 , so it has no linear factors, and since it is a quadratic polynomial, the it is irreducible. b) If p(x) = x3 + x2 + 2, then p(0) = 2,

p(1) = 4,

p(2) = 14 = 4,

p(3) = 38 = 3,

p(4) = 82 = 2.

Hence, p(x) has no roots in F5 , so it has no linear factors, and since it is a cubic polynomial, the it is irreducible. c) If p(x) = x4 + 3x3 + x2 + 3, then p(0) = 3,

p(1) = 8 = 3,

p(2) = 47 = 2,

p(3) = 174 = 4,

p(4) = 466 = 1.

Hence, p(x) has no roots in F5 , so it has no linear factors. If it does factor, then it factors into two quadratic polynomials. Furthermore, without loss of generality, we can assume that both quadratic factors are monic. Thus p(x) = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd. Thus, for equality, we need   a + c = 3  ac + b + d = 1 ad + bc = 0    bd = 3

  a = 3 + 4c  3c + 4c2 + b + d = 1 =⇒ 3d + 4cd + bc = 0    bd = 3.

The condition bd = 3 leads to four possible pairs of (b, d), namely (1, 3), (2, 4), (3, 1), and (4, 2). No matter the value for c, we can solve for a in the first equation. So we only worry about the middle two equations with the 4 possible cases of (b, d). ( ( 3c + 4c2 = 2 2=2 (b, d) = (1, 3) =⇒ =⇒ 4 + 3c = 0 c=2 The we find that a = 1. Hence, p(x) is reducible with p(x) = (x2 + x + 1)(x2 + 2x + 3). Neither of these quadratics will be reducible since otherwise, p(x) would have a root. Hence, this is a complete factorization of p(x) in F5 [x].

336

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Exercise: 4 Section 6.5 Question: For each of the following polynomials in F7 [x], decide if it is irreducible and if it is reducible give a complete factorization. a) x2 + 3x + 4 b) x3 + x2 + 2 c) x4 + 3x3 + x2 + 3 Solution: For this exercise, we do not directly use the rational root theorem to look for roots but we only need to test all the elements in the field F7 since it is finite. a) If p(x) = x2 + 3x + 4, then p(0) = 4,

p(1) = 8 = 1,

p(2) = 14 = 0,

p(3) = 22 = 1,

p(4) = 32 = 2,

p(5) = 44 = 2,

p(6) = 58 = 2.

Hence, p(x) has only one root, namely 2. Since p(x) is a quadratic, this suggests that p(x) = (x + 5)2 . It is easy to verify that this is the case, so this is the factorization into irreducible elements. b) If p(x) = x3 + x2 + 2, then p(0) = 2, p(1) = 4, p(2) = 14 = 0. So we see that p(x) has a factor of x − 2 = x + 5 and we calculate that p(x) = (x + 5)(x2 + 3x + 6). We now test the quadratic q(x) = x2 + 3x + 6 for roots: q(2) = 16 = 2,

q(3) = 24 = 3,

q(4) = 34 = 6,

q(5) = 46 = 4,

q(6) = 60 = 4.

Since q(x) is a quadratic with no roots in F7 , then q(x) is irreducible. Thus p(x) = (x + 5)(x2 + 3x + 6) is a complete factorization into irreducible polynomials. c) If p(x) = x4 + 3x3 + x2 + 3, then p(0) = 3,

p(1) = 8 = 1,

p(2) = 47 = 5,

p(3) = 174 = 6,

p(4) = 466 = 4,

p(5) = 1028 = 6,

p(6) = 1983 = 2.

Since p(x) has no roots in its field, it has no linear factors. Hence, if it does factor, then it would factor into two quadratics. Furthermore, without loss of generality, we can assume that both quadratic factors are monic. Thus p(x) = (x2 + ax + b)(x2 + cx + d) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd. Thus, for equality, we need   a + c = 3  ac + b + d = 1 ad + bc = 0    bd = 3

  a = 3 + 6c  3c + 6c2 + b + d = 1 =⇒ 3d + 6cd + bc = 0    bd = 3.

The condition bd = 3 leads to six possible pairs of (b, d), namely (1, 3), (2, 5), (3, 1), (4, 6), (5, 2), and (6, 4). No matter the value for c, we can solve for a in the first equation. So we only worry about the middle two equations with the 6 possible cases of (b, d). ( ( 2=4 3c + 6c2 = 4 (b, d) = (1, 3) =⇒ =⇒ =⇒ no solutions. 2 + 5c = 0 c=1 ( ( 3c + 6c2 = 1 6=1 (b, d) = (2, 5) =⇒ =⇒ =⇒ no solutions. 1 + 4c = 0 c=5 ( ( 3c + 6c2 = 4 2=4 (b, d) = (3, 1) =⇒ =⇒ =⇒ no solutions. 3 + 2c = 0 c=2 ( ( 3c + 6c2 = 5 2=5 (b, d) = (4, 6) =⇒ =⇒ =⇒ no solutions. 4 + 5c = 0 c=2 ( ( 3c + 6c2 = 1 4=1 (b, d) = (5, 2) =⇒ =⇒ =⇒ no solutions. 6 + 3c = 0 c=5 ( ( 3c + 6c2 = 5 2=5 (b, d) = (6, 4) =⇒ =⇒ =⇒ no solutions. 5 + 2c = 0 c=1

6.5. FACTORIZATION OF POLYNOMIALS

337

Since there are no solutions for any of the above possibilities, we find the p(x) is irreducible in F7 [x]. Exercise: 5 Section 6.5 Question: Let p be a prime number. Prove that in Z[x], the polynomial x3 + nx + p is irreducible for all but at most four values of n. Solution: Let p be a prime number in Z and consider polynomials of the form q(x) = x3 + nx + p. As a cubic, it is irreducible if and only if it does not have a linear factor, if and only if it does not have a root in Q. By the Rational Root Theorem, for any n ∈ Z, the possible roots of q(x) are ±1, ±p. We have q(1) = 1 + p + n,

q(−1) = −1 − n + p,

q(p) = p3 + p + np,

q(−p) = −p3 − np + p.

These will be 0 when n = −1 − p, p − 1, −1 − p2 , 1 − p2 . If n is not one of these four values, then q(x) does not have rational roots, and hence has no linear factors in Z[x], so is irreducible. Exercise: 6 Section 6.5 Question: Prove that in Z[x], the polynomial x3 + px + q, where p and q are odd primes, is irreducible. Solution: As a cubic, f (x) = x3 +px+q is irreducible if and only if it does not have a linear factor. Furthermore, it does not have a linear factor if and only if it does not have a rational root. According to the Rational Root Theorem, the possible rational roots are ±1 and ±q. Plugging these four options into f (x) we have f (1) = p + q + 1,

f (−1) = q − p − 1,

f (q) = q 3 + p + q,

f (−q) = −q 3 − pq + q.

The values f (1) and f (q) are not zero since they are positive; f (−1) 6= 0 because f (−1) is odd; and since p and q are odd, we also see that f (−q) is odd, so can’t be 0. Hence, f (x) has no rational roots and therefore, as a cubicis irreducible. Exercise: 7 Section 6.5 Question: Find all quadratic, cubic, and quartic irreducible polynomials in F2 [x]. Solution: If a polynomial in F2 [x] of degree two or greater has 0 for constant term, then an x factors out and hence is not irreducible. So the only quadratic polynomials that don’t have a 0 in constant term are x2 + 1 and x2 + x + 1. However, x2 + 1 = (x + 1)2 so is reducible but x2 + x + 1 has no roots so (as a quadratic) is irreducible. For cubics, we have four polynomials to consider: f31 (x) = x3 + 1, f32 (x) = x3 + x + 1, f33 (x) = x3 + x2 + 1, and f34 (x) = x3 + x2 + x + 1. We note that f31 (1) = 0, so this is not irreducible. Also f34 (x) = (x + 1)3 . We note that f32 (0) = f32 (1) = f33 (0) = f33 (1) = 1 so x3 + x + 1 and x3 + x2 + 1 has no roots so as cubics are irreducible. To find the irreducible quartics, we simply need to find quartics such that have 1 as a constant term (so 0 is not a root) and have an odd number of nonzero terms (so that 1 is not a root). Among these, we must rule out (x2 + x + 1)2 = x4 + x2 + 1, since this is the only quartic that can be obtained as the product of to irreducible quadratics. This leaves us with the polynomials x4 + x3 + 1,

x4 + x + 1,

x4 + x3 + x2 + x + 1.

Exercise: 8 Section 6.5 Question: Let F be a finite field with q elements. By determining all the reducible polynomials of degree 2, prove that there are 12 (q 2 − q) monic irreducible quadratic polynomials in F [x]. Solution: Since monic quadratic polynomials in F [x] are of the form x2 + ax + b with a, b ∈ F , then there are q 2 monic quadratic polynomials since there are q choices for a and q choices for b. We propose to count the reducible polynomials. Suppose that x2 + ax + b is reducible. Then x2 + ax + b = (x + c)(x + d) for some c, d ∈ F . Since F [x] is a UFD, then this factorization is unique up to reorganization. Also, x + c and x + d are not associates if c 6= d. If we put any total order on the elements of F , then we can assume write the factorization as (x + c)(x + d) with c ≤ d. There are q X k=1

k=

q2 + q 2

338

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

possible pairs (c, d) with c, d ∈ F and c ≤ d. This is the number of distinct reducible polynomials. Hence, the number of irreducible polynomials is q2 −

q2 + q 2q 2 − q 2 − q q2 − q = = . 2 2 2

Exercise: 9 Section 6.5 Question: List all irreducible monic quadratic polynomials in F5 [x]. Solution: From Exercise 6.5.8, there are 10 monic irreducible quadratic polynomials in F5 [x]. We consider polynomials that have a nonzero constant term so that 0 and whose coefficients do not add up to 0 so that 1 is not a root. Then we only need to check if 2, 3, and 4 are roots, though if we are doing computations mentally, it is easiest to remember that 3 = −2 and 4 = −1 in F5 . The polynomials we check are x2 + 1

2 and 3 are roots

2

irreducible

2

x +3

irreducible

2

x +x+1

irreducible

2

irreducible

2

x +x+4

2 is a double root

x2 + 2x + 1

4 is a double root

x +2

x +x+2

2

irreducible

2

irreducible

2

3 and 4 are roots

2

irreducible

2

x + 3x + 4

irreducible

x2 + 4x + 1

irreducible

x + 2x + 3 x + 2x + 4 x + 3x + 2 x + 3x + 3

2

irreducible

2

2 and 4 are roots

2

3 is a double root

x + 4x + 2 x + 4x + 3 x + 4x + 4

We have found the 10 irreducible monic quadratic polynomials in F5 [x]. Exercise: 10 Section 6.5 Question: Prove Corollary 6.5.3. Solution: Let R be a UFD and let F be its field of fractions. Let p(x) ∈ R[x] be such that its coefficients have a greatest common divisor of 1. If p(x) is irreducible as an element in F [x], then it is obviously irreducible as an element of R[x]. The converse requires Gauss’ Lemma. Suppose that p(x) is reducible in F [x]. Then p(x) = A(x)B(x) in F [x]. By Gauss’ Lemma, there exist u, v ∈ F such that a(x) = uA(x) ∈ R[x] and b(x) = vB(x) ∈ R[x] with p(x) = a(x)b(x). Furthermore, since the greatest common divisor of p(x) is 1, we cannot factor out a constant polynomial. Hence, neither a(x) nor b(x) is constant and in particular, neither is a unit. Thus p(x) is reducible in R[x]. This proves that p(x) is irreducible in F [x] if and only if it is irreducible in R[x]. Every monic polynomial is such that its coefficients have a greatest common divisor of 1 so this result applies directly. Exercise: 11 Section 6.5 Question: Let F be a field and let a be a nonzero element in F . Prove that if f (ax) is irreducible, then f (x) is irreducible.

6.5. FACTORIZATION OF POLYNOMIALS

339

Solution: Suppose that f (x) is reducible in F [x] with f (x) = p(x)q(x), with neither p(x) nor q(x) a constant polynomial. Then f (ax) = p(ax)q(ax) where neither p(ax) nor q(ax) is a constant polynomial. Thus f (ax) is reducible. Taking the contrapositive of what we just proved, if f (ax) is irreducible, then f (x) is irreducible. Exercise: 12 Section 6.5 Question: Prove a modification of Proposition 6.5.17 in which I is a prime ideal P in R and the polynomial p(x) satisfies LC(p(x)) ∈ / P. Solution: Let R be an integral domain. Suppose that I = P is a prime ideal and let p(x) ∈ R[x] such that LC(p(x)) ∈ / P. Suppose that p(x) is reducible in R[x] with p(x) = a(x)b(x), where a(x) and b(x) are non constant polynomials in R[x]. Consider the reduction homomorphism ϕ : R[x] → (R/P )[x]. Then ϕ(p(x)) = ϕ(a(x))ϕ(b(x)). However, since LC(p(x)) = LC(a(x))LC(b(x)) and simply by virtue of P being an ideal, we deduce that neither LC(a(x)) nor LC(b(x)) is in P . Thus deg ϕ(a(x)) = deg a(x) ≥ 1 and deg ϕ(b(x)) = deg b(x) ≥ 1. Since P is a prime ideal, then R/P is an integral domain, so by Proposition 5.2.6, ϕ(a(x)) and ϕ(b(x)) are not units since they each have a degree greater or equal to 1. Thus ϕ(p(x)) is reducible. The contrapositive of the implication statement that we just proved is the if ϕ(p(x)) is irreducible, then p(x) does not factor into polynomials each of degree greater than 0. We can restate what we have proven in the following way. Let P be a proper prime ideal in the integral domain R and let p(x) be a nonconstant polynomial in R[x] whose leading coefficient is not in P . If the image of p(x) in (R/P )[x] under the reduction homomorphism is an irreducible element, then p(x) = dq(x) where d is constant and q(x) is irreducible in R[x]. Exercise: 13 Section 6.5 Question: Prove that f (x) = x3 + (2 + i)x + (1 + i) is irreducible in Z[i][x]. Solution: The polynomial f (x) = x3 + (2 + i)x + (1 + i) is a cubic in Z[i][x] so by Proposition 6.5.12, it is reducible if and only if it has a root in Q[i], the field of fractions of Z[i]. But Z[i] is UFD so we can use the Rational Root Theorem, to determine the possible roots. The four units in Z[i] are ±1 and ±i and 1 + i is irreducible, to the other four possible roots to f (x) are 1 + i, −1 − i, −1 + i, and 1 − i. We calculate f (1) = 4 + 2i f (1 + i) = 6i

f (−1) = −2

f (i) = 2i

f (−1 − i) = 2 − 4if (−1 + i) = 4i

f (−i) = 2 f (1 − i) = 2 − 2i.

Since none of these are 0, we deduce that f (x) has no roots in Q[i] and hence is irreducible in Z[i][x]. Exercise: 14 Section 6.5 Question: Let R be a UFD and let a ∈ R. Then p(x) ∈ R[x] is irreducible if and only if p(x + a) is irreducible. Solution: Now if q(x) = p(x + a), then p(x) = q(x − a). Thus, in order to prove that for arbitrary a ∈ R, p(x) ∈ R[x] is irreducible if and only if p(x + a) is irreducible, we only need to prove one direction of the equivalence. Suppose that p(x) is reducible with p(x) = p1 (x)p2 (x) with neither q(x) not r(x) units in R[x]. Then p(x + a) = p1 (x + a)p2 (x + a). Since R is an integral domain, units in R[x] must be constant polynomials, which remain unchanged under a shift of variable. So if pi (x) is not a constant, then pi (x + a) is not a constant and not a unit; but if pi (x) is a constant but not a unit, then pi (x + a) = pi (x) is still not a unit. Thus pi (x + a) for i = 1, 2 is not a unit so p(x + a) is reducible. We have proven that if p(x + a) is irreducible, then so is p(x). Exercise: 15 Section 6.5 Question: Let p be a prime number in Z. Use Exercise 6.5.14 to prove that the polynomial xp−1 + xp−2 + · · · + x + 1 is irreducible in Z[x]. Solution: Let f (x) = xp−1 + xp−2 + · · · + x + 1. Then we note that f (x)(x − 1) = xp − 1. Thus f (x + 1)x = (x + 1)p − 1 so ! p p X p k−1 1 X p k x = x . f (x) = x k k k=1 k=1 However, since p is a prime, by Exercise 2.2.23, kp ≡ 0 (mod p) for 1 ≤ k ≤ p − 1. We also note that the constant term of f (x + 1) is p1 = p, which is not divisible by p2 . Thus, we can apply Eisenstein’s Criterion to

340

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

deduce that f (x + 1) is irreducible, by which we conclude that f (x) is irreducible using the result of Exercise 6.5.14. Exercise: 16 Section 6.5 Question: Consider the polynomial p(x) = x3 + 3x2 + 5x + 5 in Z[x]. Find a shift of the variable x so that you can then use Eisenstein’s Criterion to show that p(x) is irreducible. Solution: We can try a few shifts. If we try x = y − 1, then q(y) = p(y − 1) = (y − 1)3 + 3(y − 1)2 + 5(y − 1) + 5 = y 3 − 3y 2 + 3y − 1 + 3y 2 − 6y + 3 + 5y − 5 + 5 = y 3 + 2y + 2. This last polynomial is irreducible from Eisenstein’s Criterion with the prime ideal P = (2). Exercise: 17 Section 6.5 Question: Let c1 , c2 , . . . , cn ∈ Z be distinct integers. Consider the polynomial p(x) = (x − c1 )(x − c2 ) · · · (x − cn ) − 1 in Z[x]. a) Prove that if p(x) = a(x)b(x), then a(x) + b(x) evaluates to 0 at ci for i = 1, 2, . . . , n. b) Deduce that if a(x) and b(x) are nonconstant, then a(x) + b(x) is the 0 polynomial in Z[x]. c) Deduce that p(x) is irreducible. Solution: Let c1 , c2 , . . . , cn ∈ Z be distinct integers. Consider the polynomial p(x) = (x − c1 )(x − c2 ) · · · (x − cn ) − 1 in Z[x]. a) Assume that p(x) = a(x)b(x). Since p(ci ) = −1 for i = 1, 2, . . . , n, then a(ci )b(ci ) = −1. The only way for two integers to multiply together and get −1 is if one of them is 1 and the other is −1. Hence, a(ci ) + b(ci ) = 0 for all i. b) If a(x) and b(x) are nonconstant polynomials, then the degrees of a(x) and b(x) are between 1 and n − 1. Thus, deg(a(x) + b(x)) ≤ max{deg a(x), deg b(x)} ≤ n − 1. However, a(x) + b(x) evaluates to 0 at n distinct values in Z. Using Corollary 6.5.10, we deduce that a(x) + b(x) is the 0 polynomial. c) If a(x) + b(x) = 0 for all x, then b(x) = −a(x). Thus p(x) = −a(x)2 , which means that p(x) evaluates to a nonpositive number at all integers. However, if cn is the maximum of the ci , then p(cn + 2) ≥ (n + 1)! − 1, which is a positive number for all n ≥ 1. This arrives at a contradiction so we deduce that p(x) is irreducible. Exercise: 18 Section 6.5 Question: Prove the following generalization of Eisenstein’s Criterion. Let P be a prime ideal in an integral domain R and let f (x) = an xn + an−1 xn−1 + · · · a1 x + a0 be a polynomial in R[x]. Suppose that: (1) an ∈ / P ; (2) ai ∈ P for all i < n; and (3) a0 ∈ / P 2 . Then f (x) is not the product of two nonconstant polynomials. Solution: As stated, the exercise needs some adjustments as we will see. We work through the three paragraphs of the proof of Eisenstein’s Criterion. With the change of conditions, the first paragraph remains unchanged. Furthermore, in the second paragraph, f (x) = cxn , where c 6= 0 in R/P since LC(f (x)) ∈ / P . The proof that all the nonleading terms of a(x) and b(x) are in P remains the same. For the third paragraph, we only have that the leading terms of a(x) and b(x) are not in P . Hence, if we assume that both a(x) and b(x) are nonconstant polynomials, then they will have a leading term in R − P and a constant term in P . But then the constant terms of a(x)b(x) is in P 2 , which contradicts the hypotheses of the

6.5. FACTORIZATION OF POLYNOMIALS

341

proposition. Hence, one of the polynomials a(x) or b(x) is a constant polynomial. Without loss of generality, suppose that b(x) is a constant with b(x) = B. Then f (x) factors into f (x) = a(x)B. If we begin with the supposition that R is a unique factorization domain and that f (x) has a content of 1, i.e., that the greatest common divisor of its coefficients is 1, then we can deduce that B must be a unit and this would show that f (x) is irreducible. Exercise: 19 Section 6.5 Question: Let R be a UFD and let F be its field of fractions. Suppose that p(x), q(x) ∈ F [x] and that p(x)q(x) is in the subring R[x]. Prove that the product of any coefficient of p(x) with any coefficient of q(x) is an element of R. Solution: According to Gauss’ Lemma, there are elements u, v ∈ F such that P (x) = up(x) and Q(x) = vq(x) are polynomials in R[x] such that p(x)q(x) = P (x)Q(x). Obviously, uv = 1. If we write p(x) = pm xm + · · · + p1 x + p0

and q(x) = qn xn + · · · + q1 x + q0

then pj qk = u−1 v −1 Pj Qk = Pj Qk and since Pj , Qk ∈ R, we see that pj qk ∈ R. Exercise: 20 Section 6.5 Question: Let F be a finite field of order |F | = q and let p(x) ∈ F [x] with deg p(x) = n. Prove that F [x]/(p(x)) has q n elements. Solution: Suppose that p(x) = an xn + · · · + a1 x + a0 . Consider a polynomial a(x) ∈ F [x]. By performing polynomial division of a(x) by p(x), we see that a(x) = p(x)Q(x) + r(x), where r(x) = 0 or deg r(x) < deg p(x). Then in F [x]/(p(x)), we have a(x) = r(x) so coset in F [x]/(p(x)) can be expressed using a representative of degree less than n. Furthermore, suppose that r1 (x) and r2 (x) have degree less than n and r1 (x) = r2 (x). Then r2 (x) − r1 (x) ∈ (p(x)) so r2 (x) − r1 (x) = p(x)Q(x). If q(x) 6= 0, then deg p(x)Q(x) ≥ deg p(x) ≥ n, whereas deg r2 (x) − r1 (x) < n. Hence, Q(x) = 0 and r1 (x) = r2 (x) and thus every coset in F [x]/(p(x)) can be expressed uniquely using a representative of degree less than n. So the elements of F [x]/(p(x)) are an−1 xn−1 + · · · + a1 x + a0 . Since there are q options for each of the n elements a0 , a1 , . . . , an−1 , there are a total of q n elements in F [x]/(p(x)).

Exercise: 21 Section 6.5 Question: Prove that for all primes p, there is a field with p2 elements. Solution: We know that if F is a field, so that F [x] is a PID, then the ideal (p(x)) is a maximal ideal if and only if p(x) is irreducible. Using Exercise 6.5.21, we can construct a field of p2 elements by finding an irreducible quadratic polynomial f (x) in Fp [x]. By Exercise 6.5.8, there are 12 (p2 − p) > 0 irreducible quadratic monic polynomials in Fp [x]. In particular, for all primes p there does exist an irreducible quadratic polynomial. Hence, for all primes p there exists a field of p2 elements. Exercise: 22 Section 6.5 Question: Let p(x) = x4 + Ax3 + Bx2 + Cx + D be a monic quartic (degree 4) polynomial in Z[x]. a) Suppose that p(x) factors into two quadratics p(x) = (x2 + ax + b)(x2 + cx + d). Prove that if a = 0 or c = 0, then ABC = A2 D + C 2 . b) Prove that if p(x) factors into two quadratics and if ABC 6= A2 D + C 2 , then there exists an integer a 6= 0 such that C 2 − 4aD(A − a) is a square integer. c) Deduce that if D < 0, there are only a finite number of possibilities for a. d) Use the previous two parts to prove that p(x) = x4 + 3x3 − 2x − 7 is irreducible in Z[x]. Solution: Let p(x) = x4 + Ax3 + Bx2 + Cx + D be a monic quartic (degree 4) polynomial in Z[x]. a) Suppose that p(x) factors into two quadratics p(x) = (x2 + ax + b)(x2 + cx + d) with a = 0 or c = 0. Without loss of generality, suppose that a = 0. Then  c=A    b + d = B p(x) = x4 + cx3 + (b + d)x2 + bcx + bd =⇒  bc = C    bd = D.

342

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Then ABC = bc2 (b + d) = (bc)2 + bc2 d = C 2 + A2 D. b) Suppose that p(x) factors into two quadratics and ABC 6= A2 D + C 2 . Then  a+c=A    ac + b + d = B p(x) = x4 + (a + c)x3 + (ac + b + d)x2 + (ad + bc)x + bd =⇒  ad + bc = C    bd = D. This leads to   c = A − a  a(A − a)b + b2 + db = Bb adb + b2 (A − a) = Cb    bd = D

  c = A − a  bd = D =⇒ a(A − a)b + b2 + D = Bb    aD + b2 (A − a) = Cb

We consider the last equation as a quadratic equation in b. Note that A−a 6= 0 because otherwise, we would 2 2 have c = 0 and we would be in case p of the first part of the exercise where ABC = ApD + C . The quadratic 1 formula gives b = 2(A−a) (C ± c2 − 4aD(A − a)). Since b ∈ Z we must have C 2 − 4aD(A − a) ∈ Z (though that is not sufficient) for some a ∈ Z. c) Supposing we are still in the situation of part (c) but with D < 0, then C 2 − 4aD(A − a) = 4Da2 − 4ADa + C 2 = h(a) is a quadratic function in a such that h(0) = C 2 ≥ 0 and lima→±∞ h(a) = −∞. Thus, by the Intermediate Value Theorem, there are values a1 < 0 and a2 > 0 such that h(a) ≥ 0 is and only if a1 ≤ a ≤ a2 . Since there are only a finite number of integers in [a1 , a2 ], there are only a finite number of integers a for which h(a) is positive, and therefore, only a finite number of integers a for which h(a) is a square integer. d) Consider p(x) = x4 + 3x3 − 2x − 7 ∈ Z[x]. We first must test for roots using the Rational Root Theorem. The possibilities are ±1 and ±7. A quick check shows that none of these are roots. Hence, p(x) does not have a linear factor. As we decide whether p(x) is the product of two quadratics, we first note that ABC = 0 6= A2 D + C 2 = −59. Hence, if it is the product of two quadratics, both have a nontrivial x term. Furthermore, h(a) = C 2 − 4aD(A − a) = −28a2 + 84a + 4 has zeros at 21 ±

√

212 + 28 ≈ −0.038, 3.0468. 14

The integers a = 0, 1, 2, 3 are the only integers for which h(a) is positive and hence the only integers for which h(a) can possibly be a square integer. We rule out a = 0 since ABC 6= A2 D + C 2 but we also rule out a = 3 because is a = 3, then c = 0. So we only test a = 1 and a = 2. We have h(1) = 60 and h(2) = 60. Since neither of these are squares, we deduce that p(x) is irreducible.

6.6 – RSA Cryptography Exercise: 1 Section 6.6 Question: Implementing RSA over Z, take the role of Alice. Bob sends you n = 28852217 and e = 33. Suppose that you wish to send the plaintext number of m = 45678 to Bob. Calculate the corresponding ciphertext c = me in Z/nZ. Solution: Whether we have a computer that can calculate the corresponding power or perform the fast exponentiation algorithm, the result is c = me = 25318614. Exercise: 2 Section 6.6 Question: Implementing RSA over Z, take the role of Alice. Bob sends you n = 5352499 and e = 451. Suppose that you wish to send the plaintext number of m = 87542 to Bob. Calculate the corresponding ciphertext c = me in Z/nZ. Solution: Whether we have a computer that can calculate the corresponding power or perform the fast exponentiation algorithm, the result is c = me = 5131826.

6.6. RSA CRYPTOGRAPHY

343

Exercise: 3 Section 6.6 Question: Implementing RSA over Z, take the role of Alice. Bob sends you n = 5352499 and e = 451. Suppose that you wish to send the message “FLY AT NIGHT” to Bob using the string-to-number sequence encoding as given in Example 6.6.1. Show that you can use 3 blocks of 4 letters. Then compute the three ciphertext numbers in Z/nZ. Solution: Using the string encoding process as in Example 6.6.1, the sentence “FLY AT NIGHT” becomes the string of integers m1 = 6+12×29+25×292 +0×293 = 21379,

m2 = 1+20×29+0×292 +14×293 = 342027,

m3 = 9+7×29+8×292 +20×293 =

The ciphertext string is then, c = (me1 , me2 , me3 ) = (1717189, 439405, 3169133).

Exercise: 4 Section 6.6 Question: Implementing RSA over Z, take the role of Bob. You select p = 131 and q = 211 so that n = 27641. 1. Show that e = 191 is an acceptable encryption key for the RSA algorithm. 2. Find (expressed as 0 ≤ d < (p − 1)(q − 1) = 27300) the multiplicative inverse of e in Z/27300Z. 3. Suppose that Alice sends a message using 29 characters as in Example 6.6.1, also compressing strings of 3 letters in elements in Z/27641Z as in the Example. From the following ciphertext (26799, 26841, 22169, 9764, 3426) recover the plaintext and interpret the result into English. Solution: Implementing RSA over Z, take the role of Bob. You select p = 131 and q = 211 so that n = 27641. 1. Using e = 191 as an encryption key, we need to just check that gcd(e, (p − 1)(q − 1)) = 1 just so that it can have a decryption key. We have (p − 1)(q − 1) = 27300. Noticing that 191 is prime, makes it easy to tell even without a computer that gcd(e, (p − 1)(q − 1)) = 1. 2. To find the multiplicative inverse of e modulo 27300, we must perform the Extended Euclidean Algorithm. The Euclidean Algorithm goes as 27300 = 191 × 142 + 178 191 = 178 × 1 + 13 178 = 13 × 13 + 9 13 = 9 × 1 + 4 9 = 4 × 2 + 1. This is obviously the last line of the Euclidean Algorithm and shows that gcd(27300, 191) = 1. Reversing the process, we have 1=9−2×4 = 9 − 2 × (13 − 9) = −2 × 13 + 3 × 9 = −2 × 13 + 3 × (178 − 13 × 13) = 3 × 178 − 41 × 13 = 3 × 178 − 41 × (191 − 178) = −41 × 191 + 44 × 178 = −41 × 191 + 44 × (27300 − 142 × 191) = 44 × 27300 − 6289 × 191. From here, taking the final expression modulo 27300, we see that 191

−1

= −6289 = 21011.

344

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

3. Using d, given the ciphertext as shown, we can recover the plaintext as elements of Z/27641Z, (2679921011 , 2684121011 , 2216921011 , 976421011 , 342621011 ) = (4369, 13340, 10113, 4786, 801). We need to make these correspond to strings of 3 characters. Note that 293 = 24389 < 27641 so we have not lost information in performing the compression into groups of 3 characters. We have 4369 = 19 + 5 × 29 + 5 × 292 13340 = 0 + 25 × 29 + 15 × 292 10113 = 21 + 0 × 29 + 12 × 292 4786 = 1 + 20 × 29 + 5 × 292 801 = 18 + 27 × 29 + 0 × 292

→ “SEE” → “ YO” → “U L” → “ATE” → “R. ”

The plaintext is “SEE YOU LATER. “ Exercise: 5 Section 6.6 Question: Let F be a finite field. Prove that for every q(x) ∈ F [x], the order of the quotient ring F [x]/(q(x)) is |F |deg q(x) . Solution: See the solution for Exercise 6.5.20. Exercise: 6 Section 6.6 Question: Consider the Euclidean domain Z[i] and let I = (z) be a proper nontrivial ideal. (Since Z[i] is a Euclidean domain, it is a PID.) 1. Prove that {m + ni | m, n ∈ Z with 0 ≤ m, n < |z|} is a complete set of distinct representatives for the cosets in Z[i]/I. 2. Conclude that the quotient ring contains |Z[i]/I| = zz = |z|2 elements. Solution: Consider the Euclidean domain Z[i] and let I = (z) be a proper nontrivial ideal. 1. As vectors, the complex numbers z and iz form the edges of a square based at the origin with vertices 0, z, (1 + i)z, and iz. Then w, w0 ∈ Z[i] are congruent modulo (z) if and only if there exists (a + bi) ∈ Z[i] with w = (a + bi)z + w0 . So using {z, iz} as a basis of C, for all w ∈ Z[i] integers a and b can be chosen so that w − (az + biz) = pz + qiz with 0 ≤ p, q < 1. Thus, every element in Z[i] is congruent modulo (z) to an element in the square (p + qi)z with 0 ≤ p, q < 1. This square has side length |z|. We now consider the geometric process of rotating this square so that it lies in the first quadrant. Geometrically, the resulting square would consist of {m + ni | m, n ∈ Z with 0 ≤ m, n < |z|}. Note that if one edge of this square is opn, the opposite edge is closed. We wish to count how many elements in Z[i] are in the square and how this number may depend on rotation. While the elements in Z[i] are discrete, rotation is a continuous process. Consequently, as an edge of the square moves, it will pass through elements of Z[i] only at discrete values αi . Now if, as one edge moves through some αi it adds k elements of Z[i] into the square, then k elements on the opposite side leave the square. Consequently, there are the same number of elements in {m + ni | m, n ∈ Z with 0 ≤ m, n < |z|} as in the square (p + qi)z with 0 ≤ p, q < 1. 2. According to Pick’s theorem, the area |z|2 of the square pz + qiz with 0 ≤ p, q <≤ 1 is equal J + B2 − 1, where J is the number of point of Z[i] interior to the square and B is the number of Z[i] points on the boundary. Note that the four corners are all points in Z[i]. Since the elements in Z[i]/(z) are in bijective correspondence with the elements in the half open square pz + qiz with 0 ≤ p, q <≤ 1, then the number of points in Z[i]/(z) consist of the interior points, J; half of the boundary points that are not corners, 1 2 (B − 4); and one point for the corner at 0. Hence, |Z[i]/(z)| = J +

B B − 2 + 1 = J + − 1 = |z|2 . 2 2

Exercise: 7 Section 6.6 Question: Implementing RSA over Z[i], you take the role of Alice. Bob sends you n = 236 − 325i and e = 14.

6.6. RSA CRYPTOGRAPHY

345

1. What is the ciphertext c for the message m = 101 + 3i? 2. Suppose that you wish to send the message “FLY AT NIGHT” to Bob. You agree with Bob to convert strings of letters to strings of Gaussian integers by first mapping the characters in { , A,B,C, . . . , Y,Z,0 ,0 ,0 .0 } to the integers 0 through 28, and then mapping any pair of such integers (a, b) to a + bi. Convert the English message to strings of Gaussian integers and determine the string of ciphertext Gaussian integers. Solution: Implementing RSA over Z[i], you take the role of Alice. Bob sends you n = 236 − 325i and e = 14. 1. We can use the fast modular exponentiation algorithm and repeatedly do Gaussian integer division on the result to remain in Z[i]/(n). We get c = me = −76 − 204i in Z[i]/(n). 2. The plaintext becomes a string of Gaussian integers 6 + 12i,

25 + 0i,

1 + 20i,

14i,

9 + 7i,

8 + 20i.

The ciphertext is −13 − 182i,

−28 + 204i,

−63 − 143i,

12 − 75i,

−105 + 96i,

−141 + 88i.

Exercise: 8 Section 6.6 Question: Implementing RSA over F31 [x], you take the role of Bob. Suppose that you pick polynomials f (x) = x3 + x + 12 and g(x) = x2 + 4x2 + 8. 1. First prove that f (x) and g(x) are irreducible. 2. Prove that any number e that is relatively prime to 2, 3, 5, and 331 is a suitable encryption key. 3. Find the decryption key d given the encryption key e = 17. Solution: Implementing RSA over F31 [x], you take the role of Bob. Suppose that you pick polynomials f (x) = x3 + x + 12 and g(x) = x2 + 4x2 + 8. 1. Since f (x) and g(x) are both cubic or quadratic, to prove that they are irreducible simply amounts to checking that they have no roots. This is easily done so it is omitted here. 2. Since (f (x)) and (g(x)) are maximal ideals and also comaximal, then by the Chinese Remainder Theorem, F31 [x]/(f (x)g(x)) ∼ = F31 [x]/(f (x)) ⊕ F31 [x]/(g(x)). Thus,

G = U (F31 [x]/(f (x)g(x))) ∼ = U (F31 [x]/(f (x))) ⊕ U (F31 [x]/(g(x)))

so the order of the group is |U (F31 [x]/(f (x)g(x)))| = (313 − 1)(312 − 1) = 28598400 = 27 · 33 · 52 · 331. Since we need e to be relatively prime to this, we simply need e not to contain the prime divisors 2, 3, 5 and 331. 3. We use e = 17. For a decryption key, we need d to be such that ed ∼ = 1 (mod |G|). Bob needs to calculate d using the Extended Euclidean Algorithm, where |G| = 28598400. We have 28598400 = 17 × 1682258 + 14 17 = 14 × 1 + 3 14 = 3 × 4 + 2 3 = 2 × 1 + 1. Reversing the process, we have, 1=3−1×2 = 3 − (14 − 4 × 3) = −1 × 14 + 5 × 3 = −1 × 14 + 5 × (17 − 14) = 5 × 17 − 6 × 14 = 5 × 17 − 6 × (28598400 − 1682258 × 17) = −6 × 28598400 + 10093553 × 17. Taking this last line and considering it in Z/28598400Z, we get 1 = 10093553 · 17. Thus, we can use d = 10093553 as the decryption key.

346

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Exercise: 9 Section 6.6 Question: Implementing RSA over F31 [x], you take the role of Alice. Suppose that Bob has sent you n(x) = x5 + 4x4 + 9x3 + 16x2 + 25x + 3, where all the coefficients are understood to be in F31 . Let m(x) = x3 + 2x. Calculate m(x)3 in F31 [x]/(n(x)). Solution: We have m(x)3 = x9 + 6x7 + 12x5 + 8x3 . Then to find this as an element in F31 [x]/(n(x)) we need to perform polynomial division. Performing the polynomial division gives m(x)3 = (x5 + 4x4 + 9x3 + 16x2 + 25x + 3)(x4 + 27x3 + 13x2 + 30x) + 22x4 + 21x3 + 17x2 + 3x. Hence, in F31 [x]/(n(x))

3

m(x) = 22x4 + 21x3 + 17x2 + 3x.

6.7 – Algebraic Integers Exercise: 1 Section 6.7 Question: Let R be a UFD and let F be its field of fractions. Prove that R is integrally closed in its field of fractions. [This result inserts in the chain in (6.8) the the class of domains that are integrally closed between UFDs and integral domain.] Solution: Let R be a UFD and let F be its field of fractions. Let rs ∈ F be an elements such that the monic polynomial f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 has rs as a root. By the Rational Root Theorem, r | a0 and s is a unit. Thus rs ∈ R. Hence, R is its own integral closure in F . This means that R is integrally closed. Exercise: 2 Section 6.7 Question: Suppose that a1 , a2 , . . . , an ∈ C are such that the groups (Z[ai ], +) are finitely generated free groups of ranks r1 , r2 , . . . , rn . Prove that abelian group (Z[a1 , a2 , . . . , an ], +) is finitely generated of rank less than or equal to r1 r2 · · · rn . Solution: Suppose that a1 , a2 , . . . , an ∈ C are such that the groups (Z[ai ], +) are finitely generated free groups of ranks r1 , r2 , . . . , rn . We prove the result by induction on n. Obviously, if n = 1, then if (Z[a1 ], +) is a finite generated free abelian group of rank r1 , then (Z[a1 ], +) is finitely generated of rank less than or equal to r1 . Now suppose that for some n ≥ 1 and a1 , a2 , . . . , an ∈ C such that the groups (Z[ai ], +) are finitely generated free groups of ranks r1 , r2 , . . . , rn the additive group (Z[a1 , a2 , . . . , an ], +) is a finitely generated abelian group of rank r less than or equal to r1 r2 · · · rn . Let an+1 ∈ C such that (Z[an+1 ], +) is a finitely generated free group of rank rn+1 . Let {α1 , α2 , . . . , αr } be a basis of (Z[a1 , a2 , . . . , an ], +) and let {β1 , β2 , . . . , βrn+1 } be a basis of (Z[an+1 ], +). Every element in (Z[a1 , a2 , . . . , an+1 ], +) can be written as m−1 γ = pm am n+1 + pm−1 an+1 + · · · + p1 an+1 + p0

where pi ∈ Z[a1 , a2 , . . . , an ]. But then each pi can be written as ci1 α1 + ci2 α2 + · · · + cir αr with cij ∈ Z. Furthermore, every akn+1 can be written as dk1 β1 + dk2 β2 + · · · dkrn+1 βrn+1 with dk` ∈ Z. Hence, multiplying and gathering like terms, γ can be written as a Z-linear combination of the elements αi βj . There are r · rn+1 of these. Thus, (Z[a1 , a2 , . . . , an+1 ], +) is a finitely generated free abelian group with rank less than or equal to r · rn+1 , which is less than or equal to r1 r2 · · · rn+1 . The result holds by induction on n. Exercise: 3 Section 6.7 Question: This exercises guides a proof that for D = −3, −7, −11 the quadratic integer ring OQ(√D) is a Euclidean domain with its norm N .

6.7. ALGEBRAIC INTEGERS

347

√ 1. Let D ≡ 1 (mod 4) be a negative integer and set ω = (1 + D)/2. Show that as a subset of C, the quadratic integer ring Z[ω] consists of vertices of congruent parallelograms that cover the plane. 2. Consider a parallelogram with vertices 0, 1, ω, and ω + 1. Show that the furthest an interior point P can be from any of the four vertices. [Hint: Show first that P must be the circumcenter for one of the triangles of the parallelogram.] p 3. Use the previous part to prove that every element in C is at most (1 + |D|)/(4 |D|) away from some element in Z[ω]. 4. Show that (1 + |D|)2 /16|D| < 1 for D = −3, −7, −11. 5. Show that d(z) = N (z) = |z|2 is a Euclidean function on Z[ω]. [With previous results, this exercise shows that OQ(√D) is a Euclidean domain with the norm function as the Euclidean function for D = −1, −2, −3, −7, −11. It turns out, though it is more difficult to prove, that these are the only negative values of D for which OQ(√D) is a Euclidean domain.] Solution: This exercises guides a proof that for D = −3, −7, −11 the quadratic integer ring OQ(√D) is a Euclidean domain with its norm N . √ √ √ 1+ D 1. If ω = (1 + D)/2, then ω 2 = 14 (1 + D + 2 D) = D−1 + . If D ≡ 1 (mod 4), then ω 2 is of the 4 2 form a + bω such that a, b ∈ Z. Thus, all successive powers of ω are also of this form. As a subset of C, the elements of Z[ω] are at integer multiples of 1 and ω. The elements 1 and ω form the edges of a parallelogram. Then all the elements of Z[ω] consists of vertices of congruent parallelograms (each with the vector 1 and ω for edges) that tile the plane. 2. In a parallelogram with vertices 0, 1, ω, and ω + 1, the furthest an interior point is to one of the vertices is the minimum of the circumradius of any triangle formed by three of the vertices. Note that using either of the diagonals cuts the parallelogram into two pairs of congruent triangles. Hence, there are two circumradii to consider. (Recall that the circumcenter of a triangle occurs at the common intersection of the side bisectors.) The triangle with vertices 0, 1 and ω is an isosceles triangle. Some analytic geometry p 1 1 1 puts the circumcenter at 2 + i 4 |D| − √ . The circumradius is the distance between this point and |D| √ |D| ω = 12 + i 2 , which is p

|D| 1 − 2 4

p

1 |D| − p |D|

!

1 = 4

p

1 |D| + p |D|

The other triangle, 0, 1, and ω + 1, has a circumcenter located at 12 +

! =

√

|D| + 1 p . 4 |D|

|D| + √3 4 4 |D|

i. The circumradius

is s

5 D2 + 9 + . 8 16|D|

1 The difference between the second circumradius and the square of the first is 83 + 2|D| . Hence, the first p circumradius is the minimum so we conclude that every point in C is at most (|D| + 1)/4 |D|.

3. We already concluded this result in the previous part. 4. We have p √ (|D| + 1)/4 |D| < 1 ⇐⇒ (|D| + 1)2 < 16|D| ⇐⇒ |D|2 − 14|D| + 1 < 0 ⇐⇒ |D| < 7 + 4 3. p Thus, for D = −3, −7, and −11, we have (|D| + 1)/4 |D| < 1. 5. Suppose that D is −3, −7, or −11. Let a + bω and c + dω 6= 0 be in Z[ω], then the division θ=

a + bω c + dω

348

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS is an element in C. By the previous parts, there is an element γ ∈ Z[ω] such that |θ − γ|2 < 1. Let ρ = (a + bω) − γ(c + dω). Then |ρ|2 = |(a + bω) − γ(c + dω)|2 = |c + dω|2 · |θ − γ| < |c + dω|2 . This shows that there exists γ, ρ ∈ Z[ω] such that a + bω = (c + dω)γ + ρ with d(ρ) < d(c + dω). Finally, we need to check that d(β) ≤ d(αβ) for all nonzero α, β ∈ Z[ω]. Suppose that α = a1 + a2 ω. Then a2 2 |D| 2 + d(α) = a1 + a . 2 4 2 If a2 = 0 then a1 6= 0 and d(α) = a21 ≥ 1. If a2 6= 0 and a2 is odd, then d(α) ≥ 41 + |D| 4 ≥ 1, since we are |D| 2 only considering D = −3, −7, −11. If a2 is even with a2 = 2k, then d(α) ≥ 4 4k = |D|k 2 | ≥ 1. Since d(α) ≥ 1, then d(αβ) = |αβ|2 = |α|2 |β|2 ≥ |β|2 = d(β). We have shown that Z[ω] is a Euclidean domain.

Exercise: 4 Section 6.7 h √ i √ √ Question: Do the Euclidean division of (21 + 13 −7)/2 by (3 + 5 −7)/2 in Z 1+ 2 −7 . (See Exercise 6.7.3.) √ √ Solution: Setting α = (21 + 13 −7)/2 and β = (3 + 5 −7)/2, then γ=

α 259 33 √ 146 33 = − −7 = − ω, β 92 92 46 46

√

where ω = 1+ 2 −7 . We see that γ lies inside the parallelogram 3 − ω, 4 − ω, 4, and 3. We need to figure out to which vertex it is closest. Note that 8 13 γ − (3 − ω) = + ω, 46 46 which has 13 8 + ω 46 46

2

=

29 13 √ + −7 92 92

2

=

11 . 46

√

Since this is less than 1, we see that 3 − ω = 5− 2 −7 will work as a quotient to the Euclidean division. Then √ √ √ √ 3 + 5 −7 21 + 13 −7 5 − −7 = + (−2 + −7). 2 2 2 Note that the remainder is −2 +

√

−7 = −3 + 2ω and that

d(−2 +

√

−7) = 11 < 46 = d

√ 3 + 5 −7 . 2

Exercise: 5 Section 6.7 √ Question: Consider the element 16 (4 + 4 · 281/3 + 282/3 ) in the field F = Q( 3 28). Show that this is an algebraic integer in F by showing that it solves a monic cubic polynomial in Z[x]. Solution: With the hint that α = 61 (4 + 4 · 281/3 + 282/3 ) solves a monic cubic polynomial in Z[x]. We can easily calculate that 20 5 1/3 2 2/3 + 28 + 28 3 3 3 74 26 1/3 8 2/3 3 α = + 28 + 28 . 3 3 3 α2 =

6.7. ALGEBRAIC INTEGERS

349

If α is a root of x3 + ax2 + bx + c, then  74 20 2   3 + 3 a + 3b + c = 0 26 5 2 3 + 3a + 3b + c = 0  8 2 1 3 + 3 a + 6 b + c = 0. We find that α is a root of x3 − 2x2 − 8x − 6 = 0. Exercise: 6 Section 6.7 Question: Find all ways to write 91 as a sum of two squares. Solution: The integer 91 cannot be written as a sum of to squares. We have already seen that if n = a2 + b2 , then n ≡ 0, 1, 2 (mod 4). However, 91 ≡ 3 (mod 4). Exercise: 8 Section 6.7 Question: Find all ways to write 338 as a sum of two squares. Solution: In order to find ways to write 338 as a sum of two squares, we consider its factorization in Z and in Z[i]. The prime factorization of 338 is 338 = 2 · 132 . The Gaussian integer 1 + i has norm 2 while 2 + 3i and 3 + 2i have norm 13 in Z[i]. Note that all other Gaussian integers that have 2 or 13 as their norm are associates of the ones given. There are three non-associate Gaussian integers whose norm is 338: (1 + i)(2 + 3i)(2 + 3i) = −17 + 7i (1 + i)(2 + 3i)(3 + 2i) = −13 + 13i (1 + i)(3 + 2i)(3 + 2i) = −7 + 17i This leads to two distinct ways of writing 338 as a sub of two squares: 338 = 72 + 172 = 132 + 132 . The pairs (x, y) ∈ Z2 such that x2 + y 2 = 338 are (±7, ±17),

(±17, ±7),

(±13, ±13).

This accounts for 12 pairs. Since 13 is the only prime in the factorization of 338 that is congruent to 1 modulo 4, this meshes with Theorem 6.7.12. Exercise: 8 Section 6.7 Question: Find all ways to write 61,000 as a sum of two squares. Solution: The prime factorization of 61, 000 is 61, 000 = 23 · 53 · 61. There are no primes congruent to 3 modulo 4 in the factorization so by Fermat’s Theorem on the sum of two squares, this can be written as a sum of two squares and by Theorem 6.7.12, there are 4 × 2 × 4 = 32 pairs (x, y) ∈ Z2 such that x2 + y 2 = 61, 000. In Z[i], we have 2 = (1 + i)(1 − i), 5 = (2 + i)(2 − i) and also 61 = (6 + 5i)(6 − 5i). Disregarding elements that are associates, we have two elements with 5 as a norm, namely 2 + i and 1 + 2i and two elements with 61 as a norm, namely 6 + 5i and 5 + 6i. We then have the following factorizations of elements in Z[i] for elements with norm 61,000. (1 + i)3 (2 + i)3 (6 + 5i) = −66 − 238i (1 + i)3 (2 + i)2 (1 + 2i)(6 + 5i) = 90 − 230i (1 + i)3 (2 + i)(1 + 2i)2 (6 + 5i) = 210 − 130i (1 + i)(1 + 2i)3 (6 + 5i) = 246 + 22i (1 + i)3 (2 + i)3 (5 + 6i) = −22 − 246i (1 + i)3 (2 + i)2 (1 + 2i)(5 + 6i) = 130 − 210i (1 + i)3 (2 + i)(1 + 2i)2 (5 + 6i) = 230 − 90i (1 + i)(1 + 2i)3 (5 + 6i) = 238 + 66i.

350

CHAPTER 6. DIVISIBILITY IN COMMUTATIVE RINGS

Using positive integers, there are four inequivalent ways to write 61,000 as a sum of squares: 61, 000 = 222 + 2462 = 662 + 2382 = 902 + 2302 = 1302 + 2102 .

Exercise: 9 Section 6.7 √ Question: Prove that if α is an algebraic integer in C, then n α is another algebraic integer for any positive n. Solution: Suppose that α is an algebraic integer in C. Then α is the root of some monic polynomial in Z[x]. Say this polynomial is p(x). Then, for any positive integer n, the polynomial q(x) = p(xn ) is monic of degree n · deg p(x) and obviously in Z[x]. Furthermore, any nth root of α is a root of q(x). Hence, any nth root of an algebraic integer is again an algebraic integer.

7 | Field Extensions 7.1 – Introduction to Field Extensions Exercise: 1 Section 7.1 Question: Consider the ring F = F5 [x]/(x2 + 2x + 3) and call θ the element x in F . a) Prove that F is a field. b) Prove that every element of F can be written uniquely as aθ + b, with a, b ∈ F5 . c) Let α = 2θ + 3 and β = 3θ + 4. Calculate a) α + β; b) αβ; c) α/β. Solution: Consider the ring F = F5 [x]/(x2 + 2x + 3). a) We first check to see if the polynomial p(x) = x2 + 2x + 3 has any roots in F5 . p(0) = 3,

p(1) = 1,

p(2) = 1,

p(3) = 3,

p(4) = 2.

Since p(x) is a quadratic polynomial with no roots, then it is irreducible. Consequently, the ideal (p(x)) is maximal and the ring F5 [x]/(x2 + 2x + 3) is a field. b) This part is a standard result of quotient rings with polynomial rings. If we perform polynomial division by a(x) by x2 + 2x + 3, we get a(x) = (x + 2 + 2x + 3)q(x) + r(x), where deg r(x) < 2. Furthermore, with polynomial division, this remainder is unique. Hence in the quotient ring F , a(x) = r(x). This remainder with the condition that deg r(x) < 2 is unique, so the result follows. c) Note that since θ2 + 2θ + 3 = 0 in F , then θ2 = −2θ − 3 = 3θ + 2. Let α = 2θ + 3 and β = 3θ + 4. Then α + β = 5θ + 7 = 2. Also αβ = (2θ + 3)(3θ + 4) = 6θ2 + 17θ + 12 = θ2 + 2θ + 2 = 4. For the division, we have α β = aθ + b where 2θ + 3 = (aθ + b)(3θ + 4) = 3aθ2 + (4a + 3b)θ + 4b = 3a(3θ + 2) + (4a + 3b)θ + 4b = (3a + 3b)θ + (a + 4b). Hence, we need to solve in F5 , the system of equations ( ( 3a + 3b = 2 3(3 + b) + 3b = 2 =⇒ a + 4b = 3 a=3+b

( =⇒

b+4=2 a=3+b

( =⇒

b=3 a=1

.

Thus α/β = θ + 3. Exercise: 2 Section 7.1 √ √ 2 √ 2 √ Question: Let α = 1 + 3 2 − 3 3 2 and β = 3 + 3 2 in Q( 3 2). Calculate a) αβ; √ √ √ Solution: Note that Q( 3 2) has the natural basis {1 3 2, ( 3 2)2 }. a) For αβ, we have

b) α/β;

√ √ √ 2 2 3 3 3 αβ = (1 + 2 − 3 2 )(3 + 2 ) √ √ √ √ 2 2 3 3 3 3 =3+ 2 +3 2+2−9 2 −6 2 √ √ 2 3 3 =5−3 2−8 2 . √ √ 2 3 3 b) The division α β = a + b 2 + c 2 where we must have 1+

√ 3

√ √ √ √ 2 2 2 3 3 3 3 2 − 3 2 = (3 + 2 )(a + b 2 + c 2 ) √ √ 2 3 3 = (3a + 2b) + (3b + 2c) 2 + (a + 3c) 2 . 351

c) β/α.

352

CHAPTER 7. FIELD EXTENSIONS To solve the corresponding system of equations, we need to get the reduced row echelon form     3 2 0 1 1 0 0 −9/31 rref 0 3 2 1  = 0 1 0 29/31  . 1 0 3 −3 0 0 1 −28/31

2 28 √ 9 29 √ α 3 3 2− 2 . =− + β 31 31 31 √ 2 √ β c) The division α = a + b 3 2 + c 3 2 where we must have

Thus

3+

√ √ √ √ √ 2 2 2 3 3 3 3 3 2 = (1 + 2 − 3 2 )(a + b 2 + c 2 ) √ √ 2 3 3 = (a − 6b + 2c) + (a + b − 6c) 2 + (−3a + b + c) 2 .

To solve the corresponding system of equations, we need to get the reduced row echelon form     1 −6 2 3 1 0 0 −55/87 1 −6 0 = 0 1 0 −59/87 . rref  1 −3 1 1 1 0 0 1 −19/87 Thus

2 55 59 √ 19 √ β 3 3 =− − 2− 2 . α 87 87 87

Exercise: 3 Section 7.1 Question: Consider the ring F = Q[x]/(x3 + 3x + 1) and called θ the element x in F . a) Prove that F is a field (which we can write as Q(θ)). b) Prove that every element of F can be written unique as aθ2 + bθ + c, with a, b ∈ Q. c) Let α = 2θ2 − 1 and β = θ2 + 5θ − 3. Calculate (i) α + β; (ii) αβ; (iii) α/β. Solution: Consider the ring F = Q[x]/(x3 + 3x + 1) and called θ the element x in F . a) Since x3 + 3x + 1 has integer coefficients, by the Rational Root Theorem, the only rational roots of p(x) = x3 + 3x + 1 could possibly be ±1. However, p(1) = 5 and p(−1) = −3. Since p(x) does not have any rational roots, it does not have a linear factor. Since it is a cubic, this means that it is irreducible. As an irreducible element in a PID, its ideal is maximal and therefore, Q[x]/(p(x)) is a field. b) By Theorem 7.1.10, Q(θ) = Q[θ]. Since θ3 = −3θ − 1, every power θn can be rewritten as a Q-linear combination of θ2 , θ, and 1. c) Let α = 2θ2 − 1 and β = θ2 + 5θ − 3. 1. α + β = 3θ2 + 5θ − 4. 2. For αβ we calculate αβ = 2θ4 + 10θ3 − 6θ2 − θ2 − 5θ + 3 = 2(−3θ2 − θ) + 10(−3θ − 1) − 6θ2 − θ2 − 5θ + 3 = −13θ2 − 37θ − 7. 3. For α/β we need to solve for a, b, and c in α = β(aθ2 + bθ + c). But β(aθ2 + bθ + c) = a(−3θ2 − θ) + 5a(−3θ − 1) − 3aθ2 + b(−3θ − 1) + 5bθ2 − 4bθ + cθ2 + 5cθ − 3c. Thus, we need to solve   −6a + 5b + c = 2 −16a − 7b + 5c = 0   −5a − b − 3c = −1. 1 8 8 1 , 27 , 27 so α/β = 27 (−θ2 + 8θ + 8). We get (a, b, c) = − 27 Exercise: 4 Section 7.1 Question: Let F = F7 and consider the irreducible polynomial p(x) = x3 − 2. Write θ for the element x in F7 /(p(x)). Find the inverse of θ2 − θ + 3.

7.1. INTRODUCTION TO FIELD EXTENSIONS

353

Solution: We write θ for the element x in F = F7 /(x3 − 2). Since x3 − 2 is irreducible in F7 [x], then F is a field. Furthermore, θ3 = 2. The inverse of θ2 − θ + 3 is an element aθ2 + bθ + c such that 1 = (θ2 − θ + 3)(aθ2 + bθ + c) = aθ4 + bθ3 + cθ2 − aθ3 − bθ2 − cθ + 3aθ2 + 3bθ + 3c = (3a + 6b + c)θ2 + (2a + 3b + 6c)θ + (5a + 2b + 3c). Thus, we need to solve in F7 , the system of equations   3a + 6b + c = 0 2a + 3b + 6c = 0   5a + 2b + 3c = 1. We get       =0 =0 3a + 6b + c a + 2b + 5c a + 2b + 5c = 0 R1 →5R1 2a + 3b + 6c = 0 =⇒ 2a + 3b + 6c = 0 =⇒ 6b + 3c =0       5a + 2b + 3c = 1 5a + 2b + 3c = 1 6b + 6c =1       a = 1 a + 4c = 0 a + 2b + 5c = 0 =⇒ b + 4c = 0 =⇒ b + 4c = 0 =⇒ b = 1       c = 5. 3c =1 6b + 6c =1 Thus, the inverse to θ2 + 6θ + 3 is θ2 + θ + 5. Exercise: 5 Section √ 7.1 √ Question: In Q( 4 10), find the inverse of 1 + 4 10. √ √ √ √ Solution: The inverse to 1 + 4 10 in Q( 4 10) is the unique element α = a + b 4 10 + c 10 + d103/4 such that √ α(1 + 4 10) = 1. We can find this by using linear algebra methods as described in this section or with this particular element there may √ be faster ways √ √ to determine the inverse. √ √ 4 Note that (1 + 4 10)(1 − 10) = 1 − 10. Then we have (1 − 10)(1 + 10) = 1 − 10 = −9. We deduce √ that the inverse of 1 + 4 10 is √ √ 1 1 4 − (1 − 10)(1 + 10) = − (1 − 101/4 + 101/2 − 103/4 ). 9 9

Exercise: 6 Section 7.1 Question: Consider the field of order 8 constructed in Example 7.1.13. Call θ an element in F such that θ3 + θ + 1 = 0, so that we can write F = F2 [θ]. a) Let α = θ2 + 1, β = θ2 + θ + 1, and γ = θ. Calculate: (i) αγ + β; (ii) α/γ; (iii) α2 + β 2 + γ 2 . b) Solve for x in terms of y in the equation y = αx + β. c) Prove that the function f : F → F defined by f (α) = α3 is a cyclic permutation on F . Solution: a) Let α = θ2 + 1, β = θ2 + θ + 1, and γ = θ. (i) αγ + β = θ3 + θ + θ2 + θ + 1 = θ + 1 + θ + θ2 + θ + 1 = θ2 + θ. (ii) For α/γ, note that θ(θ2 + 1) = θ3 + θ = 1. Hence, 1/γ = θ2 + 1. Thus, α/γ = α2 = θ4 + 1 = θ2 + θ + 1. (iii) α2 + β 2 + γ 2 = (θ2 + θ + 1) + (θ4 + θ2 + 1) + θ2 = 0. b) If y = αx + β then y + β = αx and we saw earlier that 1/α = θ. Hence, x = θ(y + θ2 + θ + 1), so x = θy + θ3 + θ2 + θ = θy + θ2 + 1. c) We can make a chart of how the function f (α) = α3 maps the elements of F . α f (α)

0 0

1 1

θ θ+1

θ+1 θ2

θ2 θ +1 2

θ2 + 1 θ2 + θ 2 2 θ +θ θ +θ+1

θ2 + θ + 1 θ

If we label the elements in F with ordinals 1, 2, . . . , 8 following the list given in the above table, then f is in fact a permutation and f (surprisingly) permutes the elements according to the cycle (3 4 5 6 7 8).

354

CHAPTER 7. FIELD EXTENSIONS

Exercise: 7 Section 7.1 Question: Consider the field F of order 8 constructed in Example 7.1.13. Prove that U (F ) is a cyclic group. Solution: As elements in F , the set of U (F ) is {1, θ, θ + 1, θ2 , θ2 + 1, θ2 + θ, θ2 + θ + 1}. Since (U (F ), ×) is an abelian group of order 7, by FTFGAG, U (F ) ∼ = Z7 . Hence, every element except the identity will generate the group. For example, the powers of θ are successively θ,

θ2 ,

θ + 1,

θ2 + θ,

θ2 + θ + 1,

θ2 + 1,

1.

Exercise: 8 Section 7.1 √ √ Question: Prove that {1, 2, 3} is linearly independent in C as a vector space over Q. √ √ Solution: Consider the set of solutions (a, b, c) ∈ Q3 such that a + b 2 + c 3 = 0. Then √ √ √ √ 1 −a = b 2 + c 3 =⇒ a2 = 2b2 + 3c2 + 2 6bc =⇒ (a2 − 2b2 − 3c2 ) = bc 6. 2 √ If neither b nor c is 0, then we can divide this expression by bc and obtain 6 as a rational number. This is a contradiction, so we must assume that b = 0 or c = 0. √ √ If b = 0, then we have a2 − 3c2 = 0. If c 6= 0, then ac = ± 3 but this is a contradiction since 3 is irrational. Thus, we conclude that c = 0 and then also a = 0. √ √ If c = 0, then we have a2 − 2b2 = 0. Again, if b 6= 0, then ab = ± 2 but this is a contradiction since 2 is irrational. Thus, we conclude that b = 0 and then also a√= 0. √ √ √ We conclude that the only rational solution to a + b 2 + c 3 = 0 is (a, b, c) = (0, 0, 0). Hence, {1, 2, 3} is linearly independent over Q. Exercise: 9 Section 7.1 √ √ Question: Prove that {1, 3, i, i 3} is linearly independent in C as a vector space over Q. √ √ Solution: Consider the set of solutions for (a, b, c, d) ∈ Q4 satisfying a + b 3 + ci + di 3 = 0 in C. We have ( √ a2 − c2 = 3b2 − 3d2 2 2 2 2 −(a + ci) = 3(b + di) =⇒ a + 2aci − c = 3(b + 2bdi − d ) =⇒ 2ac = 6bd. Suppose that b 6= 0. Then we have d = ac 3b from the second equation. Plugging this into the first equation, and rearranging, we get to the equation that is quadratic in b2 , 9b4 − 3(a2 − c2 )b2 − a2 c2 = 0. 2

2

Using usual methods to solve quadratic equations, we find that b2 = a3 of b2 = − c3 . The first of these is √ √ 2 2 equivalent to 3 = ab2 and the second is equivalent to −3 = ac 2 , neither of which is possible since 3 and −3 are irrational numbers. Hence, we deduce that b = 0. √ √ Knowing that b √ = 0, then the real part of a + b 3 + ci + di√ 3 is just a, so we deduce that a must √ be 0. Then we must have c + d 3 = 0. If we assume that d 6= 0, then − 3 = dc , which is impossible since 3 is irrational. Thus d = 0, which √ finally brings √ us down to c = 0. √ √ Thus, if a + b 3 + ci + di 3 = 0 with a, b, c, d all rational, then (a, b, c, d) = (0, 0, 0, 0). Hence, {1, 3, i, i 3} is linearly independent over Q. Exercise: 10 Section 7.1 √ √ Question: Consider the ring K = Q[ 2, 5]. √ √ a) Prove that K = Q[ 2 + 5] and prove also that this is a field. Indicate [K : Q]. √ √ √ √ √ b) Set γ = 2 + 5. Show that B1 = {1, 2, 5, 10} and B2 = {1, γ, γ 2 , γ 3 } are two bases of K as a vector space over Q. c) Determine the change of coordinate matrix from the basis B2 to B1 coordinates. d) Use part 3) to write 2 + 3γ 2 − 7γ 3 in the basis B1 . √ √ √ e) Use part 3) to write −3 + 2 − 5 + 7 10 as a linear combination of {1, γ, γ 2 , γ 3 }. √ √ Solution: Consider the ring K = Q[ 2, 5].

7.1. INTRODUCTION TO FIELD EXTENSIONS

355

√ √ √ √ a) It is obvious that 2 + √ 5 ∈ K√so the generated √ √ ring Q[ 2 + 5] is a subring of K. Now from the main 5] = Q( √2 + 5),√which theorem in this section √ Q[ √2 + √ √ is a field. We need to prove the reverse inclusion, in particular that 2 ∈ Q[ 2 + 5] and 5 ∈ Q[ 2 + 5]. √ √ √ Note that ( 2 + 5)2 = 7 + 2 10 and √ √ √ √ ( 2 + 5)3 = 17 2 + 11 5. Then

√

√ √ √ √ √ √ √ 1 √ 1 √ 2 = (( 2 + 5)3 − 11( 2 + 5)) and 5 = − (( 2 + 5)3 − 17( 2 + 5)). 6 6 √ √ √ √ √ √ √ √ Hence 2, 5 ∈ Q[ 2 + 5], so K = Q[ 2, 5] ⊆ Q[ 2 + 5]. Therefore these rings are equal and we have seen that it is a field. √ √ √ √ Since K = Q( 2 + 5), by Theorem 7.1.10, to find [K : Q], we need to show that γ = 2 + 5 is√the root √ of some irreducible polynomial and the degree of it will be the degree of the extension. Since γ − 2 = 5, we have √ √ γ 2 − 2 2γ + 2 = 5 =⇒ γ 2 − 3 = 2 2γ =⇒ (γ 2 − 3)2 = 8γ 2 . Thus, γ is a root of the polynomial x4 − 14x2 + 9. We try to decide if this polynomial is irreducible. Using the Rational Root Theorem, it is easy to check that none of ±1, ±3, and ±9 give roots. If the polynomial factors, then it must factor into two quadratics: x4 − 14x2 + 9 = (x2 + ax + b)(x2 + cx + d). This implies that   a + c = 0  b + d + ac = −14 ad + bc = 0    bd = 9

  c = −a  b + d − a2 = −14 =⇒ a(d − b) = 0    bd = 9

The third equation leads to two cases. First case, a = 0 implies c = 0 and then b + d = 14 and bd = 9 so d = 9/b and b2 − 14b + 9 = 0. But the discriminant of this equation for b is 196 − 36 = 160 which is not a square in Q so this gives no solutions. Second cases, b = d, then b2 = 9 so b = ±3. Then in the second equation ±6 − a2 = −14 so a2 = 14 ∓ 6. However, neither of the options for the sign give perfect squares in Q so there are no solutions for a. We finally conclude that x4 − 14x2 + 9 does not factor into two quadratics. Hence it is irreducible. We conclude that [K : Q] = 4. √ √ b) We know that {1, γ, γ 2 , γ 3 } is a basis of Q(γ). Furthermore, √ √ √ since K = Q[ 2, 5], then every element in K can be written as a linear combination of {1, 2, 5, 10}. Since this set of “vector” has 4 elements, which is the dimension of K over Q and they span K over Q, then they are a basis. c) The coordinate transition desired matrix will be   1 0 7 0 0 1 0 17  [1]B1 [γ]B1 [γ 2 ]B1 [γ 3 ]B1 =  0 1 0 11 . 0 0 2 0 d) We can write 2 + 3γ 2 − 7γ 3 in the basis B1 by calculating on coordinates      1 0 7 0 2 23 0 1 0 17  0  −119      0 1 0 11  3  =  −77  . 0 0 2 0 −7 6 √ √ √ In other words, 2 + 3γ 2 − 7γ 3 = 23 − 119 2 − 77 5 + 6 10. √ √ √ e) If −3 + 2 − 5 + 7 10 = c0 + c1 γ + c2 γ 2 + c3 γ 3 , then         −1   1 0 7 0 c0 −3 c0 1 0 7 0 −3 0 1 0 17 c1   1  c1  0 1 0 17  1             0 1 0 11 c2  = −1 =⇒ c2  = 0 1 0 11 −1 . 0 0 2 0 c3 7 c3 0 0 2 0 7

356

CHAPTER 7. FIELD EXTENSIONS We calculate this with a reduced row echelon calculation and get     c0 −55/2 c1  −14/3  =  c2   7/2  . c3 1/3 √ √ √ 14 7 2 1 3 Thus −3 + 2 − 5 + 7 10 = − 55 2 − 3 γ + 2γ + 3γ .

Exercise: 11 Section 7.1 Question: Construct a field of 9 elements and write down the addition and multiplication tables for this field. Solution: We can construct a field with 9 elements as F3 [x]/(p(x)) where p(x) is an irreducible quadratic polynomial. A simple example is p(x) = x2 + 1. It is easy to check that none of the elements in F3 is a root of p(x). Since it is a quadratic without roots, it is irreducible. If we call θ the element in F = F3 [x]/(p(x)) given by x, then every element in F can be written as a + bθ with a, b ∈ F3 . The addition table is the following. + 0 1 2 θ 1+θ 2+θ 2θ 1 + 2θ 2 + 2θ

0 0 1 2 θ 1+θ 2+θ 2θ 1 + 2θ 2 + 2θ

1 1 2 0 1+θ 2+θ θ 1 + 2θ 2 + 2θ 2θ

2 2 0 1 2+θ θ 1+θ 2 + 2θ 2θ 1 + 2θ

θ θ 1+θ 2+θ 2θ 1 + 2θ 2 + 2θ 0 1 2

1+θ 1+θ 2+θ θ 1 + 2θ 2 + 2θ 2θ 1 2 0

2+θ 2+θ θ 1+θ 2 + 2θ 2θ 1 + 2θ 2 0 1

2θ 2θ 1 + 2θ 2 + 2θ 0 1 2 θ 1+θ 2+θ

1 + 2θ 1 + 2θ 2 + 2θ 2θ 1 2 0 1+θ 2+θ θ

2 + 2θ 2 + 2θ 2θ 1 + 2θ 2 0 1 2+θ θ 1+θ

The multiplication table is the following. + 0 1 2 θ 1+θ 2+θ 2θ 1 + 2θ 2 + 2θ

0 0 0 0 0 0 0 0 0 0

1 0 1 2 θ 1+θ 2+θ 2θ 1 + 2θ 2 + 2θ

2 0 2 1 2θ 2 + 2θ 1 + 2θ θ 2+θ 1+θ

θ 0 θ 2θ 2 2+θ 2 + 2θ 1 1+θ 1 + 2θ

1+θ 0 1+θ 2 + 2θ 2+θ 2θ 1 1 + 2θ 2 θ

2+θ 0 2+θ 1 + 2θ 2 + 2θ 1 θ 1+θ 2θ 2

2θ 0 2θ θ 1 1 + 2θ 1+θ 2 2 + 2θ 2+θ

1 + 2θ 0 1 + 2θ 2+θ 1+θ 2 2θ 2 + 2θ θ 1

2 + 2θ 0 2 + 2θ 1+θ 1 + 2θ θ 2 2+θ 1 2θ

Exercise: 12 Section 7.1 Question: Consider the field F3 (t) of rational expressions with coefficients in F3 . Let α=

2t + 1 , t+2

β=

1 2t2 + 1

,

γ=

t+1 . t2 + 1

Calculate (a) α + β; (b) βγ; (c) αγ/β. Solution: (2t + 1)(2t2 + 1) + (t + 2) t3 + 2t2 t2 a) α+β = = = 2 . We should have noticed this by observing 2 2 (t + 2)(2t + 1) (t + 2)(2t + 1) 2t + 1 that α is in fact equal to −1. t+1 t+1 = 4 . b) βγ = (2t2 + 1)(t2 + 1) 2t + 1 αγ γ (t + 1)(2t2 + 1) t3 + t2 + 2t + 2 c) =− =− = . β β t2 + 1 t2 + 1

7.1. INTRODUCTION TO FIELD EXTENSIONS

357

Exercise: 13 Section 7.1 Question: Prove that an automorphism of a field F leaves the prime subfield of F invariant. Solution: Let F be a field and let ψ : F → F be an automorphism. By properties of ring homomorphisms, ψ(0) = 0. Since 1 = 12 , we have ψ(1) = ψ(1)2 . Hence ψ(1)(ψ(1) − 1) = 0 so ψ(1) is equal to 0 or 1. Since ψ is injective, we must have ψ(1) = 1. By induction on properties of homomorphisms, ψ(n · 1) = n · ψ(1) = n · 1 for all positive integers n. Again, using the additive property of homomorphisms, we can extend ψ(n · 1) = n · 1 to all integers. Now, let m ∈ Z such that m · 1 6= 0. Every element in the prime subfield of F has the form (m · 1)−1 (n · 1). We have ψ (m · 1)−1 (n · 1) = ψ (m · 1)−1 ψ(n · 1) = ψ (m · 1)−1 (n · 1). However, for all nonzero elements a ∈ F , we have 1 = ψ(1) = ψ(a−1 a)) = ψ(a−1 )ψ(a). Hence, ψ(a−1 ) = ψ(a)−1 . Thus, ψ (m · 1)−1 (n · 1) = ψ (m · 1)−1 ψ(n · 1) = (m · 1)−1 (n · 1). Consequently, every element in the prime subfield of F is invariant under an automorphism. Exercise: 14 Section 7.1 √ √ √ √ Question: Prove that the function f : Q[ 5] → Q[ 5] defined by f (a + b 5) = a − b 5 is an automorphism. √ √ √ Solution: Let a + b 5 and c + d 5 be two elements in Q[ 5. Then √ √ f ((a + b 5) + (c + d 5)) √ √ = f ((a + c) + (b + d) 5) = (a + c) − (b + d) 5 √ √ √ √ = a − b 5 + c − d 5 = f (a + b 5) + f (c + d 5). Furthermore, √ √ √ √ f ((a + b 5)(c + d 5)) = f ((ac + 5bd) + (ad + bc) 5) = (ac + 5bd) − (ad + bc) 5 √ √ √ √ = (a − b 5)(c − d 5) = f (a + b 5)f (c + d 5). √ √ √ The above show that f is a ring homomorphism 5 ∈ Q[ 5],√we see that √ two identities √ √ of Q[ 5].√For all a + b √ f (a − b 5) = a + b 5, so f is √ surjective. Finally if f (a + b 5) = f (c + d 5), then a − b 5 = c − d 5, which is equivalent to a − c = (b − d) 5. Since a, b, c, d ∈ Q, the only way this is possible is for a = c and b = d. Thus f is injective. We conclude that f is an automorphism. Exercise: 15 Section 7.1 Question: Prove that there exists an isomorphism of fields f : R → R that maps π to −π. Solution: Apologies. There does not exist such an isomorphism. Exercise 11.1.7 shows that this is not true. Exercise: 16 Section 7.1 Question: Let K = F (α) where α is the root of some irreducible polynomial p(x) ∈ F [x]. Suppose that p(x) = an xn + · · · + a1 x + a0 . Show that the function fα : K → K defined by fα (x) = αx is a linear transformation and that the matrix of fα with respect to the ordered basis (1, α, α2 , . . . , αn−1 ) is   0 0 0 ··· 0 −a0 /an 1 0 0 · · · 0 −a1 /an    0 1 0 · · · 0 −a2 /an   .  .. .. .. . .  .. .. . . .  . . . 0

0

0

···

1

−an−1 /an

Solution: Let K = F (α) where α is the root of some irreducible polynomial p(x) ∈ F [x]. Suppose that p(x) = an xn + · · · + a1 x + a0 . From Theorem 7.1.10, F (α) = F [α] and so K is a vector space over F with ordered basis (1, α, α2 , . . . , αn−1 ), where n = deg p(x). The function fα (x) = αx satisfies fα (β + γ) = α(β + γ) = αβ + αγ = fα (β) + fα (γ) fα (cβ) = α(cβ) = cαβ = cfα (β)

for all β, γ ∈ K,

for all β ∈ K and c ∈ F.

358

CHAPTER 7. FIELD EXTENSIONS

Thus fα is a linear transformation on K. If 0 ≤ k < n − 1, then fα (αk ) = αk+1 and this is the next basis vector in the ordered basis we use. But if k = n − 1, then a0 an−1 n−1 a1 fα (αn−1 ) = αn = − α α− . − ··· − an an an This expresses αn as a linear combination in the basis vectors we use. This shows that the matrix of fα with respect to this ordered basis is   0 0 0 ··· 0 −a0 /an 1 0 0 · · · 0 −a1 /an    0 1 0 · · · 0 −a2 /an  .    .. .. .. . . . ..  . . . . .. . 0

0

0

···

1

−an−1 /an

Exercise: 17 Section 7.1 Question: Prove that the only continuous automorphism on the field of real numbers is the identity function. Solution: Note that the prime subfield of R is Q. Furthermore, any automorphism ψ of a field fixes the prime subfield of the field. Hence, ψ must fix Q. Let x0 ∈ R − Q. Since ψ is continuous, taking ε, there exists δ > 0 such that 0 < |x − x0 | < δ implies |ψ(x) − ψ(x0 )| < 2ε . Since there are rational numbers arbitrarily close to every real number, there exists q ∈ Q such that |x0 − q| < min(δ, 2ε ). Then since ψ(q) = q, |ψ(x0 ) − x0 | = |ψ(x0 ) − q + q − x0 | < |ψ(x0 ) − q| + |q − x0 | <

ε ε + = ε. 2 2

Since ε is arbitrary and in particular arbitrarily small, we deduce that ψ(x0 ) = x0 . Exercise: 18 Section 7.1 Question: Let ϕ : F → F 0 be an isomorphism of fields. Let p(x) ∈ F [x] be an irreducible polynomial and let p0 (x) be the polynomial obtained from p(x) by applying ϕ to the coefficients of p(x). Let α be a root of p(x) in some extension of F and let β be a root of p0 (x) in some extension of F 0 . Prove that there exists an isomorphism Φ : F (α) → F 0 (β) such that Φ(α) = β and Φ(c) = ϕ(c) for all c ∈ F . Solution: By Theorems 7.1.10 and 7.1.12, F (α) = F [α] = F [x]/(p(x)). Furthermore, every element in F (α) can be written uniquely as c0 + c1 α + · · · + cn−1 αn−1 , where n = deg p(x) and ci ∈ F . Define Φ : F (α) → F 0 (β) by Φ(c0 + c1 α + · · · + cn−1 αn−1 ) = ϕ(c0 ) + ϕ(c1 )β + · · · + ϕ(cn−1 )β n−1 This expression is well-defined since each element in F (α) is unique defined by this type of linear combination. Also, it satisfies the conditions Φ(α) = β and Φ(c) = ϕ(c) for all c ∈ F . Since ϕ is an isomorphism, {1, α, . . . , αn−1 } is a basis of F (α), and {1, β, . . . , β n−1 } is a basis of F 0 (β), then Φ is a bijection. Furthermore, Φ((c0 + c1 α + · · · + cn−1 αn−1 ) + (d0 + d1 α + · · · + dn−1 αn−1 )) = Φ((c0 + d0 ) + (c1 + d1 )α + · · · + (cn−1 + dn−1 )αn−1 ) = ϕ(c0 + d0 ) + ϕ(c1 + d1 )β + · · · + ϕ(cn−1 + dn−1 )β n−1 = ϕ(c0 ) + ϕ(c1 )β + · · · + ϕ(cn−1 )β n−1 + ϕ(d0 ) + ϕ(d1 )β + · · · + ϕ(dn−1 )β n−1 = Φ(c0 + c1 α + · · · + cn−1 αn−1 ) + Φ(d0 + d1 α + · · · + dn−1 αn−1 ). Proving the multiplicativity property is harder. As mentioned above, every element in F (α) can be written uniquely as a(α) where a(x) ∈ F [x] and deg a(x) < n. As defined, Φ(a(α)) = a0 (β), where a0 (x) be the polynomial obtained from a(x) by applying ϕ to the coefficients of a(x). Now if deg b(x) < n, then a(α)b(α) = r(α), where

7.1. INTRODUCTION TO FIELD EXTENSIONS

359

r(x) is the remainder of a(x)b(x), when divided by p(x). Suppose that a(x)b(x) = p(x)q(x) + r(x) expresses the polynomial division of a(x)b(x) by p(x). Now   n−2 X X  a(x)b(x) = ai bj  xk . k=0

i+j=k

Since ϕ is an isomorphism, we have       n−2 n−2 n−2 X X X X X X   ϕ ai bj  xk = ϕ(ai bj ) xk = ϕ(ai )ϕ(bj ) xk = a0 (x)b0 (x). k=0

i+j=k

k=0

i+j=k

k=0

i+j=k

Hence, a0 (x)b0 (x) is the polynomial obtained from a(x)b(x) by applying ϕ to the coefficients of a(x)b(x). This happens for all polynomial multiplication and also addition. In particular, a0 (x)b0 (x) = p0 (x)q 0 (x) + r0 (x). Then we have Φ(a(α)b(α)) = Φ(r(α)) = r0 (β) = r0 (β) + p0 (β)q 0 (β) = a0 (β)b0 (β) = Φ(a(α))Φ(b(α)), because p0 (β) = 0. Thus, we have also shown that Φ is multiplicative and we conclude that Φ is a field automorphism satisfying the desired conditions. Exercise: 19 Section 7.1 √ √ Question: Let D be a square-free integer and let K = Q[ D]. Prove that the function f : Q[ D] → M2 (Q) defined by √ a Db f (a + b D) = b a √ is an injective ring homomorphism. Conclude that M2 (Q) contains a subring isomorphic to Q[ D]. √ Solution: Consider the function f : Q[ D] → M2 (Q) defined by √ a Db f (a + b D) = . b a Then √

√

√

a+c f ((a + b D) + (c + d D)) = f ((a + c) + (b + d) D) = b+d a Db c Dd = + b a d c √ √ = f (a + b D) + f (c + d D).

D(b + d) a+c

For multiplication, we have first √ √ √ f ((a + b D)(c + d D)) = f ((ac + Dbd) + (ad + bc) D) = while

√ √ f (a + b D)f (c + d D) =

(ac + Dbd) D(ad + bc) , (ad + bc) (ac + Dbd)

a b

Db c Dd ac + Dbd adD + Dbc = . a d c bc + ad Dbd + ac √ √ √ √ Hence, f ((a + b D)(c + d D)) = f (a + b D)f (c + d D). We have shown that f is a ring homomorphism. To see that f is injective, ( √ √ a=c a Db c Dd f (a + b D) = f (c + d D) =⇒ = =⇒ . b a d c b=d √ √ Since f in injective, Ker f = {0}. By the first isomorphism theorem of rings, Q[ D] ∼ = Q[ D]/ Ker f ∼ = Im f , which is a subring of M2 (Q). Exercise: 20 Section 7.1 √ Question: Consider the field Q[ 3 2].

360

CHAPTER 7. FIELD EXTENSIONS

√ a) Prove that the function ϕ : Q[ 3 2] → M3 (Q) defined by   a 2c 2b √ 2 3 ϕ(a + b 2 + c 2 ) =  b a 2c c b a √ 3

√ is an injective homomorphism. Conclude that M3 (Q) contains a subring that is a field isomorphic to Q[ 3 2]. √ √ 2 b) Use this homomorphism and matrix inverses to find the inverse of 3 − 3 2 + 5 3 2 . √ Solution: Consider the field Q[ 3 2]. √ a) The product of two generic elements in Q( 3 2) is √ √ √ √ √ √ 2 2 3 3 3 3 3 3 (a + b 2 + c 2 )(d + e 2 + f 2 ) = (ad + 2bf + 2ce) + (ae + bd + 2cf ) 2 + (af + be + cd) 2. To check the homomorphism properties, the addition has √ √ √ √ √ √ 2 2 2 3 3 3 3 3 3 ϕ((a + b 2 + c 2 ) + (d + e 2 + f 2 )) = ϕ((a + d) + (b + e) 2 + (c + f ) 2 )   (a + d) 2(c + f ) 2(b + e) =  (b + e) (a + d) 2(c + f ) (c + f ) (b + e) (a + d)     d 2f 2e a 2c 2b =  b a 2c +  e d 2f  f e d c b a √ √ √ √ 2 2 3 3 3 3 = ϕ(a + b 2 + c 2 ) + ϕ(d + e 2 + f 2 ) and the multiplication is    d 2f 2e a 2c 2b √ √ √ 2 2 3 3 3 ϕ(a + b 2 + c 2 )ϕ(d + e 2 + f 2 )) =  b a 2c  e d 2f  f e d c b a   ad + 2ce + 2bf 2af + 2cd + 2be 2ae + 4cf + 2bd 2bf + ad + 2ce 2be + 2af + 2cd =  bd + ae + 2cf cd + be + af 2cf + bd + ae 2ce + 2bf + ad √ √ 2 3 3 = ϕ((ad + 2bf + 2ce) + (ae + bd + 2cf ) 2 + (af + be + cd) 2 ) √ √ √ √ 2 2 3 3 3 3 = ϕ((a + b 2 + c 2 )(d + e 2 + f 2 )). √ 3

√ √ 2 We have shown that ϕ is a ring homomorphism. Furthermore, if ϕ(a + b 3 2 + c 3 2 ) is the zero matrix then we must have a = b = c = 0. Hence ker ϕ = {0} and ϕ is injective. By the First Isomorphism Theorem, √ 3 Im ϕ ∼ = Q( 2). b) We have   3 10 −2 √ √ 2 3 3 10  . ϕ(3 − 2 + 5 2 ) = −1 3 5 −1 3 The matrix inverse is 

3 A = −1 5

10 3 −1

 −1  19 −28 106 −2 1  10  = 53 19 −28 615 3 −14 53 19

Since ϕ is injective, we determine the inverse to the field element by (3 −

√ 3

√ 2 2 19 53 √ 14 √ 3 3 3 2 + 5 2 )−1 = ϕ−1 (A) = + 2− 2 . 615 615 615

Exercise: 21 Section 7.1 Question: Consider the field of rational expressions K1 = Q(x) with coefficients in Q and also the field

7.1. INTRODUCTION TO FIELD EXTENSIONS

361

√ K2 = Q( p | p is prime). Prove that K1 and K2 are extensions of Q of infinite degree. Prove also that K1 and K2 are not isomorphic. Solution: The set of elements {1, x, x2 , x3 , . . .} in Q(x) are linearly independent over Q. Hence, Q(x) as a vector space over Q, cannot have finite dimension. Thus [Q(x) : Q] is infinite. The biggest difficulty in this exercise is to prove that K2 is infinite. Call pi the i prime in N. Every element √ √ √ √ √ √ in Q( p1 , p2 , . . . , pn ) = Q[ p1 , p2 , . . . , pn ] is a Q-linear combination of the 2n elements ω=

q

p 11 p 22 · · · p nn ,

where i = 0, 1. (We have not shown that this set is linearly independent, but it is. We could use that result for the subsequent portion of the proof but we offer an alternate proof that does not need linear independence of these ω.) First, it is not hard to show using methods as in Exercise 7.1.8 or 7.1.9 that if p and q are distinct primes, √ √ that q ∈ / Q( p). √ √ √ √ Now assume that pn+1 ∈ Q[ p1 , p2 , . . . , pn ]. Write √

pn+1 =

X

cω ω

ω

in such a way as to use the fewest primes necessary occurring under the square roots of the ω spanning elements. √ Call C the set of ω elements used in this linear combination. Suppose that these primes used to express pn+1 are q1 , q2 , . . . , qk . Let A be the subset of ω ∈ C that involve the prime qk and let B = C − A. Note that since √ √ q∈ / Q( p) for any primes p and q, then k ≥ 2 and A is a nonempty proper subset of C. Then √

pn+1 =

X

cω ω +

ω∈A

so

√

qk

X

cω

ω∈A

ω √ qk

X

cω ω

ω∈B

=

√

X

pn+1 −

cω ω.

ω∈B

Squaring both sides gives qk

X

cω

ω∈A

ω √ qk

!2

!2

X √ cω ω + = pn+1 − 2 pn+1

X

ω∈B

Then we have √

pn+1 + pn+1 =

P

ω∈B cω ω

2

2

− qk

P

P

ω∈B cω ω

ω∈A cω

cω ω

.

ω∈B

√ω qk

2 .

√ Since √ωqk for all ω ∈ A does not have the prime qk in it, then √ωqk ∈ B. Hence, this expression for pn+1 gives it as an element in Q(q1 , q2 , . . . , qk−1 ) = Q[q1 , q2 , . . . , qk−1 ], which contradicts the minimality the primes √ √ of the √ √ used in occurring under the square roots of the ω spanning elements. Thus pn+1 ∈ / Q[ 2, 3, . . . , pn ]. Hence, K2 has an infinite degree. Finally, K1 and K2 cannot be isomorphic as fields since in Q(x) there is an element x such that no polynomial with coefficients in Q has x as a root. However, every element in K2 is a root of a polynomial. (It is not hard to observe that if α ∈ K2 is written as a linear combination that involve k of the primes roots, √under √ the square √ then there is a polynomial of degree at most 2k with α as a root. For example, if α = 2 − 3 + 2 5, then √ √ √ √ √ √ α − 2 + 3 = 2 5 =⇒ α2 + 2 + 3 − 2 2α + 2 3α − 2 6 = 20 √ √ √ √ √ =⇒ α2 − 2 2α − 15 = −2 3α + 2 6 = 3(−2α + 2 2) √ √ √ =⇒ α4 + 8α2 + 225 − 4 2α3 − 30α2 + 50 2α = 12(α2 − 2 2α + 2) √ √ √ =⇒ α4 − 34α2 + 201 = 4 2α3 − 62 2α = 2(4α3 − 74α) Squaring both sides again will produce a polynomial of degree 8 that has α as a root.)

362

CHAPTER 7. FIELD EXTENSIONS

7.2 – Algebraic Extensions Exercise: 1 Section 7.2 √ √ Question: Find the minimal polynomial of 10 + 11 over Q. √ √ Solution: If α = 10 + 11, then √ √ √ α − 10 = 11 =⇒ α2 − 2 10α + 10 = 11 √ =⇒ α2 − 1 = 2 10α =⇒ α4 − 2α2 + 1 = 40α2 =⇒ α4 − 42α2 + 1 = 0. Thus α is a root of p(x) = x4 − 42x2 + 1. We need to show that this polynomial is irreducible in Q[x]. By the Rational Root Theorem, the only possible rational roots are ±1 but neither of these is a root. So if p(x) is reducible, we would have p(x) = (x2 + ax + b)(x2 + cx + d). This would then require that  a+c=0    ac + b + d = −42  ad + bc = 0    bd = 1

 c = −a    d = 1 b =⇒ 2  −a + b + 1b = −42    1 a b − b = 0.

√ √ If a = 0, then c = 0 and b solves b2 + 42b + 1 = 0 so b = 21 (−42 ± 422 − 4) = −21 ± 2 110 but this is not a rational number. If a 6= 0, then we must have b − 1b = 0 so b = ±1. If b = −1, then d = −1 and −a2 = −40 √ √ but a = ± 40 is not a rational. Also, if b = 1, then d = 1 and −a2 = −44 but a = ± 44 is not rational. We conclude that p(x) does not factor √ into √ two quadratic either so p(x) is irreducible. Being irreducible in Q[x], this is the minimal polynomial of 10 + 11. Exercise: 2 Section 7.2 √ √ Question: Find the minimal polynomial of 5 + 3 2 over Q. √ √ √ Solution: Let α = 5 + 3 2. Then (α − 5)3 = 2 so √ √ α3 − 3 5α2 + 15α − 5 5 = 3 √ =⇒ α3 + 15α − 3 = 5(3α2 − 5) =⇒ (α3 + 15α − 3)2 = 5(3α2 − 5)2 =⇒ α6 + 225α2 + 9 + 30α4 − 6α3 − 90α = 45α4 − 150α2 + 125 =⇒ α6 − 15α4 − 6α3 + 375α2 − 90α − 116 = 0 This shows that α is a root of the degree 6 polynomial p(x) = x6 − 15x4 − 6x3 + 375x2 − 90x − 116. It is crucial to show that this is the minimal polynomial mα,Q (x). Since α is a root of p(x), then the minimal polynomial mα,Q (x) must divide p(x). Q(α) = Q[x]/(mα,Q (x)), it suffices to show that [Q(α) : Q] = 6. √ Since √ 3 However, by Theorem 7.2.13, Q( 5, 2) must be less than by 6. (Since the degree √ √ √ √6 and must be divisible √ √ of 5 is 2 and the degree of 3 2 over Q is 3.) Since α ∈ Q( 5, 3 2), then Q(α) ⊆ Q( 5, 3 2). By Theorem 7.2.9

√ √ √ √ 3 3 [Q( 5, 2) : Q] = [Q( 5, 2) : Q(α)][Q(α) : Q] √ √ √ √ √ √ √ √ √ so [Q(α) : Q] divides 6. Consider the elements 1, α, α2 , α3 in Q( 5, 3 2) and with respect to the basis (1, 3 2, 3 4, 5, 5 3 2, 5 3 4) √ 2+ 5 √ √ √ 3 3 α2 = 5 + 4 + 2 5 2 √ √ √ √ 3 3 α3 = 2 + 15 2 + 5 5 + 3 5 4 α=

√ 3

7.2. ALGEBRAIC EXTENSIONS

363

√ √ Now, putting these four “vectors” in coordinates with respect to the ordered basis above of Q( 5, 3 2), we have     1 0 0 0 1 0 5 2 0 1 0 15 0 1 0 0     0 0 1 0  0 0 1 0  = rref  0 1 0 5  0 0 0 1     0 0 2 0  0 0 0 0 0 0 0 0 0 0 0 3 This shows that {1, α, α2 , α3 } is a linearly independent set. Hence [Q(α) : Q] ≥ 4. However, since this degree must divide 6, we deduce that [Q(α) : Q] = 6 and hence that deg mα,Q (x) = 6. Hence, mα,Q (x) = p(x) = x6 − 15x4 − 6x3 + 375x2 − 90x − 116.

Exercise: 3 Section 7.2 √ √ 2+√7 Question: Find the minimal polynomial of 1+ over Q( 2). 2 √ √ √ √ √ Solution: Let α = (2+ 7)/(1+ 2). Now α is not in Q( 2) since otherwise we would deduce that 7 ∈ Q( 2), √ but that is not true. Hence, α must be the root of a polynomial in Q( 2)[x] of degree at least 2. We have √ √ √ √ √ √ (1 + 2)α = 2 + 7 =⇒ (1 + 2)α − 2 = 7 =⇒ (1 + 2)2 α2 − 4(1 + 2)α + 4 = 7. √ √ Hence, we conclude that α is the root of m(x) = (3 + 2 2)x2 − (4 + 4 2)x − 3.√Since this polynomial must be of minimal degree having α as a root, we deduce that m(x) is irreducible in Q( 2)[x] and that it is the desired √ minimal polynomial of α over Q( 2). Exercise: 4 Section 7.2 √ √ 2 Question: Determine the degree of 1 − 2 3 7 + 10 3 7 over Q. √ √ 2 √ √ √ Solution: The element α = 1 − 2 3 7 + 10 3 7 is in Q( 3 7). We know that [Q( 3 7) : Q] = 3 because 3 7 has degree 3 over Q with minimal polynomial x3 − 7 over Q. By Theorem 7.2.9, √ √ 3 3 [Q( 7) : Q] = [Q( 7) : Q(α)][Q(α) : Q] Thus, [Q(α) : Q] divides 3 so is equal to 1 or 3, where [Q(α) : Q] = 1 if and only if α ∈ Q. √ 2 √ √ Now α ∈ Q if and only if 3 7 − 5 3 7 ∈ Q, which is equivalent to 3 7 solving a quadratic polynomial with coefficients in Q. This does not occur so α ∈ / Q. Hence, [Q(α) : Q] 6= 1 so [Q(α) : Q] = 3. Exercise: 5 Section 7.2 Question: Consider the field F = F2 [x]/(x3 + x + 1) with θ an element in F such that θ3 + θ + 1 = 0. Find the minimal polynomial of θ2 + 1 over F2 . Solution: Consider the field F = F2 [x]/(x3 + x + 1) with θ an element in F such that θ3 + θ + 1 = 0. The element α = θ2 + 1 is not in F2 but is in F . By Theorem 7.2.9, 3 = [F : F2 ] = [F : F2 (α)][F2 (α) : F2 ]. Since, α ∈ / F2 , we deduce that [F2 (α) : F2 ] 6= 1 and hence that [F2 (α) : F2 ] = 3. We know that α will be the root of some cubic polynomial. We calculate α2 = θ4 + 1 = θ(θ + 1) + 1 = θ2 + θ + 1 α3 = θ6 + θ4 + θ2 + 1 = (θ + 1)2 + (θ2 + θ) + θ2 + 1 = θ2 + θ. From this, we observe that α is the root of x3 + x2 + 1 ∈ F2 [x]. Exercise: 6 Section 7.2 Question: Consider the field F = F2 [x]/(x3 + x + 1) with θ an element in F such that θ3 + θ + 1 = 0. In F [x], the polynomial x + θ divides x3 + x + 1. Find the polynomial q(x) ∈ F [x] such that x3 + x + 1 = (x + θ)q(x).

364

CHAPTER 7. FIELD EXTENSIONS

Solution: By definition, θ is a root of the polynomial x3 + x + 1 in F so in F [x], then linear polynomial x + θ is a factor of x3 + x + 1, viewed as a polynomial in F [x]. So x3 + x + 1 = (x + θ)(x2 + ax + b) = x3 + (a + θ)x2 + (b + aθ)x + bθ. Hence a + θ = 0 so a = −θ = θ. Then we also have b + aθ = 1 so b = 1 − θ2 = 1 + θ2 . We should also have bθ = 1 so we check it: bθ = θ(θ2 + 1) = θ3 + θ = 1 because θ3 + θ + 1 = 0. Thus x3 + x + 1 = (x + θ)(x2 + θx + (1 + θ2 )).

Exercise: 7 Section 7.2 Question: Consider the field F = F5 [x]/(x3 + 2x + 4) with θ an element in F such that θ3 + 2θ + 4 = 0. In F [x], the polynomial x − θ divides x3 + 2x + 4. Find the polynomial q(x) ∈ F [x] such that x3 + 2x + 4 = (x − θ)q(x). Solution: Consider the field F = F5 [x]/(x3 + 2x + 4) with θ an element in F such that θ3 + 2θ + 4 = 0. Since θ is a root of p(x) = x3 + 2x + 4, if we view p(x) as an element of F [x], then it has (x − θ) as a linear factor. Setting q(x) = x2 + ax + b, with a, b ∈ F , we must have x3 + 2x + 4 = (x + 4θ)(x2 + ax + b) = x3 + (4θ + a)x2 + (4θa + b)x + 4θb and thus

  4θ + a = 0 4θa + b = 2   4θb = 4

  a = θ =⇒ 4θ2 + b = 2   θb =

  a = θ =⇒ b = 2 + θ2   θb = 1.

It is easy to check that the third equation is satisfied by (θ2 + 2)θ = θ3 + 2θ = (3θ + 1) + 2θ = 1. Hence, we have deduced that q(x) = x2 + θx + (θ2 + 2). Exercise: 8 Section 7.2 Question: Let K/F be a field extension and let α ∈ K. Prove that a polynomial p(x) ∈ F [x] satisfies p(α) = 0 if and only if mα,F (x) divides p(x) in F [x]. Solution: By definition, α is a root of mα,F (x). Consider the polynomial division of p(x) by mα,F (x) in the Euclidean ring F [x]. We have p(x) = mα,F (x)q(x) + r(x), where r(x) = 0 or deg r(x) < deg mα,F (x). If r(x) 6= 0, then r(α) = p(α) − mα,F (α)q(α) = 0 − 0q(α) = 0. Hence α is a root of r(x). However, mα,F (x) is a polynomial of minimum degree that has α as a root. Hence, this leads to a contradiction. Hence, r(x) = 0 and we deduce that mα,F (x) divides p(x). Exercise: 9 Section 7.2 Question: (Palindromic polynomials) A palindromic polynomial in Q[x] is a polynomial p(x) = an xn + · · · + a1 x + a0 ∈ Q[x] such that ai = an−i for all i = 0, 1, . . . , n.

1 a) Let q(x) ∈ Q[x] be a polynomial of degree n. Prove that x q x + x degree 2n. n

is a palindromic polynomial of

1 b) Show that every palindromic polynomial p(x) ∈ Q[x] of even degree can be written x q x + for some x q(x) ∈ Q[x]. c) Use this to solve the equation x4 − 3x3 − 2x2 − 3x + 1 = 0. n

Solution:

7.2. ALGEBRAIC EXTENSIONS

365

a) Let q(x) ∈ Q[x] be a polynomial of degree n with q(x) = bn xn + · · · + b1 x + b0 . Then ! n n−1 1 1 1 1 n n x q x+ =x bn x + + bn−1 x + + · · · + b1 x + + b0 x x x x     ! n n−1 X X n − 1 n 1 + b0  = xn bn xn−2i + bn−1  xn−1−2j  + · · · b1 x + j x i i=0 j=0   ! n n−1 X X n − 1 n 2n−2i = bn x + bn−1  x2n−1−2j  + · · · b1 xn+1 + xn−1 + b0 xn i j i=0 j=0 Let us call xn q x + x1 = c2n x2n + · · · + c1 x + c0 . If k is even with k = 2m, then ck is c2m = bn

min(m,n−m) X n n−2 n − 2` + bn−2 + ··· = bn−2` . n−m m−1 m−` `=0

Then we also have min(n−m,m)

X

c2n−2m =

bn−2`

`=0 min(n−m,m)

X

=

bn−2`

`=0 min(n−m,m)

X

=

n − 2` n−m−` n − 2` (n − 2`) − (n − m − `)

bn−2`

`=0

n − 2` = c2m . m−`

If k is odd with k = 2m + 1, then ck is min(m,n−m−1)

c2m+1 =

X

bn−1−2`

`=0

n − 1 − 2` . m−`

Then c2n−k = c2n−(2m+1) = c2(n−m−1)+1 . Thus, min(n−m−1,m)

c2n−k =

X

n − 1 − 2` n−m−1−`

n − 1 − 2` (n − 1 − 2`) − (n − m − 1 − `)

bn−1−2`

`=0 min(n−m−1,m)

=

X

bn−1−2`

`=0 min(m,n−m−1)

=

X `=0

n − 1 − 2` bn−1−2` = ck . m−`

The above two calculations show that xn q x + x1 is a palindromic polynomial. b) Let p(x) be a palindromic polynomial of even degree 2n. We will show by induction on n the every such polynomial can be written as xn q(x + 1/x) for some appropriate q(x) of degree n. This is obvious with n = 0 since then p(x) = q(x) work as constant polynomial. Suppose now that every palindromic polynomial of degree even degree 2m less than 2n can be written as xm q(x + 1/x) for some q(x) of degree m. Now let p(x) be a palindromic polynomial of degree 2. Write p(x) = a2n x2n + · · · + a1 x + a0 . Note that

n X n 1 n 2n−2j = x . x x+ x j j=0 n

366

CHAPTER 7. FIELD EXTENSIONS This is obviously a palindromic polynomial of degree 2n + 2 because

n j

n n−j

=

. Then, since a2n = a0 ,

n 1 p(x) − a2n xn x + = b2n x2n + · · · + b1 x + b0 x has a 0 coefficient for x0 and for x2n but still has the property that bk = b2n−k . Suppose that b2n x2n + · · · + b1 x + b0 in fact has degree 2n − d for some d ≥ 1. This means that bk = 0 for 2n − d + 1 ≤ k ≤ 2n and thus also for 0 ≤ k ≤ d − 1. This means that we can factor out xd from b2n x2n + · · · + b1 x + b0 . We get b2n x2n + · · · + b1 x + b0 = xd (c2n−2d x2n−2d + · · · + c1 x + c0 ), where ck = bk+d = b2n−(k+d) = c2n−2d−k . Hence, c2n−2d x2n−2d + · · · + c1 x + c0 is a palindromic polynomial of even degree that is less than 2n. By the induction hypothesis, there exists some polynomial q(x) of degree n − d such that c2n−2d x2n−2d + · · · + c1 x + c0 = xn−d q(x + 1/x). But then, n 1 + xd (c2n−2d x2n−2d + · · · + c1 x + c0 ) p(x) = a0 xn x + x n 1 1 = a0 xn x + + xn q x + x x 1 , = xn q x + x where q(x) = q(x) + a0 xn . The proposition follows by induction. c) We now proceed to solve x4 − 3x3 − 2x2 − 3x + 1 = 0. First, we note that 2 1 x4 − 3x3 − 2x2 − 3x + 1 − x2 x + = −3x3 − 4x2 − 3x = x(−3x2 − 4x − 3). x Then

1 −3x − 4x − 3 + 3x x + x 2

= −4x = x(−4).

Putting the pieces together, we deduce that 4

3

2

x − 3x − 2x − 3x + 1 = x

2

1 x+ x

2

1 −3 x+ x

! −4 ,

so that the relevant q(x) as discussed in part (b) is q(x) = x2 − 3x − 4. Now a palindromic polynomial never has a trivial constant term and hence never has 0 as a solution. Thus, x4 − 3x3 − 2x2 − 3x + 1 = 0, where q(x + 1/x) = 0. Hence, we can first solve for the two roots α1 and α2 of q(x) and then obtain all four roots of x4 − 3x3 − 2x2 − 3x + 1 = 0 by solving the two quadratics x + x1 = αi . In this example x2 − 3x − 4 = 0 if and only if x = −1, 4. Now using α1 = −1, √ 1 −1 ± 3i 2 x + = −1 =⇒ x + x + 1 = 0 =⇒ x = . x 2 Using α2 = 4, we have x+

√ 1 = 4 =⇒ x2 − 4x + 1 = 0 =⇒ x = 2 ± 3. x

These give the four roots of x4 − 3x3 − 2x2 − 3x + 1 = 0. Exercise: 10 Section 7.2 Question: (Even Quartic Polynomials)

p p √ √ a) Consider a complex number of the form a + b c, where a, b, c ∈ Q. Prove that a + b c is the root of an even (all powers are even) quartic polynomial. p √ b) Prove that the roots of of any even quartic polynomial are of the form ± a ± b c with a, b, c ∈ Q. p √ c) Let α = a + b c. Show that all the roots of mα,Q (x) are in Q(α) if and only if a2 − cb2 ∈ Q(α).

7.2. ALGEBRAIC EXTENSIONS

367

Solution: p √ a) Consider an element α = a + b c, where a, b, c ∈ Q. Then, √ √ α2 = a + b c =⇒ α2 − a = b c =⇒ α4 − 2aα2 + a2 − b2 c = 0. Thus α is the root of a quartic even polynomial. b) Consider now a polynomial of the form Ax4 + Bx2 + C. Using the quadratic formula, we can solve first for x2 to get r B B 1 p 2 1 p 2 2 x =− B − 4AC =⇒ x = ± − B − 4AC. ± ± 2A 2A 2A 2A Thus all the solutions have the desired form. p p p √ √ √ c) Let α = a + √ b c. The four roots of mα,Q (x) are ±α and ± a − b c. Call β = a − b c. We note that β 2 = a − b √c = 2a − α2 . Thus, if β ∈ / Q(α), then √ β is the root of a quadratic polynomial in Q(α)[x]. However, αβ = a2 − b2 c so β ∈ Q(α) if and only if a2 − b2 c ∈ Q(α). Exercise: 11 Section 7.2 Question: Let L be an algebraic extensions of a field F and let K1 , K2 be two subfields of L containing F . Prove that K1 ∩ K2 is a field extension of F and that [K1 ∩ K2 : F ] divides gcd([K1 : F ], [K2 : F ]). Solution: As additive subgroups of (L, +), the intersection K1 ∩ K2 is an additive subgroup of L. Furthermore, since F ⊆ K1 and F ⊆ K2 , then F ⊆ K1 ∩K2 . Let a, b ∈ K1 ∩K2 −{0}, which is nonempty since 1 ∈ K1 ∩K2 −{0}. Then ab−1 ∈ K1 − {0} since (K1 − {0}, ×) is a subgroup of (L, ×) and similarly for K2 − {0}. Thus, ab−1 ∈ K1 ∩ K2 − {0}. Therefore K1 ∩ K2 − {0} is a multiplicative subgroup of L. For all elements a, b, c ∈ K1 ∩ K2 , the condition of distributivity of × over + is also satisfied. Hence, K1 ∩ K2 is a subfield of L and it contains F so K1 ∩ K2 is a field extension of L. By Theorem 7.2.9, [K1 : F ] = [K1 : K1 ∩ K2 ][K1 ∩ K2 : F ] [K2 : F ] = [K2 : K1 ∩ K2 ][K1 ∩ K2 : F ]. In particular, [K1 ∩ K2 : F ] divides both [K1 : F ] and [K2 : F ], so it divides gcd([K1 : F ], [K2 : F ]). Exercise: 12 Section 7.2 √ √ √ Question: Let F = Q( r1 , r2 , . . . , rn ) where ri ∈ Q. a) Prove that [F : Q] = 2k for some nonnegative k. √ b) Deduce that 3 7 ∈ / F. √ √ √ Solution: Let F = Q( r1 , r2 , . . . , rn ) where ri ∈ Q. √ a) Consider the chain of nested field extensions of F0 = Q, defined inductively by Fi = Fi−1 ( ri ) for 1 ≤ i ≤ n. Note that F = Fn . By Theorem 7.2.9, [F : Q] = [Fn : Fn−1 ][Fn−1 : Fn−2 ] · · · [F2 : F1 ][F1 : F0 ] For each i with 1 ≤ i ≤ n, either

√

ri ∈ Fi−1 or

√

ri ∈ / Fi−i . If

√

(7.1)

ri ∈ Fi−1 , then

√ [Fi : Fi−1 ] = [Fi−1 ( ri ) : Fi−1 ] = 1. √ √ On the other hand, if ri ∈ / Fi−1 , then [Fi : Fi−1 ] > 1. However, in Fi−1 [x], the element ri is a root of the polynomial x2 − ri . Hence, [Fi : Fi−1 ] = 2. We deduce that [Fi : Fi−1 ] is equal to 1 or 2. Hence, by (7.1), [F : Q] is a power of 2. √ √ √ b) The minimal polynomial of 3 7 over Q is x3 − 7 so [Q( 3 7) : Q] = 3. Assume that Q( 3 7) is a subfield of F , then √ √ √ 3 3 3 [F : Q] = [F : Q( 7)][Q( 7) : Q] =⇒ 2k = 3[F : Q( 7)]. √ This is a contradiction so 3 7 ∈ / F.

368

CHAPTER 7. FIELD EXTENSIONS

Exercise: 13 Section 7.2 Question: Prove Proposition 7.2.21. Solution: Let K1 and K2 be extensions of a field F , both subfields of an extension L. Let {α1 , α2 , . . . , αm } be a basis of K1 over F and let {β1 , β2 , . . . , βn } be a basis of K2 over F . We claim that the set B = {αi βj | 1 ≤ i ≤ m, 1 ≤ j ≤ n} spans K1 K2 as a vector space over F . We consider the set S ⊆ L of linear combinations γ=

m X n X

such that cij ∈ F.

cij αi βj

i=1 j=1

The set S is obviously nonempty since it contains K1 , K2 , and F among other subsets. We also note that S ⊆ K1 K2 . Furthermore, S is closed under subtraction so by the One-Step Subgroup Criterion it is an additive subgroup of L. The product of any two linear combinations is a sum of the form m X n X m X n X

cij dkl αi αk βj β` .

i=1 j=1 k=1 `=1

However, each product αi αk is in K1 so is a linear combination of the αi and likewise, each βj β` is again a linear combination of the βj . Hence, a product of these linear combinations is again a linear combination in B. Thus S is closed under multiplication. By properties inherited from the field L, in S multiplication is distributive over addition. We now show that S is closed under taking inverses as well. We can write  !  n m X X γ= cij αi βj  j=1

i=1

so that γ is a K1 linear combination of the βj . As such, γ ∈ K1 (β1 , β2 , . . . , βn ). Since βj are all algebraic over F (otherwise [K2 : F ] would not be finite) and therefore they are algebraic over K1 , we know that K1 (β1 , β2 , . . . , βn ) = K1 [β1 , β2 , . . . , βn ]. Then 1/γ ∈ K1 [β1 , β2 , . . . , βn ]. However, since products of the βj and hence powers thereof are again linear combinations of {β1 , β2 , . . . , βn }, we deduce that 1/γ can again be written as K1 -linear combinations of {β1 , β2 , . . . , βn } and therefore 1/γ can be written as an F -linear combination of B. Thus for all γ ∈ S, we also have 1/γ ∈ S. We have proven that S is a subfield of L. It is a field containing both K1 and K2 so by the definition of the composite, we see that K1 K2 ⊆ S. We already have the reverse inclusion, so we deduce that S = K1 K2 . We can conclude that B spans K1 K2 as a vector space over F . However, the above proof does not show that B is linearly independent. (And indeed, in general, B is not linearly independent.) Thus, the dimension of K1 K2 over F is less than mn. This means that [K1 K2 : F ] ≤ [K1 : F ][K2 : F ].

Exercise: 14 Section 7.2 Question: Suppose that L/F is a field extension of degree p, with p prime. Prove that any subfield K of L containing F is either L or F . Solution: By Theorem 7.2.9, for any field extension K of F in L we have p = [L : F ] = [L : K][K : F ]. Thus, either [K : F ] = 1, in which cases K = F , or [L : K] = 1, in which cases L = K. Exercise: 15 Section 7.2 Question: Let F be a field and consider a simple extension F (α) such that [F (α) : F ] is odd. Prove that F (α) = F (α2 ). Solution: Let F be a field and consider a simple extension F (α) such that [F (α) : F ] is odd. The element α is the root of the x2 − α2 . Hence, α is a root of a quadratic polynomial in F (α2 )[x]. Consequently, [F (α) : F (α2 )] is equal to 1 or 2. But F ⊆ F (α2 ) ⊆ F (α) so by Theorem 7.2.9, [F (α) : F ] = [F (α) : F (α2 )][F (α2 ) : F ].

7.2. ALGEBRAIC EXTENSIONS

369

But since [F (α) : F ] is odd, then [F (α) : F (α2 )] cannot be 2. Thus [F (α) : F (α2 )] = 1 and so F (α) = F (α2 ). Exercise: 16 Section 7.2 √ √ √ Question: Prove that the composite field of Q( 2) and Q( 3 3) is Q( 6 72). √ √ Solution: Consider K1 = Q( 2) and K2 = Q( 3 3) as field extensions of Q. The composite field will need to √ √ contain 2 3 3. We can write this as √ √ √ √ √ 3 6 6 6 2 3 = 8 9 = 72. √ √ √ √ √ √ √ 1 6 So Q( 6 72) ⊆ K1 K2 . On the other hand, ( 6 72)4 = 22 · 3 3 3. Thus, 3 3 = 12 ( 72)4 and so 3 3 ∈ Q( 6 72). √ √ √ √ √ √ √ √ Thus, K2 ⊆ Q( 6 72). Since 3 √ 3 ∈ Q( 6 72), we also have 2 = 6 72/ 3 3 ∈ Q( 6 72). Hence, K1 ⊆ Q( 6 72). √ We conclude that since Q( 6 72) is a field extension that contains both K1 and√K2 , we have K1 K2 ⊆ Q( 6 72). We already had the reverse inclusion which establishes the equality K1 K2 = Q( 6 72). Exercise: 17 Section 7.2 Question: Let K/F be a field extension and let α, β ∈ K with degrees n1 and n2 respectively over F . Show that if gcd(n1 , n2 ) = 1, then [F (α, β) : F ] = n1 n2 . Solution: By Theorem 7.2.9, we have [F (α, β) : F ] = [F (α, β) : F (α)][F (α) : F ] [F (α, β) : F ] = [F (α, β) : F (β)][F (β) : F ]. Since n1 = [F (α) : F ] and n2 = [F (β) : F ], we see that n1 and n2 both divide [F (α, β) : F ]. Thus, [F (α, β) : F ] is a multiple of lcm(n1 , n2 ) = n1 n2 , where the latter equality holds because gcd(n1 , n2 ) = 1. But by Theorem 7.2.13, [F (α, β) : F ] ≤ n1 n2 so we deduce that [F (α, β) : F ] = n1 n2 . Exercise: 18 Section 7.2 √ Question: Prove that [Q(x, 1 − x2 ) : Q(x)] = 2. √ Solution: Setting α = 1 − x2 , we see that α2 + (x2 − 1) = 0, so α is the root of the polynomial y 2 + (x2 − 1) in Q(x)[y]. Since Q[x] is a UFD with field of fractions Q(x). We can use the Rational Root Theorem to show that the possible solutions to y 2 + (x2 − 1) in Q(x) are c, c(x − 1), c(x + 1) and c(x2 − 1) with c ∈ Q − {0} as these the possible combination of units multiplied by factors of (x2 − 1) divided by factors of 1. Obviously for no nonzero rational c do we have c2 +(x2 −1) = 0, as a 0-polynomial. With y = c(x−1), we have 2 c (x − 1)2 + (x2 − 1). This cannot be the zero polynomial since the x2 component is (c2 + 1), a nonzero rational. The same holds for c(x + 1). if Y = c(x2 − 1), we have c2 (x4 − 2x2 − 1) + (x2 − 1) = c2 x4 + (1 − 2c2 )x2 − 1 − c2 . Again, this cannot be the 0 polynomial. Since y 2 + (x2 − 1) in Q(x)[y] has no roots in Q(x), it is an irreducible √ quadratic. Consequently, it is the unique monic irreducible polynomial with α as a root. Thus, [Q(x, 1 − x2 ) : Q(x)] = degQ(x) α = 2. Exercise: 19 Section 7.2 Question: Let pn be the nth prime number (so p1 = 2, p2 = 3, p3 = 5, and so on). √ √ √ √ √ √ √ a) Prove that pn ∈ / Q( p1 , p2 , . . . , pn−1 ). [Hint: Use a proof by induction on n. Use F−1 = Q( p1 , p2 , . . . , pn−1 ) √ √ √ and F0 = Q( p1 , p2 , . . . , pn ).] √ √ √ b) Deduce that [Q( p1 , p2 , . . . , pn ) : Q] = 2n for all positive integers n. √ c) Deduce that Q( p | p ≥ 2 is a prime) is an algebraic extension of Q of infinite degree. Solution: See the solution provided in 7.1.23. Exercise: 20 Section 7.2 √ √ √ √ Question: Show that 3 2 ∈ / Q( 3 3). Prove also that [Q( 3 2, 3 3) : Q] = 9. √ √ √ √ √ Solution: Showing that 3 2 ∈ / Q( 3 3) is rather challenging. Assuming that 3 2 ∈ Q( 3 3) means that 3 2 = √ √ 2 a + b 3 3 + c 3 3 for some a, b, c ∈ Q. Cubic this gives √ √ 2 3 3 2 = (a3 + 3b3 + 9c3 + 14abc) + (3a2 b + 5b2 c + 9c2 a) 3 + (3a2 c + 3b2 a + 5c2 b) 3 .

370

CHAPTER 7. FIELD EXTENSIONS

√ √ 2 √ Thus, since {1, 3 3, 3 3 } is a basis of Q( 3 3) over Q, then we deduce that  3 3 3  a + 3b + 9c + 14abc = 2 2 2 3a b + 5b c + 9c2 a = 0   2 3a c + 3b2 a + 5c2 b = 0. It now takes some effort to eliminate variables, but it is possible. (Quick techniques to simplify this system of equations are described in Chapter 12.) However, from this system, it is possible to show that a must satisfy the polynomial equation 125000000 − 682500000a3 + 1554012000a6 − 1103851736a9 + 1258345608a12 − 774438264a15 + 132863447a18 . Hence, we see that a3 must be a root of the degree 6 polynomial p(x) = 132863447x6 − 774438264x5 + 1258345608x4 − 1103851736x3 + 1554012000x2 − 682500000x + 125000000. We are looking for rational values of a, b, c. If a is rational then so is a3 . So we are looking for roots of p(x) that are cubes of rational numbers. Note that the prime factorization of 132863447 is 132863447 = 19 × 6992813, whereas the prime factorization of 125000000 is 125000000 = 26 × 59 . By the Rational Root Theorem, we see that the possible rational roots of p(x) that a rational cubes are ±1, ±8, ±64, ±125, ±1000, ±8000, ±15625, ±125000, ±1000000, ±1953125, ±15625000, and ±125000000. We can test all these numbers and we find that none of them are roots of p(x). This proves that the degree 18 polynomial in a has no rational roots. Thus √ √ 3 2∈ / Q( 3 3). √ √ √ √ Since degQ 3 2 = degQ 3 3, we know by Theorem 7.2.13 that [Q( 3 2, 3 3) : Q] is a multiple of 3 and divides √ √ √ √ √ 9. From the result that 3 2 ∈ / Q( 3 3), we√ deduce that [Q( 3 2, 3 3) √ : Q( 3 3)] > 1 so it can be either 2 or 3. √ √ √ Obviously, 3 2 is the root of x3 − 2. If [Q( 3 2, 3 3) : Q( 3 3)] = 2, the 3 2 would need to be a root of a quadratic √ polynomial, which would mean that x3 − 2 would factor into a√quadratic and a linear polynomial in Q( 3 3)[x]. √ This would mean that one of the other roots of x3 − 2 besides√3 2 √ would be√in Q( 3 3). However, the other roots √ 3 3 3 3 of x − 2 are complex roots, whereas Q( 3) is real. Thus [Q( 2, 3) : Q( 3 3)] 6= 2 and so this degree has to be 3. Then √ √ √ √ √ √ 3 3 3 3 3 3 [Q( 2, 3) : Q] = [Q( 2, 3) : Q( 3)][Q( 3) : Q] = 3 × 3 = 9.

Exercise: 21 Section 7.2 √ √ Question: Let S = { n 2 | n ∈ Z with n ≥ 2}. Prove that 3 ∈ / Q[S]. √ √ √ √ √ n Solution: Let S = { 2 | n ∈ Z with n ≥ 2}. √For n ≥ 2, let us call Kn = Q( 2, 3 2, . . . , n 2). If 3 ∈ Q[S], then it is a finite Q-linear combination of some ni 2 with n1 , n2 , . . . , nk integers greater or equal to 2 and therefore it is in Kn , where n = max(n1 , n2 , . . . ,√ nk ). It is not hard to see that Kn = Q( m 2), where m √ = lcm(2, 3, . . . , n). Consequently, if n is not a prime power, then Kn = Kn−1 . √ We prove by √ induction on n that 3∈ / Kn for all n ≥ √ √ 2. We know that 3 ∈ / K√ 3p= a + b 2, then 3 = a2 + 2b2 + 2ab 2, which forces a = 0 or b = 0, neither 2 . If of which is possible since 3 and 3/2 are irrational numbers. √ √ Suppose now that 3 ∈ / Kn . If n + 1 is not a prime power, then K 3∈ / Kn+1 . If n is an odd n+1 = Kn√so √ √ p n n/p k prime power with say n = p , then [K : K ] = p since 2 − 2. Since 3 solves a quadratic equation n√ n−1 √ x2 − 3 = 0 so [Kn ( 3) : Kn ] = 2. Then 3 ∈ / Kn+1 because if we assume otherwise, then √ √ √ p = [Kn+1 : Kn ] = [Kn+1 : Kn ( 3)][Kn ( 3) : Kn ] = 2[Kn+1 : Kn ( 3)], √ k which is a contradiction since p is odd. If n + 1 = 2k for some k, then Kn+1 = Kn ( 2 2) and [Kn+1 : Kn ] = 2 √ √ √ k by the same comment as for odd prime powers. Assuming 3 ∈ Kn+1 , we can write 3 = α + β 2 2 with α, β ∈ Kn . Then √ √ 2k 2k α2 + 2β 2 + 2αβ 2 = 3 + 0 2. √ Thus, either α or β is 0. If β = 0,√ then α2 = 3, which implies that 3 ∈ Kn which contradicts √ √ √ the induction hypothesis. If α = 0, then 2β 2 = 3, which implies that 2β = 3, which implies again that 3 ∈ Kn , again is a contradiction. √ √ We conclude that 3 ∈ / Kn for all n ≥ 2. Hence, 3 ∈ / Q[S].

7.2. ALGEBRAIC EXTENSIONS

371

Exercise: 22 Section 7.2 Question: Prove that cos(kπ/n) is algebraic for all positive integers k and n. Solution: The function cos(x) has properties that cos(mx) = Tm (cos(x)), where Tm (y) is a polynomial (called the mth Tchebyshev polynomial). For example, since cos(2x) = 2 cos2 x − 1, then T2 (x) = 2x2 − 1. Also

cos(3x) = cos(2x) cos x−sin(2x) sin x = (2 cos2 x−1) cos x−2 sin x cos x sin x = 2 cos3 x−cos x−2(1−sin2 x) cos x = 4 cos3 x−3 c Thus, T3 (x) = 4x3 − 3x. In fact, using trigonometric addition formulas, we can determine Tm (x) recursively as T0 (x) = 1, T1 (x) = x, and Tm (x) = 2xTm−1 (x) − Tm−2 (x) for all m ≥ 2. We see that Tm (x) ∈ Q[x] and in fact in Z[x]. We note that Tn cos kπ = cos(kπ), which has a value of 1 or −1. Hence, cos(kπ/n) is a root of the n polynomial Tn (x) − cos(kπ). This is a polynomial of degree n. Exercise: 23 Section 7.2 Question: In this exercise, we prove that the set of algebraic numbers Q is countable. Recall that Q is countable. a) Prove that A1 , A2 , . . . , An are n countable sets, then the Cartesian product A1 ×A2 ×· · ·×An is a countable set. b) Prove that the set of polynomials Q[x] is a countable set. c) Since algebraic numbers consist of all the roots polynomials in Q[x], deduce with a proper proof that Q is countable. Solution: In this exercise, we prove that the set of algebraic numbers Q is countable. a) It is a known result (see Exercise 1.2.7) that if A and B are countable sets, then A × B is countable. Let A1 , A2 , . . . , An be n countable sets. The above result establishes a base case of A1 × A2 is countable. Now call B i = A1 × A2 × · · · × Ai . If Bi is countable, then Bi+1 = Bi × Ai+1 and hence Bi+1 is countable. By induction, Bn is countable for all integers n and any collection of countable sets A1 , A2 , . . . , An . b) Define the set S0 as all constant polynomials, including 0. This is in bijection with Q so is countable. For any positive integer n, define the set Sn of polynomials of degree exactly n. Note that Sn is countable since p(x) = pn xn + · · · + p1 x + p0 with pn ∈ Q∗ and pi ∈ Q for 0 ≤ i ≤ n − 1 and thus Sn is in bijection with Q∗ × Qn . Suppose that gn : N∗ → Sn is a bijection from N∗ to Sn that gives an enumeration of polynomials in Sn . Now let f : N∗ → N × N∗ be a bijection, where we write f (n) = (f1 (n), f2 (n)) in components. We create a bijection between F : N∗ → Q[x] via by F (m) is the polynomial F (m) = gf1 (m) (f2 (m)). This function is obviously surjective onto Q[x] and we can also see that it is injective as well. Hence, it is a bijection. c) We construct an enumeration of all elements in Q as follows. For each positive integer n, successively list the distinct roots of the polynomial f (n) (where f is the function in part (b)) unless that complex number has already been listed. In so doing, we eventually arrive at every polynomial in Q[x] and hence we eventually arrive at every algebraic element over C. Exercise: 24 Section 7.2 Question: Prove that the field of fractions of the algebraic integers (introduced in Section 6.7) is the field of algebraic numbers Q. Solution: Recall that the ring of algebraic integers OC consist of the subring of C of elements that are roots of monic polynomial equations in Z[x]. The ring OC is obviously a subring of Q, which consists of all algebraic elements over Q. The field of fractions of OC is therefore a subfield of Q. To show that every algebraic number is a fraction of algebraic integers, we will show that every algebraic number is in fact a fraction of an algebraic integer by a regular integer. Let α ∈ Q be an algebraic number and let mα,Q (x) ∈ Q[x] be its minimal polynomial. Suppose that mα,Q (x) = xn + cn−1 xn−1 + · · · + c1 x + c0

372

CHAPTER 7. FIELD EXTENSIONS

and let d be the least common multiple of all the denominators of c0 , c1 , . . . , cn−1 . Consider the polynomial x f (x) = dn mα,Q d n x n−1 x n x =d + dn c0 + dn cn−1 + · · · + dn c1 d d d = xn + dcn−1 xn−1 + d2 cn−2 xn−2 + · · · + dn−1 c1 x + dn c0 . Since d clears all the denominators, we note that f (x) is a monic polynomial in Z[x]. Furthermore, f (dα) = dn mα,Q (α) = 0. Hence, dα is an algebraic integer. Thus, α is an algebraic integer divided by the integer d. Thus, Q is a subfield of the field of fractions of OC , and thus, since we already have the reverse inclusion, they are equal. Exercise: 25 Section 7.2 Question: Let F be a field and let α be an algebraic element over F for which we may not know the minimal polynomial mα,F (x). Consider the linear transformation fα : F (α) → F (α) defined by fα (x) = αx in Exercise 7.1.16. a) Prove that α is an eigenvalue of fα . b) Deduce that α is a root of the characteristic polynomial for the linear transformation fα . Solution: Let F be a field and let α be an algebraic element over F . Consider the linear transformation fα : F (α) → F (α) defined by fα (x) = αx in Exercise 7.1.16. a) It is obvious that α is an eigenvalue of fα (x) since fα (x) = αx for all x ∈ F (α). b) The point then is that according to Exercise 7.1.16, fα (x) is an F -linear transformation on F (α), a vector space over F . An eigenvalue must solve the characteristic equation of fα (x). Exercise: 26 Section 7.2 √ √ 2 Question: Use the result of Exercise 7.2.25 to find the minimal polynomial of β = 1 + 3 7 − 3 7 over Q. √ √ √ 2 Solution: Using the basis B = {1, 3 7, 3 7 } of Q( 3 7) as a vector space over Q, we note that √ √ 2 3 3 β·1=1+ 7− 7 √ √ √ 2 3 3 3 β · 7 = −3 + 7 + 7 √ √ √ 2 2 3 3 3 β· 7 =3−3 7+ 7 . Consequently, the matrix of fβ (x) with respect to the basis B is   1 −3 3 1 1 −3 . −1 1 1 The characteristic equation for this is x−1 −1 1

3 x−1 −1

−3 3 = x3 − 3x2 + 12x − 4 x−1

The Rational Root Theorem shows that this does not have any rational roots. Thus, since it is a cubic, it is irreducible. This is the minimal polynomial of β over Q. Exercise: 27 Section 7.2 √ √ Question: Use the result of Exercise 7.2.25 to find the minimal polynomial of β = 2 − 3 2 + 4 2 over Q. √ √ √ 3 √ Solution: Using the basis B = {1, 4 2, 2, 4 2 } of Q( 4 2) as a vector space over Q, we note that √ √ 2 4 4 β·1=2+ 2−3 2 √ √ √ √ 2 3 4 4 4 4 β· 2=2 2+ 2 −3 2 √ √ √ 2 2 3 4 4 4 β · 2 = −6 + 2 2 + 2 √ √ √ 3 3 4 4 4 β· 2 =2−6 2+2 2

7.2. ALGEBRAIC EXTENSIONS

373

Consequently, the matrix of fβ (x) with respect to the basis B is   2 0 −6 2 1 2 0 −6  . −3 1 2 0 0 −3 1 2 The characteristic equation for this is x4 − 8x3 − 12x2 + 136x + 146. Using Eisenstein’s Criterion with p = 2, we immediately see that this polynomial is irreducible. This is the minimal polynomial of β over Q. Exercise: 28 Section 7.2 Question: Prove Corollary 7.2.26. Solution: Let b > 2 be an integer and let {ak }k≥1 be a sequence of integers with values from {0, 1, 2, . . . , b}. Consider the series ∞ X ak . bk! k=1

We would like to apply the Ratio Test but some of the terms ak may be 0. Instead, we apply the Ratio Test on +1 . We have the terms ck = akbk! k!k̇ ak+1 +1 ak+1 + 1 1 (k+1)! lim bak +1 = lim k→∞ ak + 1 k→∞ b bk! P +1 1 Since b+1 ≤ aak+1 ≤ b + 1, we see that limk→∞ |ck+1 /ck | = 0 so the series k ck converges. However, k +1 ∞ X ak k=1

and is a series of nonnegative terms the series Let us call sn =

n X ak k=1

bk!

b

< k!

∞ X ak k=1

bk!

∞ X

ck

k=1

also converges to real number α.

this is a rational number that in reduced form has a denominator of bn! . Then

α − sn =

X k=n+1

X ak ak 1 ∞ k!−n·n! ∞ k! = n·n! b b b k=n+1

But, with the supposition that n ≥ 2, ∞ (n+k)!−n·n! X X ak 1 b ∞ k!−n·n! < ∞ k!−n·n! = b b b b i=1 k=n+1 k=n+1 k ∞ ∞ (n+k)(n+k−1)···(n+1)−n X X 1 1 b =b <b = n! . n! n! b b b −1 i=1 i=1

X

Thus,

n 1 b . bn! bn! − 1 Now by Liouville’s Theorem, if α is algebraic of degree m, there exists a real number A > 0 such that for all integers p, q p A α− > m. q q Consider the element α defined above. For all integers m ≥ 2 and all positive real numbers A, there exists an integer n such that n ≥ m and bn!b−1 < A. Then

α − sn <

α − sn < where q = bn! , the denominator of sn .

A qm

374

CHAPTER 7. FIELD EXTENSIONS

7.3 – Solving Cubic and Quartic Equations Exercise: 1 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 15x + 10 = 0 Solution: The discriminant is ∆ = −27 · 102 − 4 · (−15)3 = 10800. By Theorem 7.3.2, the equation has 3 real roots. We calculate that q q √ √ 3 3 u0 = −5 + −100 and v0 = −5 − −100 √ √ ⇐⇒u0 = 3 −5 + 10i and v0 = 3 −5 − 10i. The three roots of the equation are √ √ x1 = 3 −5 + 10i + 3 −5 − 10i √ ! −1 + 3i √ 3 x2 = −5 + 10i + 2 √ ! −1 − 3i √ 3 −5 + 10i + x3 = 2

√ ! −1 − 3i √ 3 −5 − 10i 2 √ ! −1 + 3i √ 3 −5 − 10i. 2

In order to write these without reference to complex numbers, note first that √ √ −5 + 10i = 25 + 100eiθ = 5 5eiθ where θ = tan−1 (−2) + π so that

√ 3

−5 + 10i =

√

1 1 5 cos( tan−1 (−2) + π/3) + i sin( tan(−2) + π/3) . 3 3

√

1 1 5 cos( tan−1 (−2) + π/3) − i sin( tan(−2) + π/3) . 3 3

Then we also have √ 3

−5 − 10i =

√ This gives us x1 = 2 5 cos( 31 tan−1 (−2) + π/3). For the other roots, we have √ 1 1 −1 x2 = (cos(2π/3) + i sin(2π/3) 5 cos( tan (−2) + π/3) + i sin( tan(−2) + π/3) 3 3 √ 1 1 −1 + (cos(2π/3) − i sin(2π/3) 5 cos( tan (−2) + π/3) − i sin( tan(−2) + π/3) 3 3 √ 1 1 −1 −1 = 2 5 cos(2π/3) cos( tan (−2) + π/3) − sin(2π/3) sin( tan (−2) + π/3) 3 3 √ 1 = 2 5 cos( tan−1 (−2) + π), 3 and similarly

√ 1 x3 = 2 5 cos( tan−1 (−2) + 5π/3). 3

Exercise: 2 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 + 6x − 2 = 0 Solution: The discriminant is ∆ = −27 · (−2)2 − 4 · 63 = −972. By Theorem 7.3.2, the equation has 1 real root and 2 complex roots. We calculate that q q √ √ 3 3 u0 = 1 + 9 and v0 = 1 − 9 √ √ 3 3 ⇐⇒u0 = 4 and v0 = − 2.

7.3. SOLVING CUBIC AND QUARTIC EQUATIONS

375

The three roots of the equation are x1 =

√ 3

x2 = x3 =

4−

√ 3

2 √ ! −1 + 3i √ 3 4− 2 √ ! −1 − 3i √ 3 4− 2

√ ! −1 − 3i √ 3 2 2 √ ! −1 + 3i √ 3 2. 2

We can write these more simply as √ 3

√ 3 4− 2 √ √ √ 1 √ 3 √ 3 3 3 3 x2 = (− 4 + 2) + ( 4 + 2)i 2 2 √ √ √ 3 √ 1 √ 3 3 3 3 x3 = (− 4 + 2) − ( 4 + 2)i. 2 2 x1 =

Exercise: 3 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 9x + 10 = 0 Solution: The discriminant is ∆ = −27 · 102 − 4 · (−9)3 = 216. By Theorem 7.3.2, the equation has 3 real roots. We calculate that q q √ √ 3 3 u0 = −5 + −2 and v0 = −5 − −2. The three roots of the equation are q q √ √ 3 3 x1 = −5 + 2i + −5 − 2i √ !q √ −1 + 3i 3 x2 = −5 + 2i + 2 √ !q √ −1 − 3i 3 x3 = −5 + 2i + 2

√ !q √ −1 − 3i 3 −5 − 2i 2 √ !q √ −1 + 3i 3 −5 − 2i. 2

In order to write these without reference to complex numbers, note first that −5 +

√

2i =

√

25 + 2e

iθ

√ = 3 3eiθ

−1

where θ = tan

√ ! 2 +π − 5

which we can also write as −5 +

√

2i = −(5 −

√

√

where α = tan−1

2i) = 3 3eiα

√ ! 2 − . 5

This means that q √ √ 3 u0 = −5 + 2i = − 3 cos

1 tan−1 3

√ = − 3 cos

1 tan−1 3

√ !! √ !!! 1 2 2 −1 − + i sin tan − 5 3 5 √ !! √ !!! 2 1 2 − i sin tan−1 5 3 5

Then we also have q √ √ 3 v0 = −5 − 2i = − 3 cos

1 tan−1 3

√ !! 2 i sin 5

1 tan−1 3

√ !!! 2 . 5

376

CHAPTER 7. FIELD EXTENSIONS

√ √ This gives us x1 = u0 + v0 = −2 3 cos 31 tan−1 52 . For the other roots, we have √ x2 = −2 3 cos

1 tan−1 3

√ x3 = −2 3 cos

1 tan−1 3

! √ ! 2π 2 + 5 3 ! √ ! 2 4π + . 5 3

Exercise: 4 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 6x + 4 = 0 Solution: The discriminant is ∆ = −27 · 42 − 4 · (−6)3 = 432. By Theorem 7.3.2, the equation has 3 real roots. We calculate that q q √ √ 3 3 u0 = −2 + −4 and v0 = −2 − −4 √ √ ⇐⇒u0 = 3 −2 + 2i and v0 = 3 −2 − 2i. The three roots of the equation are √ √ x1 = 3 −2 + 2i + 3 −2 − 2i √ ! −1 + 3i √ 3 −2 + 2i + x2 = 2 √ ! −1 − 3i √ 3 x3 = −2 + 2i + 2

√ ! −1 − 3i √ 3 −2 − 2i 2 √ ! −1 + 3i √ 3 −2 − 2i. 2

In order to write these without reference to complex numbers, note first that √ √ −2 + 2i = 4 + 4eiθ = 8ei3π/4 so that

π π √ √ u0 = 3 −2 + 2i = 2 cos + i sin = 1 + i. 4 4 √ Then we also have v0 = 3 −2 − 2i = 1 − i. Thus, the three roots are x1 = (1 + i) + (1 − i) = 2 √ ! −1 + 3i x2 = (1 + i) + 2 √ ! −1 − 3i x3 = (1 + i) + 2

√ ! √ −1 − 3i (1 − i) = −1 − 3 2 √ ! √ −1 + 3i (1 − i) = −1 + 3. 2

(We would have been able to solve this equation without the Cardano-Ferrari method by first noticing that the equation has a rational root of 2.) Exercise: 5 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 12x + 8 = 0 Solution: The discriminant is ∆ = −27 · 82 − 4 · (−12)3 = 5184. By Theorem 7.3.2, the equation has 3 real roots. We calculate that q q √ √ 3 3 u0 = −4 + −48 and v0 = −4 − −48 q q √ √ 3 3 ⇐⇒u0 = −4 + 4 3i and v0 = −4 − 4 3i p p 3 3 ⇐⇒u0 = 8e2πi/3 = 2e2πi/9 and v0 = 8e−2πi/3 = 2e−2πi/9 .

7.3. SOLVING CUBIC AND QUARTIC EQUATIONS

377

The three roots of the equation are x1 = 2e2πi/9 + 2e−2πi/9 = 4 cos

2π 9

8π 9 4π . x3 = e−2πi/3 2e2πi/9 + e2πi/3 2e−2πi/9 = 4 cos 9

x2 = e2πi/3 2e2πi/9 + e−2πi/3 2e−2πi/9 = 4 cos

Exercise: 6 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 12x + 16 = 0 Solution: The discriminant is ∆ = −27 · 162 − 4 · (−12)3 = 0. By Theorem 7.3.2, the equation has a double root, which will be real, along with another real root. We calculate that √ √ u0 = 3 −8 + 0 and v0 = 3 −8 − 0 ⇐⇒u0 = −2

and v0 = −2.

The three roots of the equation are x1 = −2 + (−2) = −4 √ ! −1 + 3i x2 = −2 −2 2 √ ! −1 − 3i −2 x3 = −2 2

√ ! −1 − 3i =2 2 √ ! −1 + 3i = 2. 2

We could have solved this cubic equation using the Rational Root Theorem. Exercise: 7 Section 7.3 Question: For the equation x3 + 3x2 + 12x + 4 = 0, solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. Solution: The first step is to do a variable shift of x = y − 1. So we have (y − 1)3 + 3(y − 1)2 + 12(y − 1) + 4 = 0 =⇒ y 3 + 9y − 6 = 0. This equation has the discriminant of ∆ = −27q 2 − 4p3 = −27 · 36 − 4 · 729 = −3888. The u and v terms in Cardano’s method are q q p p 3 3 u = 3 + −∆/108 and v = 3 − −∆/108 so u=

√ 3

9

and

√ 3 v = − 3.

So the roots of the polynomial in y are 1 real and 2 complex roots: √ ! √ ! √ √ √ √ √ −1 + 3i −1 − 3i 3 3 3 3 3 y1 = 9 − 3, y2 = 9 − 3 , y3 = 9 2 2

√ ! √ −1 − 3i 3 − 3 2

So the three roots the the original equation are √ √ 3 3 x1 = 9 − 3 − 1 √ √ √ 1 √ 3 √ 3 3 3 3 x2 = (− 9 + 3) + ( 9 + 3)i − 1 2 2 √ √ √ 1 √ 3 √ 3 3 3 3 x3 = (− 9 + 3) − ( 9 + 3)i − 1. 2 2

√ ! −1 + 3i 2

378

CHAPTER 7. FIELD EXTENSIONS

Exercise: 8 Section 7.3 Question: Solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. x3 − 9x2 + 24x − 16 = 0 Solution: In order to approach this equation, we first perform the variable shift with x = y + 3. The equation becomes (y + 3)3 − 9(y + 3)2 + 24(y + 3) − 16 = 0 ⇐⇒ y 3 − 3y + 2 = 0 The discriminant is ∆ = −27 · 22 − 4 · (−3)3 = 0. By Theorem 7.3.2, the equation has a double root and hence two distinct real roots. We calculate that √ √ u0 = 3 −1 + 0 and v0 = 3 −1 − 0 ⇐⇒u0 = −1

and v0 = −1.

The three roots of the equation for y are y1 = −1 + (−1) = −2 √ ! −1 + 3i y2 = − − 2 √ ! −1 − 3i y3 = − − 2

√ ! −1 − 3i =1 2 √ ! −1 + 3i = 1. 2

Consequently, the roots for the original equation are x = 1 and a double root at x = 4. Exercise: 9 Section 7.3 Question: For the equation x4 + 4x2 + 12x + 7 = 0, solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. Solution: This equation has no cubic term so we do not need to shift the variable. We start to work with x4 = −4x2 − 12x − 7 Cardano’s method requires us to add to both side the expression tx2 +

t2 , where t solves the equation 4

t3 − 4t2 − 28t − 32 = 0 After looking for rational roots, we find that t = −2 is a root of this polynomial. Hence, we will add −2x2 + 1 to both sides of the quartic in x to obtain x4 − 2x2 + 1 = −6x2 − 12x − 6 =⇒(x2 − 1)2 = −6(x + 1)2 =⇒(x2 − 1)2 + 6(x + 1)2 = 0 √ √ =⇒(x2 − 1 + 6i(x + 1))(x2 − 1 − 6i(x + 1)) = 0 √ √ √ √ =⇒(x2 + 6ix + ( 6i − 1))(x2 − 6ix + (− 6i − 1)) = 0

We get the following four roots q q √ √ 1 √ 1 √ (− 6i ± −2 − 4 6i), ( 6i ± −2 + 4 6i). 2 2 √ 2 √ We observe the (2 ± 6i) = −2 ± 4 6i so the four roots become √ √ 1 √ 1 √ (− 6i ± (2 − 6i)), ( 6i ± (2 + 6i)) 2 2

7.3. SOLVING CUBIC AND QUARTIC EQUATIONS or in other words 1+

√

6i, −1, 1 −

379 √

6i, −1.

It is possible to verify that x4 + 4x2 + 12x + 7 = (x + 1)2 (x2 − 2x + 7), which justifies the result that we obtained from Cardano’s method. Exercise: 10 Section 7.3 Question: For the equation x4 + 4x2 − 3x + 1 = 0, solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. Solution: This equation has no cubic term so we do not need to shift the variable. We start to work with x4 = −4x2 + 3x − 1. Cardano’s method requires us to add to both side the expression tx2 +

t2 , where t solves the equation 4

t3 − 4t2 − 4t + 7 = 0 We notice that t = 1 is a rational root so we can use this. (Otherwise, the exercise could get very hard.) This is equivalent to the equation 2 1 1 1 1 = −3 x2 − x + x4 + x2 + = −4x2 + 3x − 1 + x2 + ⇐⇒ x2 + 4 4 2 4 2 2 1 1 = −3 x − ⇐⇒ x2 + 2 2 2 2 1 1 ⇐⇒ x2 + +3 x− 2 2 √ ! √ ! √ √ 1 3i 1 3i 2 2 ⇐⇒ x + 3ix + − x − 3ix + + . 2 2 2 2

This gives us the four roots q √ √ 1 − 3i ± −5 + 2 3i 2

and

1 2

√

q 3i ±

√ −5 − 2 3i .

Exercise: 11 Section 7.3 Question: For the equation x4 − 4x3 + 4x2 − 8x + 4 = 0, solve the equation using Cardano-Ferrari methods. For the solutions to a cubic equation, if all the roots are real, then write the solutions without reference to complex numbers. Solution: To solve x4 − 4x3 + 4x2 − 8x + 4 = 0, we first need to shift the variable with x = y + 1. The equation becomes y 4 − 2y 2 − 8y − 3 = 0. This equation has no cubic term so we do not need to shift the variable. We start to work with y 4 = 2y 2 + 8y + 3 Cardano’s method requires us to add to both side the expression ty 2 +

t2 , where t solves the equation 4

t3 + 2t2 + 12t − 40 = 0 We note that t = 2 is a rational root of this resolvent equation so we use this value. The equation for y becomes y 4 = 2y 2 + 8y + 3 ⇐⇒ y 4 + 2y 2 + 1 = 2y 2 + 8y + 3 + 2y 2 + 1 ⇐⇒ (y 2 + 1)2 = 4(y + 1)2 ⇐⇒ (y 2 + 1)2 − (2y + 2)2 = 0 ⇐⇒ (y 2 + 2y + 3)(y 2 − 2y − 1) = 0.

380

CHAPTER 7. FIELD EXTENSIONS

So the four roots for the y equation are −1 ±

√

2i,

1±

√

2.

Consequently, the four roots of the original equation are √ √ 2 ± 2. ± 2i,

Exercise: 12 Section 7.3 Question: Consider the polynomial p(x) = x3 − 6x2 + 11x − 6. 1. Solve the equation via Cardano’s method. 2. Find the rational roots of this polynomial by the Rational Root Theorem. 3. Decide which rational root corresponds to which solution via Cardano’s method. Solution: Consider the polynomial p(x) = x3 − 6x2 + 11x − 6. a) We solve p(x) = 0 via Cardano’s method. This requires a variable shift with x = y + 2. The equation in y becomes y 3 − y = 0. This is a very simple equation that has roots y = −1, 0, 1 so also x = 1, 2, 3. The discriminant is ∆ = 4 and then we have sr s r s 1 i i 1 i 3 3 u0 = − and v0 = − − = 3 − √ =√ = −√ . 27 27 3 3 3 3 By Cardano’s method, the solutions are i i x1 = 2 + u0 + v0 = 2 + √ − √ = 2 3 3 √ ! √ ! −1 + 3i −1 − 3i i i √ − √ x2 = 2 + 2 2 3 3 √ ! √ ! −1 − 3i i i −1 + 3i √ − √ . x3 = 2 + 2 2 3 3 b) We already found the rational roots of 1, 2, and 3. c) We easily found that x1 = 2. Now, i x2 = 2 + √ 3 i x3 = 2 + √ 3

√ √ ! −1 + 3i −1 − 3i − =1 2 2 √ √ ! −1 − 3i −1 + 3i − = 3. 2 2

Exercise: 13 Section 7.3 Question: Let n be a real number and consider the polynomial p(x) = x3 − (3n + 3)x2 + (3n2 + 6n + 2)x − (n3 + 3n2 + 2n). Apply Cardano’s method to solve this equation. After finding the roots, explain why it was so easy to solve. Solution: In order to apply Cardano’s method, we begin with a shift of variable x = y + n + 1. Then the polynomial p(x) becomes p(y + (n + 1)) = y 3 + (3(n2 + 2n + 1) − (3n + 3)(2n + 2) + (3n2 + 6n + 2))y ((n3 + 3n2 + 3n + 1) − 3(n + 1)3 + (3n2 + 6n + 2)(n + 1) − (n3 + 3n2 + 2n)) = y3 − y

7.3. SOLVING CUBIC AND QUARTIC EQUATIONS

381

Hence, the roots of the equation for y are −1, 0, 1 and those for x are n, n + 1, n + 2. This is easy to solve because after the shift of the variable x, we obtain a simple cubic that easily factors into a linear and quadratic equation.

Exercise: 14 Section 7.3 Question: Prove that a palindromic polynomial of odd degree has −1 as a root. Use this and Exercise 7.2.9 to find all the roots to x5 + 2x4 + 3x3 + 3x2 + 2x + 1 = 0. Solution: Let p(x) = an xn + · · · + a1 x + a0 be a palindromic polynomial of odd degree. Since ai = an−1 we have p(−1) = a0 ((−1)n + 1) + a1 ((−1)n−1 + (−1)) + · · · + a(n−1)/2 ((−1)(n+1)/2 + (−1)(n−1)/2 ) = 0 Thus −1 is a root. We now solve x5 + 2x4 + 3x3 + 3x2 + 2x + 1 = 0 by first factoring out x + 1. We get x5 + 2x4 + 3x3 + 3x2 + 2x + 1 = (x + 1)(x4 + x3 + 2x2 + x + 1). We could proceed by solving the quartic using methods from this section but we can also use technique described in Exercise 7.2.9 for palindromic polynomials. We quickly see that 2 ! 1 1 4 3 2 2 + x+ . x + x + 2x + x + 1 = x x+ x x So the roots of x4 + x3 + 2x2 + x + 1 are the four roots resulting from 1 1 = −1 or x+ =0 x x ⇐⇒x2 + x + 1 = 0 or x2 + 1 = 0 √ −1 ± 3i ⇐⇒x = ±i, . 2 x+

√

So the five roots of p(x) are x = −1, ±i, −1±2 3i . Exercise: 15 Section 7.3 Question: Consider the polynomial p(x) = x6 + 4x4 + 4x2 + 1. Use either the strategy provided by Exercise 7.2.9 to find the roots or use Cardano’s method to solve the equation in x2 to find all the roots. Which do you think is easier? Solution: We propose to use Cardano’s method to solve for x2 . Let z = x2 and solve z 3 + 4z 2 + 4z + 1. 4 11 Cardano’s method requires a shift of variables with z = y − 34 . This leads to the equation y 3 − y + . The 3 27 discriminant is then 2 3 11 4 ∆ = −27 −4 − = 5. 27 3 The cubic equation in y (and hence in z) will have 3 real roots. We calculate r q q 11 5 11 1 √ 11 1 √ u0 = − + − =− + 53i and v0 = − − 53i. 54 108 54 6 54 6 The three roots of the equation for z are r r q q 11 1 √ 11 1 √ 4 3 3 53i + − − 53i z1 = + − + 3 54 6 ! 54 6 r √ √ !r q q 4 −1 + 3i 3 11 1 √ −1 − 3i 3 11 1 √ z2 = + − + 53i + − − 53i 3 2 54 6 2 54 6 √ !r √ !r q q 4 −1 − 3i 3 11 1 √ −1 + 3i 3 11 1 √ z3 = + − + 53i + − − 53i 3 2 54 6 2 54 6

382

CHAPTER 7. FIELD EXTENSIONS

√ √ √ The six roots of p(x) are ± z1 , ± z2 , ± z3 . It is unfortunate that it is not obvious that these roots are real, let alone what they are. We now try the alternate method taking advantage of the fact that p(x) is palindromic. We find that 3 ! 1 1 + x+ p(x) = x6 + 4x4 + 4x2 + 1 = x3 x+ x x so that q(x) = x3 + x. The roots of q(x) are easily found to be 0, i, −i. To solve for x, now we have 1 = 0 =⇒ x = ±i x √ 1 1 x + = i =⇒ x2 − ix + 1 = 0 =⇒ x = (1 ± 5)i x 2 √ 1 1 x + = −i =⇒ x2 + ix + 1 = 0 =⇒ x = (−1 ± 5)i x 2 x+

The question of which method is easiest may be subjective, but either way requires solving a cubic equation. Using Cardano’s method to solve for x2 does not require us to find the polynomial q(x) such that p(x) = x3 q(x + 1/x) and we do not have so solve three quadratic equations once we find the roots of q(x). So at the outset, we might suspect that solving for x2 and then using Cardano’s method may be easiest, it turns out for this example that using properties of palindromic polynomials turns out to be√easiest. What √ is not obvious from the first of the above two methods is that z1 , z2 , z3 mus be equal to −1, 21 (1 ± 5), 21 (−1 ± 5) and it is not clear which corresponds to which.

7.4 – Constructible Numbers Exercise: 1 Section 7.4 p √ Question: Find a compass and straightedge construction for the number 3 + 5. p √ Solution: A compass and straightedge construction for the number 3 + 5: • Start with the unit segment OR. ←→ • Construct a point B on the line OR at a distance 3 from O by: (a) first constructing the circle Γ1 of center ←→ R and radius RO; (b) taking A the intersection of Γ1 and OR that is opposite O; (c) constructing the ←→ circle Γ2 of center A and radius AR; define B as the intersection pf Γ2 and OR that is not R. We have OB = 3OR. ←→ • Construct (according to Euclidean geometry constructions) the line L1 perpendicular to OR and through R. √ √ • Define C as one of the two intersections of L1 and Γ1 . By Pythagoras, BC = RA2 + RB 2 = OR2 + 4OR2 = √ 5OR. • Construct the circle Γ3 of center B and radius BC. ←→ • Construct D √ the intersection point of Γ3 and OR that is not in the segment OB. Note that we have OD = (3 + 5)OR. ←→ • Construct the point E on OR that is a distance of OR from O but not opposite R. ←→ • Construct the circle Γ4 of diameter ED and construct the line L2 perpendicular to OR through O. • Define F as one intersections of Γ4 and L2 . According to the strategy given in the proof of Proposition p of the √ 7.4.2, OF = 3 + 5OR.

7.4. CONSTRUCTIBLE NUMBERS

383

Exercise: 2 Section 7.4 Question: Discuss how to construct a regular dodecagon (12 sides) with a compass and straightedge. Solution: Recall the Euclid’s Proposition 1 states that it is possible to construct an equilateral triangle with a given side as a starting segment. (If the segment is AB, then construct a circle Γ1 of center A and radius AB and construct another circle Γ2 of center B and radius BA. Let C be one of the intersection points of Γ1 and Γ2 . Then AB = AC = BC so 4ABC is equilateral.) Let O be the circumcenter of 4ABC (the intersection of the edge bisectors of the triangle) and let Γ be the circumcircle. We also know that it is possible to bisect any angle (Proposition 9). If we bisect ∠AOB, the − → resulting ray O` intersects the circle Γ in a point P1 . Repeating the same process with the angles ∠BOC and ∠COA we produce points P2 and P3 on Γ. Since 4OAP1 ≈ 4OBP1 , we deduce that AP1 = P1 B and similarly BP2 = P2 C and CP3 = P3 A. Thus, AP1 BP2 CP3 is a regular hexagon on the circle Γ. We can then again bisect all the angles of the hexagon through the center O. The rays of each angle bisector intersect Γ at six more points, which along with AP1 BP2 CP3 , for the vertices of a regular dodecagon. Exercise: 3 Section 7.4 Question: This exercise guides a proof that it is impossible to construct a regular heptagon with a compass and straightedge. Set α = 2π 7 . a) Prove that cos(3α) = cos(4α). b) Deduce that cos α solves 8x3 + 4x2 − 4x − 1 = 0. c) Deduce that [Q(cos(α)) : Q] = 3 and explain clearly why this means that the regular heptagon is not constructible with a compass and a straightedge. Solution: Set α = 2π 7 . a) For any angle β, cos β = cos(−β) = cos(2π − β). Hence, cos(3α) = cos(2π − 6π/7) = cos(8π/7) = cos(4α). b) We will use repeatedly the angle addition formula of cos(x + y) = cos x cos y − sin x sin y. With the tripling angle, we have cos(3α) = cos(2α + α) = cos(2α) cos(α) − sin(2α) sin(α) = (2 cos2 α − 1) cos α − 2 sin2 α cos α = 2 cos3 α − cos α − 2(1 − cos2 α) cos α = 4 cos3 α − 3 cos α. For the quadrupling angle, we have cos(4α) = 2 cos2 (2α) − 1 = 2(2 cos2 α − 1)2 − 1 = 8 cos4 α − 8 cos2 α + 1 Since α satisfies cos(4α) = cos(3α) so cos α satisfies 4x3 − 3x = 8x4 − 8x2 + 1 =⇒ 8x4 − 4x3 − 8x2 + 3x + 1. We notice that x = 1 is a root of this equation (which makes sense since 1 = cos 0 and as an angle 4 · 0 = 3 · 0). So x − 1 factors out of the polynomial with 8x4 − 4x3 − 8x2 + 3x + 1 = (x − 1)(8x3 + 4x2 − 4x − 1). Since cos(2π/7) 6= 1, then cos α is a root of 8x3 + 4x2 − 4x − 1. c) By the rational root theorem, the cubic polynomial p(x) = 8x3 + 4x2 − 4x − 1 has for possible rational roots ±1, ± 21 , ± 14 , ± 18 . We have 1 1 1 p(1) = 7, p(−1) = −1, p( ) = −1, p(− ) = 1, p( ) = −13/8, 2 2 4 1 1 1 p(− ) = 1/8, p( ) = −91/64, p(− ) = −29/64. 4 8 8

384

CHAPTER 7. FIELD EXTENSIONS This shows that p(x) has no rational roots. Since p(x) is cubic with no roots in Q, then p(x) is irreducible. Thus [Q(cos(2π/7)) : Q] = 3. Since this degree is not a power of 2, cos(2π/7) is not constructible, so it is impossible to construct a regular heptagon with a compass and a straightedge.

Exercise: 4 Section 7.4 Question: Prove that it is impossible to construct a regular 9-gon with a compass and straightedge. Solution: Let α = 2π/9. Because cos(2π − x) = cos x¡ we deduce that cos(3α) = cos(6α). Since cos(2x) = cos2 x − 1, then cos(3α) solves 2x2 − 1 = x =⇒ 2x2 − x − 1 = 0 =⇒ (2x + 1)(x − 1). Since cos(3α) 6= 1, then cos(3α) = − 21 . By angle addition formulas, we determine that cos(3α) = 4 cos3 α−3 cos α. Hence cos(2π/9) is a solution to the equation 4x3 − 3x = −

1 =⇒ 8x3 − 6x + 1. 2

Using the Rational Root Theorem, we can determine that this cubic has no roots in Q and hence it is irreducible over Q. Thus cos(2π/9) has degree 3 over Q. Consequently, cos(2π/9) is not a constructible number and hence it is impossible to construct a regular 9-gon with compass and straightedge.

Exercise: 5 Section 7.4 Question: Given a circle Γ, is it possible to construct a circle with double the area? If so, provide a compass and straightedge construction. Solution: Suppose that Γ is a circle of center O and radius OR. The length of the radius √ is r = OR √ and the area of the circle is A = πr2 . A circle of double the area must have a radius of length 2r. Since 2 is a constructible number, it is possible to construct a circle of double the area. ←→ Construct a line perpendicular to OR and and through the point O. (Euclid’s Elements Proposition 10). Let ←→ A be one of the points √ between OR and Γ. Since ∠ROA is a right angle and OR = √OA, then by √ of intersection Pythagoras, AR = 1 + 1OR = 2OR. Hence, the circle of center R and radius RA has area π( 2r)2 = 2πr2 , double the area of Γ.

Exercise: 6 Section 7.4 Question: For each nonnegative integer n, let Tn (x) be the Chebyshev polynomial defined by Tn (cos θ) = cos(nθ) for all θ. a) Prove Tmn (x) = Tm (Tn (x)). b) Use this to find the four roots of T4 (x). c) Deduce that T4 (x) is irreducible in Q[x]. d) Also use (a) to find all 6 roots of T6 (x). Solution: For each nonnegative integer n, let Tn (x) be the Chebyshev polynomial defined by Tn (cos θ) = cos(nθ) for all θ. a) Observe that by definition of the functions Tn (x), for all θ ∈ R, Tm (Tn (cos θ)) = Tm (cos(nθ)) = cos(mnθ) = Tmn (cos θ). Thus, as polynomials, Tm (Tn (x)) and Tmn (x) are equal for an infinite value of x, namely all x ∈ [−1, 1]. Thus the polynomial Tm (Tn (x)) − Tmn (x) has an infinite number of zeros and thus must be 0. Therefore, the polynomials Tmn (x) and Tm (Tn (x)) are equal for all x.

7.4. CONSTRUCTIBLE NUMBERS

385

b) From part (a), we deduce that T4 (x) = T2 (T2 (x)). Now, T2 (x) comes from the angle doubling formula of cos(2θ) = 2 cos2 θ − 1, so that T2 (x) = 2x2 − 1. Consequently, T4 (x) = 0 ⇐⇒T2 (T2 (x)) = 0 ⇐⇒2(T2 (x))2 − 1 = 0 1 ⇐⇒T2 (x) = ± √ 2 1 ⇐⇒2x2 − 1 = ± √ 2 s s 1 1 1 1 ⇐⇒x = ± + √ , or ± − √ . 2 2 2 2 2 2 c) The polynomial T4 (x) is T4 (x) = 2(2x2 − 1)2 − 1 = 8x4 − 8x2 − 1. Since T4 (x) ∈ Z[x], by Gauss’ Lemma, if T4 (x) factors in Q(x), then it factors in Z[x]. Hence, if T4 (x) factors, we know that it must have the form (ax2 + bx + 1)(cx2 + dx − 1) = acx4 + (ad + bc)x3 + (a − c + bd)x2 + (d − a)x − 1 with a, b, c, d ∈ Z. This means that we need   ac = 8     ac = 8 ad + bc = 0 =⇒ b(a + c) = 0  a − c + bd = −8    a − c + b2 = −8d = b.  d−b=0 It is impossible to have a + c = 0 and ac = 8, since a + c = 0 means a and c have opposite signs. Thus, from the second equation we deduce that d = b = 0. Then we are left with ac = 8 and a − c = −8. Thus c = 8/a so a − a8 = −8 which means that a2 + 8a − 8 = 0. It is easy to check that this equations as no rational solutions. Hence, T4 (x) is irreducible. d) From part (a), we deduce that T4 (x) = T2 (T3 (x)). We saw that T2 (x) = 2x2 − 1. Using trigonometric addition formulas, we can also find that T3 (x) = 4x3 − 3x. Consequently, T6 (x) = 0 ⇐⇒T3 (T2 (x)) = 0 ⇐⇒4(T2 (x))3 − 3T2 (x) = 0 √ 3 ⇐⇒T2 (x) = 0, ± 2√ 3 ⇐⇒2x2 − 1 = 0, ± s2 √ s √ √ 2 1 3 1 3 ,± − ,± + ⇐⇒x = ± 2 2 4 2 4 √ q q √ √ 1 2 1 ⇐⇒x = ± 2 − 3, ± 2+ 3 ,± 2 2 2

Exercise: 7 Section 7.4 Question: Is it possible to construct, using a compass and a straightedge, a triangle with sides in the ratio of 2 : 3 : 4 that has the same area of a given square? If so, suppose that one side of the square is a segment OR; describe a compass and straightedge construction for the desired triangle. [Hint: Use Heron’s Formula.] Solution: Heron’s Formula for the area A of a triangle with sides of length a, b, and c is p A = s(s − a)(s − b)(s − c), where s is the semiperimeter, s = (a + b + c)/2. If a triangle has side ratios of 2 : 3 : 4, then there exists a positive real number r such that a = 2r, b = 3r, and c = 4r, assuming that a < b < c. Then s = 29 r and the area of the triangle satisfies 9 5 3 1 135 4 A2 = r r r r = r . 2 2 2 2 16

386

CHAPTER 7. FIELD EXTENSIONS

If a square has side length of `, then its area is A = `2 . Thus if such a square has the same area of a triangle with sides in proportion of 2 : 3 : 4, then 135 4 `4 = A2 = r 16 so that r 2 24 3 r= √ `= `. 4 3 5 135 q Since 32 4 35 is a constructible number, we conclude that it is possible to construct the desired triangle. q A construction of 23 4 35 , using the segment OR as the unit segment, i.e., the segment of length r. ←−→ ←→ • Construction a point R0 so that OR0 is perpendicular to OR and OR = OR0 . −−→ −−→ • Construct a point A on the ray OR such that OA = 3OR. Also construct a point B on the ray OR0 such that OB = 5OR0 . ←→ ←→ • Construct the line L parallel to AB through the point R0 and let C be the intersection of L and OR. By Thales Theorem, OC = 53 OR. ←→ • Let D be the point on OR such that O is the midpoint of CD. ←−→0 −−→0 • Construct the circle Γ of diameter CD. Let E be the q intersection of Γ and OR opposite the ray OR . By comments in this section, we conclude that OE =

3 5 OR.

←→ 0 0 • Construct the circle Γ0 of diameter q R E. Let F be an intersection of Γ and OR. By comments in this section, we conclude that OF = 4 35 OR.

←→ • By similar methods as at the beginning, using Thales Theorem, we can construct a point G on OR such q that OG = 23 4 35 OR.

←→ • Now to construct the desired triangle, we first construct a point H on the line OG such that OH = 2OG. The segment OH corresponds to the side of the triangle with the ratio of 2. • Use methods described in the section to construct any point J such that OJ = 32 OH and let Γ1 be the circle of center O and radius OJ. • Use methods described in the section to construct any point K such that HK = 2HO and let Γ2 be the circle of center H and radius HK. • Let N be a point of intersection between Γ1 and Γ2 . By construction, OH : ON : HN = 2 : 3 : 4. Thus, 4OHN is the desired triangle.

7.5 – Cyclotomic Extensions Exercise: 1 Section 7.5 Question: Show explicitly that 2 is a primitive root modulo 11. Use this to find all primitive roots modulo 11. Solution: In the field F11 , we have 21 = 2,

22 = 4,

23 = 8,

24 = 5,

25 = 10

26 = 9,

27 = 7,

28 = 3,

29 = 6,

210 = 1

This shows that 2 is a generator of U (F11 ), so 2 is a primitive root modulo 11. We obtain all the generators of U (F11 ) by taking 2a , where gcd(a, 10) = 1. Thus, all the primitive roots modulo 11 are 2, 8, 7, and 6.

7.5. CYCLOTOMIC EXTENSIONS

387

Exercise: 2 Section 7.5 Question: Calculate Φ14 (x). Solution: We calculate Φ14 (x) by the recursion formula that x14 − 1 = Φ1 (x)Φ2 (x)Φ7 (x)Φ14 (x). Note that Φ1 (x)Φ7 (x) = x7 − 1 so that Φ2 (x)Φ14 (x) = Hence Φ14 (x) =

x14 − 1 = x7 + 1. x7 − 1

x7 + 1 = x6 − x5 + x4 − x3 + x2 − x + 1. x+1

Exercise: 3 Section 7.5 n−1 Question: Prove that Φ2n (x) = x2 +1 Solution: Note that all the divisors of 2n are 2m , where 0 ≤ m ≤ n. Thus the recursion formula for the cyclotomic polynomials is n X n x2 − 1 = Φ2m (x). m=0

Then n

x2 − 1 = Φ2n (x)

n−1 X

n−1

Φ2m (x) = Φ2n (x)(x2

− 1).

m=0

Thus 2n

Φ2n (x) =

x −1 = x2n−1 − 1

n−1

x2 x

2

2n−1

−1

−1

n−1

= x2

+ 1.

Exercise: 4 Section 7.5 Question: Calculate Φ14 (x). Solution: By the recursive formula for cyclotomic polynomials, we have x14 − 1 = Φ14 (x)Φ7 (x)Φ2 (x)Φ1 (x). Since x7 − 1 = Φ7 (x)Φ1 (x), we have Φ14 (x)Φ2 (x) =

x14 − 1 = x7 + 1. x7 − 1

Thus, Φ14 (x) =

x7 + 1 = x6 − x5 + x4 − x3 + x2 − x + 1. x+1

Exercise: 5 Section 7.5 n−1 Question: Prove that Φ2n (x) = x2 + 1. Solution: By the recursive formula for cyclotomic polynomials, we have n

n−1

x2 − 1 = Φ2n (x)Φ2n−1 · · · Φ22 (x)Φ2 (x)Φ1 (x) = Φ2n (x)(x2 Thus, 2n

Φ2n (x) =

x −1 = x2n−1 − 1

n−1

x2

2

−1

x2n−1 − 1

n−1

= x2

+ 1.

− 1).

388

CHAPTER 7. FIELD EXTENSIONS

Exercise: 6 Section 7.5 Question: Prove that if n is odd, then Φ2n (x) = Φn (−x). Solution: We note that if n is odd, then ζna with 0 ≤ a ≤ n − 1 are all the nth roots of units, and furthermore, 2πi

2πia

−ζna = eπi e2πia/n = e 2 + n = e2πi(n+2a)/2n . Since n + 2a is odd, then the denominator of n+2a never reduces to an odd number. The points −ζna with 2n 0 ≤ a ≤ n − 1 form the n points on the regular 2n-gon that are not already obtained by the points ζna with 0 ≤ a ≤ n − 1. Now suppose that n is odd and that gcd(a, n) = 1, i.e., that a and n are relatively prime. We prove that gcd(n + 2a, 2n) = 1. There exist s, t ∈ Z such that sn + ta = 1. Suppose that t is even. Then t t(n + 2a) + s − (2n) = tn + 2(ta + sn) − tn = 2. 2 Since n + 2a is odd, then there exists k, ` ∈ Z such that k(n + 2a) + ` · 2 = 1. Then t t (2n) = (k + t`)(n + 2a) + ` s − (2n). 1 = k(n + 2a) + ` t(n + 2a) + s − 2 2 This shows that of t is even, then gcd(n + 2a, 2n) = 1. Now suppose that t is odd. Then t t(n + 2a) + s − (2n) = n + 2 2 and

t − 1 (2n) = −n + 2 t(n + 2a) + s − 2

so

t − 1 (2n) = 4. 2t(n + 2a) + 2s − 2 2

Since n + 2a is odd, then gcd(n + 2a, 4) = 1 so there exist k 0 , `0 ∈ Z such that k 0 (n + 2a) + `0 · 4 = 1. Then t 0 0 1 = k (n + 2a) + ` 2t(n + 2a) + 2s − 2 − 1 (2n) 2 t = (k 0 + 2`0 t)(n + 2a) + `0 2s − 2 − 1 (2n). 2 We have done all of this to show that if n is odd and gcd(n, a) = 1, then gcd(n + 2a, 2n) = 1, which in turns shows that the primitive roots of unity mod 2n are the negatives of the primitive roots of unity modulo n. (There may be other ways to show this, perhaps more geometrically.) Recall that if n is odd, the formula for the Euler totient formula shows that φ(n) is even. Thus Y Y Y b Φn (−x) = (−x − ζna ) = (x − (−ζna )) = (x − ζ2n ) = Φ2n (x). a : gcd(a,n)=1

a : gcd(a,n)=1

b : gcd(b,2n)=1

As an example, note that if n is an odd prime, then from the recursive formula x2n − 1 = Φ2n (x)Φn (x)Φ2 (x)Φ1 (x), we have (xn )2 − 1 = Φ2n (x)(xn − 1)(x + 1). This leads to Φ2n (x) =

(xn )2 − 1 xn + 1 = = xn−1 − xn−2 + · · · − x + 1 = Φn (−x), (x + 1)(xn − 1) x+1

7.5. CYCLOTOMIC EXTENSIONS

389

since Φn (x) = xn−1 + xn−2 + · · · + x + 1. Exercise: 7 Section 7.5 Question: Show that if ζn is a primitive nth root of unity, then all the roots of Φn (x) are in Q(ζn ). Solution: Let ζn be a primitive nth root of unity. The roots of Φn (x) are precisely all the primitive nth roots of unity. Then the set of all the primitive nth roots of unity if {ζna | gcd(a, n) = 1}. This shows that all the roots of Φn (x) have the form ζna for some a ∈ Z. Hence, all of the roots of Φn (x) are in the field extension Q(ζn ). Exercise: 8 Section 7.5 Question: Suppose that m and n are positive relatively prime integers. Let ζn be a primitive nth root of unity and let ζm be a primitive mth root of unity. Prove that ζn ζm is a primitive mnth root of unity. Solution: Let m and n be relatively prime. By the Fundamental Theorem of Arithmetic, since m and n are relatively prime, a divisor d of mn must have a prime factorization in which the primes in common with m are distinct from the primes in common with n. Therefore, if d | mn, then d = d1 d2 , where d1 | m and d2 | n. Let ζn be a primitive nth root of unity and let ζm be a primitive mth root of unity. Then (ζn ζm )mn = m n n m (ζm ) (ζn ) = 1. Thus ζm ζn is a mnth root of unity. Assume that ζm ζn is not a primitive root of mnth unity. Then ζm ζn is a dth root of unity where d is a strict divisor of mn. Then d = d1 d2 such that d1 min m and d2 | n and d1 6= m or d2 6= n. Then d2 d1 d1 d2 d1 d2 d1 d2 ) (ζn ) . = (ζm ζn 1 = (ζm ζn )d = (ζm ζn )d1 d2 = ζm d2 is a primitive mth root of unity and ζnd1 is a primitive nth Since gcd(m, d2 ) = 1 and gcd(n, d1 ) = 1, then ζm root of unity. We also have d2 d1 (ζm ) = (ζnd1 )−d2 d2 and ζnd1 is respectively an mth roots and nth roots of unity. Since gcd(m, n) = 1, then we must have Since ζm d2 d1 (ζm ) = (ζnd1 )−d2 = 1

since both are mth roots of unity and nth roots of unity. Since d1 is a strict divisor of m or d2 is a strict divisor d2 is not a primitive mth root of unity or ζnd1 is not a primitive nth root of unity. of n, then we conclude that ζm This is a contradiction so we must conclude that ζm ζn is a primitive mnth root of unity. Exercise: 9 Section 7.5 Question: Prove that if ζn is a primitive nth root of unity, then ζnd is a primitive (n/d) root of unity. Solution: Let n be a positive integer and d a divisor of n. Let ζn be a primitive nth root of unity. Then (ζnd )n/d = ζnn = 1 so ζnd is a (n/d)th root of unity. Assume that d 6= 1 and that ζnd is not a primitive (n/d)th root of unity. Then there exists a strict divisor m of n/d such that (ζnd )m = 1. But then ζndm = 1 and dm is a strict divisor of n so ζn is not a primitive nth root of unity. This is a contradiction so ζnd must be a primitive (n/d)th root of unity. Exercise: 10 Section 7.5 Question: Prove that in the ring Z[x], the sequence of polynomial {xn − 1}∞ n=1 satisfies the identity that gcd(xm − 1, xn − 1) = xgcd(n,m) − 1, where in the first greatest common divisor we take the monic polynomial. Solution: The ring Z[x] is a UFD and since the cyclotomic polynomials are irreducible, then the recursive definition Y xn − 1 = Φd (x) d|n

gives factorizations into irreducibles. So by a result in the proof of Proposition 6.4.14, a greatest common divisor of xm − 1 and xn − 1 is the product of all the irreducibles occurring in the factorizations of xm − 1 and xn − 1. Hence, a greatest common divisor of xm − 1 and xn − 1 is Y Y Φd (x) = Φd (x) = xgcd(m,n) − 1. d|m and d|n

d|gcd(m,n)

390

CHAPTER 7. FIELD EXTENSIONS

Exercise: 11 Section 7.5 Question: Suppose that p | n. Prove that Φpn (x) = Φn (xp ). Solution: Suppose that p divides n. As a consequence of Proposition 2.1.27, if p | n, then φ(pn) = pφ(n), where φ is Euler’s totient function. In particular, there are pφ(n) distinct primitive pnth roots of unity. Let ζ ∈ C be a primitive nth root of unity, i.e., a root of Φn (x). Let ω be any of the p possible pth roots of ζ. Then ω pn = ζ n = 1 so ω is a pnth root of unity. β1 β2 βt αs 1 α2 Assume that ω is not a primitive pnth root of unity. Suppose that p = pα 1 p2 · · · ps and n = p1 p2 · · · pt with s ≤ t and αi ≤ βi for i = 1, 2, . . . , s. Since ω is not a primitive pnth root of unity, there exists a strict divisor m of pn such that ω m = 1. By multiplying m by an appropriate factor if necessary, we can suppose that m has one fewer prime factors than pn in its prime factorization. If m = n/pi , then 1 = ω np/pi = ζ n/pi since pi | n for all i with 1 ≤ i ≤ t. This means that ζ is not a primitive nth root of unity. This contradiction shows that every ω as defined is a primitive pnth root of unity. Since every pth root of every primitive nth root of unity is distinct, we have found all primitive pnth roots of unity. Call this set of primitive pnth roots of unity µpn . Now Φn (xp ) is a polynomial of order pφ(n). Furthermore, for all ω ∈ µnp , we have Φn (ω p ) = 0. Thus, Φnp (x) =

Y

(x − ω)

ω∈µpn

is a polynomial of degree pφ(n) in C[x] such that Φnp (x) divides Φn (xp ). Hence, Φpn (x) = Φn (xp ). Exercise: 12 Section 7.5 Question: Determine Φ60 (x). Solution: To determine, Φ60 (x), we note that the prime decomposition of 60 is 60 = 22 · 3 · 5. We first calculate Φ15 (x) using the recursive formula x15 − 1 = Φ1 (x)Φ3 (x)Φ5 (x)Φ15 (x) 3 x5 − 1 x15 − 1 = = x10 + x5 + 1 =⇒Φ15 (x)Φ3 (x) = 5 x −1 x5 − 1 x10 + x5 + 1 =⇒Φ15 (x) = 2 = x8 − x7 + x5 − x4 + x3 − x + 1. x +x+1 From the result that Φ2n (x) = Φn (−x) if n is odd, we get Φ30 (x) = x8 + x7 − x5 − x4 − x3 + x + 1. Finally, from the result that Φpn (x) = Φ(xp ), whenever p | n and p is prime, then we deduce that Φ60 (x) = x16 + x14 − x10 − x8 − x6 + x2 + 1.

Exercise: 13 Section 7.5 Question: Use Exercise 7.5.12 to determine Φ225 (x) Solution: The prime decomposition of 225 is 225 = 32 × 52 . Thus, by Exercise 7.5.12, Φ225 (x) = Φ15 (x15 ). So we now need to determine Φ15 (x). By the recursive formula for cyclotomic polynomials, x15 − 1 = Φ15 (x)Φ5 (x)Φ3 (x)Φ1 (x) = Φ15 (x)Φ3 (x)(x5 − 1).

7.5. CYCLOTOMIC EXTENSIONS

391

Thus, since Φ3 (x) = x2 + x + 1, we have Φ15 (x) =

x15 − 1

=

(x5 − 1)(x2 + x + 1)

x10 + x5 + 1 = x8 − x7 + x5 − x4 + x3 − x + 1. x2 + x + 1

We deduce that Φ225 (x) = x120 − x105 + x75 − x60 + x45 − x15 + 1.

Exercise: 14 Section 7.5 Question: Determine Φ60 (x). Solution: The prime decomposition of 60 is 60 = 22 × 3 × 5. By Exercise 7.5.12, Φ60 (x) = Φ30 (x2 ). By Exercise 7.5.6, Φ60 (x) = Φ15 (−x2 ). By the work in Exercise 7.5.13, Φ15 (x) = x8 − x7 + x5 − x4 + x3 − x + 1 so we deduce that Φ60 (x) = x16 + x14 − x10 − x8 − x6 + x2 + 1.

Exercise: 15 Section 7.5 Question: In 1883, Migotti proved that if p and q are distinct primes then Φpq (x) only involves the coefficients of −1, 0, and 1. Use this theorem and previous exercises to prove that if Φn (x) has coefficients besides −1, 0, or 1, then it must be divisible by at least three distinct odd primes. Solution: If n = pα with α ∈ N∗ , then by Exercise 7.5.12, α−1

Φn (x) = Φp (xp

) = xp

α−1

(p−1)

+ xp

α−1

(p−2)

+ · · · + xp

α−1

+1

and therefore involves only coefficients that are 0 or 1. If n is divisible by only two distinct primes with n = pα q β with α, β ∈ N∗ , then by Exercise 7.5.12, α−1 β−1

Φn (x) = Φpq (xp

q

),

so by Migotti’s result only involves −1, 0, or 1 for coefficients. If n is of the form n = 2α pβ q γ , where p and q are odd primes and α, β, γ ∈ N∗ . Then α−1 β−1 γ−1

Φn (x) = Φ2pq (x2

p

q

α−1 β−1 γ−1

) = Φpq (−x2

p

q

),

where the last equality holds by Exercise 7.5.6. By Migotti’s result, Φpq (x) only involves coefficients of −1, 0, or 1. So though it is possible for some coefficients to change since going from those of Φpq (x) to those of Φn (x), the coefficients of Φn (x) still only involve −1, 0, and 1. Consequently, for Φn (x) to possibly have coefficients that are not −1, 0, or 1, n must be divisible by at least three distinct odd primes. Exercise: 16 Section 7.5 Question: Suppose that p is a prime number such that p | Φn (2). Prove that p | Φpn (2). Deduce that p | Φpα n (2) for all positive integers α. Solution: Note that since Φn (x) is palindromic and monic, then the constant coefficient of Φn (x) is 1. Hence, Φn (2) is odd and therefore if p | Φn (2), then p is odd. Let p be an odd prime number and let n be the order of 2 in U (Fp ). Then p | (2n − 1) and p - (2d − 1) for all d < n. In particular, this implies that p | Φn (2) because Y p | 2n − 1 = Φd (2) d|n

and if p | Φd (2), then p | 2d − 1, which is not the case. Let ordp (Φn (2)) = ordp (2n − 1) = α so that 2n = 1 + pα k where p - k.

392

CHAPTER 7. FIELD EXTENSIONS Now suppose that for some integer ordp (2m − 1) = β with β ≥ 1 so that 2m = 1 + pβ k where p - k. Then 2pm = (1 + pβ k)p =

p X p i=0

We know that

i

piβ k i .

( 1 if 1 ≤ i ≤ p − 1 p ordp = i 0 ifi = 0 orp.

So pm

2

p X p

p(p − 1) 2β 2 p k + · · · + p · pβ(p−1) k p−1 + ppβ k p 2 i i=1 p(p − 1) β 2 β β(p−2) p−1 (p−1)β−1 p =p·p k+ p k + ··· + p k +p k . 2p

−1=

piβ k i = p · pβ k +

All the terms in the parentheses are divisible by p except for k. Hence, ordp (2pm − 1) = β + 1. This tells us the power of p that divides 2pm − 1 but we wish to know the power of p dividing Φk (2) for various k. We know from assumption that ordp (Φn (2)) = α. We have shown that ordp (2pn − 1) = α + 1. Since n is the order of 2 modulo p, then p | (2k − 1) if and only if n | k. We have Y pα+1 | (2n − 1) Φdp (2) d|n

so p|

Y

Φdp (2),

d|n

but since p | 2k − 1 implies n | k, we deduce that n | dp. Thus we must have d = n. So ordp Φpn (2) = 1. γ Furthermore, by induction, if n is the order of 2 in U (Fp ), and if ordp (2n −1) = α, then ordp (2p n −1) = 2α +γ and ordp Φpγ n (2) = 1. Exercise: 17 Section 7.5 Question: Let p be an odd prime dividing n. Suppose that a ∈ Z satisfies Φn (a) ≡ 0 (mod p). Prove that a is relatively prime to p and the that order of a in U (p) is precisely n. Solution: Let p be an odd prime dividing n. Suppose that a ∈ Z satisfies Φn (a) ≡ 0 (mod p). Since Y an − 1 = Φn (a), d|n

then since p | Φn (a) we also have p | (an − 1). This implies in particular that p - a so a and p are relatively prime. Furthermore, an = 1 in U (p) so n is a multiple of the order of a in U (p). As the previous exercise showed, it does not have to be precisely the order of a in U (p). Exercise: 18 Section 7.5 Question: Let m and n be positive integers and let l = lcm(m, n) and d = gcd(m, n). Prove that a) Q(ζm )Q(ζn ) = Q(ζl ) as a composite field; b) Q(ζm ) ∩ Q(ζn ) = Q(ζd ). Solution: Let m and n be positive integers and let l = lcm(m, n) and d = gcd(m, n). a) The composite field Q(ζm )Q(ζn ) is the smallest field that contains both ζm and ζn , where here by ζk we mean e2πi/k . Let l = ma and l = nb for positive integers a, b. Then ζla = ζm and ζlb = ζn . Hence, ζm , ζn ∈ Q(ζl ) so Q(ζm )Q(ζn ) ⊆ Q(ζl ). Now suppose that sm + tn = d for some integers s and t. Then t

s

t s ζm ζn = e2πit/m e2πis/n = e2πi( m + n ) = e2πi(tn+sm)/mn = e2πid/mn = e2πi/l = ζl .

Hence, ζl ∈ Q(ζm )Q(ζn ) so Q(ζl ) ⊆ Q(ζm )Q(ζn ). Since we already have the other inclusion, we conclude that Q(ζl ) = Q(ζm )Q(ζn ).

7.5. CYCLOTOMIC EXTENSIONS

393

b) Now consider the intersection field Q(ζm ) ∩ Q(ζn ) as an extension of Q. Let m = ad and n = bd for positive integers a, b. We have a ζd = e2πi/d = e2πia/m = ζm = ζnb . Hence, ζd ∈ Q(ζm ) ∩ Q(ζn ) so Q(ζd ) ⊆ Q(ζm ) ∩ Q(ζn ). The proof of the reverse inclusion is the most challenging part of the exercise. We first need to prove a lemma about Euler’s totient function, namely that if for positive integers m and n, with l = lcm(m, n) and d = gcd(m, n), we have ϕ(m)ϕ(n) = ϕ(l)ϕ(d). Recall that mn = ld. Suppose that and n = pβ1 1 pβ2 2 · · · pβt t ,

αt 1 α2 m = pα 1 p2 · · · pt

where pi are distinct primes for i = 1, 2, . . . , t (the primes dividing m or n) and αi , βi ≥ 0. We also know from results in Chapter 2 that min(α1 ,β1 ) min(α2 ,β2 ) min(αt ,βt ) p2 · · · pt

max(α1 ,β1 ) max(α2 ,β2 ) max(αt ,βt ) p2 · · · pt .

d = p1

and l = p1

Proposition 2.1.27 gives a formula for the totient function. Suppose that we order the primes pi such that βi = 0 for 1 ≤ i ≤ r and that αi = 0 for s ≤ i ≤ t. Note that 1 ≤ r, s ≤ t, that either r ≤ s or r = s + 1 when gcd(m, n) = 1, that for 1 ≤ i ≤ r the primes pi only divide m, and that for s ≤ i ≤ t the primes pi only divide n. Then    Y Y i −1 pα (pi − 1)  pβi i −1 (pi − 1) φ(m)φ(n) =  i 1≤i≤s−1

r+1≤i≤t



 =

Y

i −1 pα (pi − 1)  i

piαi +βi −2 (pi − 1)2  

r+1≤i≤s−1

1≤i≤r



 Y

Y

pβi i −1 (pi − 1) ,

s≤i≤t

where the middle product does not occur (so is 1) if gcd(m, n) = 1, and    Y Y min(αi ,βi )−1 max(αi ,βi )−1 pi (pi − 1)  pi (pi − 1) φ(d)φ(l) =  r+1≤i≤s−1

 =

Y

1≤i≤r

 =

1≤i≤t

 i −1 pα (pi − 1)  i

 min(αi ,βi )+max(αi ,βi )−2

Y

pi

r+1≤i≤s−1

1≤i≤r

i −1 pα (pi − 1)  i

 piαi +βi −2 (pi − 1)2  

Y

pβi i −1 (pi − 1)

s≤i≤t

 Y

(pi − 1)2  

 Y

r+1≤i≤s−1

 Y

pβi i −1 (pi − 1) .

s≤i≤t

This shows that φ(m)φ(n) = φ(d)φ(l). By Corollary 7.5.10, [Q(ζa ) : Q] = φ(a). In the collection of subfields Q(ζl ) Q(ζm )

Q(ζn ) Q(ζm ) ∩ Q(ζn ) Q(ζd ) Q

we have [Q(ζl ) : Q] = [Q(ζl ) : Q(ζn )][Q(ζn ) : Q] so φ(l) = [Q(ζl ) : Q(ζn )]φ(n). Thus [Q(ζl ) : Q(ζn )] = φ(l)/φ(n) = φ(m)/φ(d). But by a similar reasoning, we have [Q(ζm ) : Q(ζd )] = φ(m)/φ(d). Now [Q(ζm )Q(ζn ) : Q(ζn )] = [Q(ζm ) : Q(ζm ) ∩ Q(ζn )]. Hence, [Q(ζm ) : Q(ζm ) ∩ Q(ζn )] = φ(l)/φ(n) = φ(m)/φ(d). But then we have the chain Q(ζd ) ⊆ Q(ζm ) ∩ Q(ζn ) ⊆ Q(ζm ),

394

CHAPTER 7. FIELD EXTENSIONS so [Q(ζm ) : Q(ζd )] = [Q(ζm ) : Q(ζm ) ∩ Q(ζn )][Q(ζm ) ∩ Q(ζn ) : Q(ζd )] ⇐⇒(φ(m)/φ(d)) = (φ(m)/φ(d))[Q(ζm ) ∩ Q(ζn ) : Q(ζd )]. Thus [Q(ζm ) ∩ Q(ζn ) : Q(ζd )] = 1 and hence Q(ζm ) ∩ Q(ζn ) = Q(ζd ).

Exercise: 19 Section 7.5 Question: Prove that the Möbius inversion formula can also be written as Y Y µ(n/d) an = bd ⇒ bn = ad . d|n

d|n

Solution: There is a bijection on the set of divisors of n defined by f (d) = nd . In the Möbius inversion formula, if we apply this bijection so that d0 = f (d), then d = n/d0 and an =

Y

bd ⇒ bn =

d|n

Y µ(d) Y µ(n/d0 ) an/d = ad0 . d0 |n

d|n

Exercise: 20 Section 7.5 Question: Prove that polynomial iteration satisfies (a) P n (P m (x)) = P m+n (x); (b) (P n )m (x) = P mn (x); and (c) deg P n (x) = k n where deg P (x) = k. Solution: Let P (x) ∈ F [x], where F is a field. a) The iterated polynomial P n (P m (x)) is equal to the polynomial P (x) iterated m + n times, i.e., m times

n times

m+n times

z }| {z }| { z }| { P (P (x)) = P (· · · (P P (· · · (P (x))) · · · )) = P (· · · (P (x))) · · · )). n

m

b) The iterated polynomial (P n )m (x) can be expanded out as m times

z }| { (P ) (x) = P n (P n (· · · (P n (x)) · · · )), n m

which shows that P is iterated mn times. Thus, (P n )m (x) = P mn (x). c) Suppose that deg P (x) = k. We prove that deg P n (x) = k n for all polynomials P (x) and all n ≥ 1 by induction on n. If n = 1, then the result is obvious for all P (x) ∈ F [x]. Suppose that For all P (x) ∈ F [x] we have deg P n (x) = k n for some positive integer n. Let P (x) = ak xk + · · · + a1 x + a0 . Then P n+1 (x) = P (P n (x)) = ak (P n (x))k + ak−1 (P n (x))k−1 + · · · a1 P n (x) + a0 . Since deg P n (x) = k n , the term of highest degree in P n+1 (x) occurs in ak (P n (x))k when the kth power is applied to the term of top degree of P n . Hence, the term of highest degree has degree k n · k = k n+1 . Thus, deg P n (x) = k n for all polynomials P (x) ∈ F [x] and all n. Exercise: 21 Section 7.5 Question: Suppose that deg P (x) = m > 1. Prove that deg ΦP,n (x) =

X

µ(d)mn/d .

d|n

Deduce that a sequence satisfying xn+1 = P (xn ) can have at most n1

n/d primitive n-cycles. d|n µ(d)m

P

7.5. CYCLOTOMIC EXTENSIONS

395

Solution: From the recursive definition of dynatomic polynomials and he Möbius inversion formula applied to Y P n (x) − x = ΦP,n (x), d|n

we get that ΦP,n (x) =

Y (xn/d − 1)µ(d) . d|n

Hence, the degree of polynomials has logarithmic-type properties so X deg ΦP,n (x) = µ(d)(deg P (x))n/d . d|n

If a sequence in F is defined as xn+1 = P (xn ), then an n-cycle occurs when xn = x0 , that is to say when P n (x0 ) = x0 . Thus, a point on an n-cycle is a root of the equation P n (x) − x = 0 of degree mn . A primitive n-cycle is an n-cycle that is not also a d-cycle for some d | n. If and n-cycle were also a d cycle with d | n, then a value on a d would be the root of the polynomial P d (x) − x. By the recursive definition of the dynatomic polynomials, a d-cycle would be the root of ΦP,d (x) for d | n. Hence, points on primitive n-cycles must be roots of ΦP,n (x). Since points on primitive n-cycles are roots of ΦP,n (x) there at most deg ΦP,n (x) such points. However, there are exactly n points on primitive n-cycles. Furthermore, since a sequence is completely determined by the one previous term (i.e., via xn+1 = P (xn )) then cycles are disjoint. Hence, we deduce that X µ(d)(deg P (x))n/d d|n

is divisible by n and that there are at most 1X µ(d)(deg P (x))n/d n d|n

primitive n-cycles. Exercise: 22 Section 7.5 Question: Let Q(x) = x2 − 2. Calculate ΦQ,2 (x), ΦQ,3 (x), and ΦQ,4 (x). Solution: Let Q(x) = x2 − 2. Then ΦQ,1 (x) = Q(x) − x = x2 − x − 2. To calculate ΦQ,2 (x), we use the recursive relation ΦQ,1 (x)ΦQ,2 (x) = Q(Q(x)) − x = (x2 − 2)2 − 2 − x = x4 − 4x2 − x + 2. Thus, x4 − 4x2 − x + 2 = −x2 + x + 2. x2 − x − 2 To calculate ΦQ,3 (x), we use again use the recursive relation ΦQ,2 (x) =

ΦQ,1 (x)ΦQ,3 (x) = Q(Q(Q(x))) − x = ((x2 − 2)2 − 2)2 − 2 − x = x8 − 8x6 + 20x4 − 16x2 − x + 2. Thus, x8 − 8x6 + 20x4 − 16x2 − x + 2 = x6 + x5 − 5x4 − 3x3 + 7x2 + x − 1. x2 − x − 2 To calculate ΦQ,4 (x), we use the recursive relation ΦQ,3 (x) =

ΦQ,1 (x)ΦQ,2 (x)ΦQ,4 (x) = (Q2 (x) − x)ΦQ,4 (x) = Q4 (x) − x, Thus, x16 − 16x14 + 104x12 − 352x10 + 660x8 − 672x6 + 336x4 − 64x2 − x + 2 x4 − 4x2 − x + 2 12 10 9 8 = x − 12x + x + 54x − 8x7 − 111x6 + 20x5 + 100x4 − 15x3 − 30x2 + 1.

ΦQ,4 (x) =

396

CHAPTER 7. FIELD EXTENSIONS

Exercise: 23 Section 7.5 Question: Let Q(x) = x3 − 2. Calculate ΦQ,2 (x), ΦQ,3 (x), and ΦQ,4 (x). Solution: Let Q(x) = x3 − 2. Then ΦQ,1 (x) = Q(x) − x = x3 − x − 2. To calculate ΦQ,2 (x), we use the recursive relation ΦQ,1 (x)ΦQ,2 (x) = Q(Q(x)) − x = (x3 − 2)3 − 2 − x = x9 − 6x6 + 12x3 − x − 10. Thus, ΦQ,2 (x) =

x9 − 6x6 + 12x3 − x − 10 = x6 + x4 − 4x3 + x2 − 2x + 5. x3 − x − 2

To calculate ΦQ,3 (x), we use again use the recursive relation ΦQ,1 (x)ΦQ,3 (x) = Q(Q(Q(x))) − x = ((x3 − 2)3 − 2)3 − 2 − x = x27 − 18x24 + 144x21 − 678x18 + 2088x15 − 4392x12 + 6348x9 − 6120x6 + 3600x3 − x − 1002. Thus, ΦQ,3 (x) = x24 + x22 − 16x21 + x20 − 14x19 + 113x18 − 12x17 + 85x16 − 464x15 + 61x14 − 294x13 + 1221x12 − 172x11 + 633x10 − 2122x9 + 289x8 − 856x7 + 2393x6 − 278x5 + 681x4 − 1612x3 + 125x2 − 250x + 501 To calculate ΦQ,4 (x), we use the recursive relation ΦQ,1 (x)ΦQ,2 (x)ΦQ,4 (x) = (Q2 (x) − x)ΦQ,4 (x) = Q4 (x) − x, Thus, ΦQ,4 (x) =

Q4 (x) − x . Q2 (x) − x

This is a polynomial of degree 72 and is ugly. At this point, it is useful to use a computer algebra system to calculate polynomial division. We list the polynomial for completeness: ΦQ,4 (x) = x72 − 48x69 + 110x66 + x64 − 16208x63 − 42x61 + 170688x60 + 840x58 − 1373568x57 + x56 − 10654x55 + 8784496x54 − 36x53 + 96264x52 − 45837504x51 + 612x50 − 659736x49 + 198767809x48 − 6540x47 + 3564372x46 − 725740004x45 + 49320x44 − 15571800x43 + 2251965144x42 − 279216x41 + 55967185x40 − 5976499994x39 + 1231836x38 − 167431574x37 + 13618840108x36 − 4336992x35 + 420051480x34 − 26693581872x33 + 12371041x32 − 887340376x31 + 44989922632x30 − 28843064x29 + 1579864352x28 − 65043760544x27 + 55170684x26 − 2365796448x25 + 80275425697x24 − 86489094x23 + 2963367960x22 − 83927590498x21 + 110450940x20 − 3075351888x19 + 73488491056x18 − 113514840x17 + 2604934369x16 − 53007044060x15 + 91976700x14 − 1760082028x13 + 30738992768x12 − 56804208x11 + 915267580x10 − 13809851792x9 + 25240321x8 − 345110554x7 + 4523044708x6 − 7222606x5 + 84298204x4 − 963703480x3 + 1006012x2 − 10060120x + 100601201.

Exercise: 22 Section 7.5 Question: Let Q(x) = x2 − 2. Calculate ΦQ,2 (x), ΦQ,3 (x), and ΦQ,4 (x). Solution: Let Q(x) = x2 − 2x + 2. Then ΦQ,1 (x) = Q(x) − x = x2 − 3x + 2.

7.5. CYCLOTOMIC EXTENSIONS

397

To calculate ΦQ,2 (x), we use the recursive relation ΦQ,1 (x)ΦQ,2 (x) = Q(Q(x)) − x = x4 − 4x3 + 6x2 − 5x + 2. Thus, ΦQ,2 (x) =

x4 − 4x3 + 6x2 − 5x + 2 = x2 − x + 1. x2 − 3x + 2

To calculate ΦQ,3 (x), we use again use the recursive relation ΦQ,1 (x)ΦQ,3 (x) = Q(Q(Q(x))) − x = x8 − 8x7 + 28x6 − 56x5 + 70x4 − 56x3 + 28x2 − 9x + 2. Thus, ΦQ,3 (x) =

x8 − 8x7 + 28x6 − 56x5 + 70x4 − 56x3 + 28x2 − 9x + 2 = x6 − 5x5 + 11x4 − 13x3 + 9x2 − 3x + 1. x2 − 3x + 2

To calculate ΦQ,4 (x), we use the recursive relation ΦQ,1 (x)ΦQ,2 (x)ΦQ,4 (x) = (Q2 (x) − x)ΦQ,4 (x) = Q4 (x) − x, The polynomial division of Q4 (x) − x by Q2 (x) − x gives ΦQ,4 (x) = x12 − 12x11 + 66x10 − 219x9 + 486x8 − 756x7 + 841x6 − 672x5 + 384x4 − 155x3 + 42x2 − 6x + 1.

Exercise: 25 Section 7.5 Question: Let Q(x) = x2 − 45 . Prove by direct calculation that for this particular polynomial, ΦQ,2 (x) divides ΦQ,4 (x). Solution: Let Q(x) = x2 − 54 . We first calculate that 5 5 1 Q2 (x) − x = x4 − x2 − x + = (2x)4 − 10(2x)2 − 8(2x) + 5 . 2 16 16 Then ΦQ,2 (x) =

1 Q(Q(x)) − x = x2 + x − . Q(x) − x 4

We also can calculate that Q4 (x) − x = Q2 (Q2 (x)) − x = x16 − 10x14 + =

155 12 575 10 7755 8 1275 6 13725 4 7375 2 5105 x − x + x − x − x + x −x+ 4 8 128 128 1024 2048 65536

1 (2x)16 − 40(2x)14 + 620(2x)12 − 4600(2x)10 + 15510(2x)8 − 10200(2x)6 65536 −54900(2x)4 + 59000(2x)2 − 32768(2x) + 5105 .

But then ΦQ,4 (x) = x12 −

15 10 315 8 309 6 55 5 295 4 9 2985 2 615 1021 x + x9 + x − 5x7 − x + x + x − x3 + x − x+ . 2 16 16 8 256 16 512 256 4096

Finally, it is tedious, but it is possible to perform the polynomial division and show that 25 89 115 5 69 4 87 3 1235 2 1439 1021 ΦQ,4 (x) = ΦQ,2 (x) x10 − x9 − x8 + 7x7 + x6 − x − x + x − x + x− . 4 8 8 32 16 256 256 1024

398

CHAPTER 7. FIELD EXTENSIONS

7.6 – Splitting Fields and Algebraic Closure Exercise: 1 Section 7.6 Question: Find the splitting field of x4 − 3x2 + 1 ∈ Q[x]. Solution: We can solve the equation x4 − 3x2 + 1 = 0 by s s √ √ √ √ 1 1 3 − 5 3+ 5 2 x = (3 ± 9 − 4) = (3 ± 5) =⇒ x = ± or ± . 2 2 2 2 q q √ √ The splitting field of the polynomial is Q( (3 − 5)/2, (3 + 5)/2). Exercise: 2 Section 7.6 Question: Find the splitting field of x6 − 2x3 − 1 ∈ Q[x]. Solution: We can find the roots of the polynomial x6 − 2x3 − 1 by first solving for x3 as follows: q √ √ 3 j 3 x = 1 ± 2 =⇒ x = ω 1± 2 where ω = e2πi/3 and j = 0, 1, 2. This gives us six roots of the polynomial. If we call K the splitting field of p √ 3 x6 − 2x3 − 1,pthen K must contain 1 + 2, which is one of the real roots, and ω, which is a complex number √ 3 so not in Q( 1 + 2). Notice also that q q √ 3 √ √ 3 1 − 2 1 + 2 = 3 1 − 2 = −1. So the root

p 3

1−

√

2 = −1/

p 3

1+

√

2 ∈ Q(

p 3

1+

√

2). Hence we find that K = Q(

p √ 3 1 + 2, ω).

Exercise: 3 Section 7.6 Question: Find the splitting field of (x2 − 2)(x3 − 2) ∈ Q[x] √ √ j 3 Solution: This polynomial has 5 roots,√namely ±√ 2 and ω√ 2, where = e2πi/3 and j = 0, 1, 2. The splitting √ √ √ √ ω√ √ 3 6 3 6 3 field √ needs to contain Q( 2, 2). Since 2 = 2/ 2, then 2 ∈ Q( 2, 2). However, it is easy to see that 2 √ √ and 3√2 are both√in Q( 6 2). So the splitting field contains Q( 6 2). Furthermore, since the splitting √ field contains both 3 2 and ω 3 2, it must also contain ω. However, all the roots of the polynomial are in Q( 6 2, ω) so this is the splitting field. Exercise: 4 Section 7.6 Question: Find the splitting field of (x3 − 2)(x3 − 7) ∈ Q[x].

√ √ 3 Solution: The six solutions to the equation (x3 − 2)(x√ − 7) =√0 are 3 2ζ3j and 3 7ζ3j , where ζ3 = e2πi/3 = √ / Q( 3 7) so the splitting field of the polynomial is (−1√+ √3i)/2 and j = 0, 1, 2. It is not hard to show that 3 2 ∈ 3 3 Q( 2, 7, ζ3 ). Exercise: 5 Section 7.6 Question: Find the splitting field of x6 − 5 ∈ Q[x] and determine the degree of the splitting field over Q. √ √ Solution: The solutions to the equation x6 − 5 = 0 are 6 5ζ6j , where ζ6 = e2πi/6 = (1 √ + 3i)/2. The splitting √ field√of x6 − 5 over Q is K = Q( 6 5, ζ6 ). We note that√the minimal polynomial of 6 5 over Q √ is x6 − 5 so 6 6 [Q( 5) √ : Q] = 6. We also note that every element in Q( 5) is real while ζ6 is not. Thus ζ6 ∈ / Q( 6 5). √ Hence, 6 6 2 [K : Q( 5)] > 1. However, ζ is the root of x − x + 1, a quadratic polynomial in Q[x] and thus also in Q( 5)[x]. 6 √ √ 6 6 We conclude that [K : Q( 5)] ≤ 2 and therefor that [K : Q( 5)] = 2. Hence, √ √ 6 6 [K : Q] = [K : Q( 5)][Q( 5) : Q] = 2 × 6 = 12.

Exercise: 6 Section 7.6 Question: Describe the splitting field of x4 + 2x2 + 1 ∈ F7 [x] as a quotient ring of F7 [x].

7.6. SPLITTING FIELDS AND ALGEBRAIC CLOSURE

399

Solution: The polynomial x4 + 2x2 + 1 = (x2 + 1)2 over F7 has for roots the roots of x2 + 1. These elements are not in F7 . Note that if θ is one root of x2 + 1 in some field extension of F7 , then the other root is −θ. So the four roots of x4 + 2x2 + 1 are θ, θ, −θ, −θ. Hence, the splitting field of the polynomial is F7 [x]/(x2 + 1). Exercise: 7 Section 7.6 Question: Let F be a field and let a ∈ F . Show that the splitting field of a polynomial p(x) is the same as the splitting field of the polynomial q(x) = p(x − a). Solution: Let α1 , α2 , . . . , αn be the distinct roots of p(x). The splitting field of p(x) is F (α1 , α2 , . . . , αn ). Then the distinct roots of q(x) = p(x − a) are a + α1 , a + α2 , . . . , a + αn and the splitting field of q(x) is F (a + α1 , a + α2 , . . . , a + αn ). However, since a ∈ F , then F (a + α1 , a + α2 , . . . , a + αn ) = F (α1 , α2 , . . . , αn ), the splitting field of p(x). Exercise: 8 Section 7.6 Question: Let p be a prime number and let q be a prime number in N. Prove that the splitting field of xp − q is an extension of degree p(p − 1) for Q. √ Solution: The roots of xp − q over Q are p qζpj , where ζp = e2πi/p and j = 0, 1, . . . , p − 1. By Eisenstein’s √ p criterion, we see that x − q is irreducible, so this is the minimal polynomial of p q over Q. In particular, √ [Q( p q) : Q] = p. Furthermore, the minimal polynomial if ζp is the cyclotomic polynomial Φp (x), which has degree p − 1. √ The splitting field of xp − q over Q is K = Q( p q, ζp ), where these two generators have degrees n1 = p and n2 = p − 1 over Q. By Proposition 7.2.13, lcm(p, p − 1) divides [K : Q] and [K : Q] ≤ p(p − 1). Since gcd(p, p − 1) = 1 for all primes p, we have lcm(p, p − 1) = p(p − 1). We deduce that [K : Q] = p(p − 1). Exercise: 9 Section 7.6 Question: Let F be a field and let f (x) be an irreducible cubic polynomial in F [x]. Prove that the splitting field K of f (x) has degree ( 3 if the discriminant ∆ is the square of an element in F , [K : F ] = 6 otherwise. Solution: Since f (x) is irreducible and a cubic polynomial, then any root α of f (x) has degree 3 over F , i.e., [F (α) : F ] = 3. Let α1 , α2 , α3 be the three roots of f (x). We know that f (x) = (x − α1 )q(x), where q(x) is a quadratic polynomial in F (α1 )[x]. Either q(x) factors (so α2 , α3 ∈ F (α1 )) in F (α1 )[x] in which cases α2 , α3 ∈ F (α1 ) and the splitting field of f (x) is F (α1 ) or q(x) does not factor (so α2 , α3 ∈ / F (α1 )) and is therefore irreducible, in which case degF (α1 ) α2 = 2 and the splitting of f (x) is F (α1 , α2 ). √ The strategy of the solution is to try to show that ∆ is in the splitting field K = F (α1 , α2 ). We will use the formula for the solutions of the cubic as given on page 341, with αi = yi − a3 . The shift of − a3 is somewhat tedious. It would be nice to try to take the cubic power of α1 but this gets messy because of that shift. Taking differences √ of the roots gets rid of this variable shift. Hence, we could consider α1 − α2 , take the cube power, and see if ∆ emerges. This turns out not to be ideal. Instead, following symmetry, it is useful to calculate d = (α1 − α2 )(α2 − α3 )(α3 − α1 ), which is obviously in K. This is d = ((1 − ω)u0 + (1 − ω 2 )v0 )((ω − ω 2 )u0 + (ω 2 − ω)v0 )((ω 2 − 1)u0 + (ω − 1)v0 ). Expanding this expression gives d = (1 − ω)(ω − ω 2 )(ω 2 − 1)u30 + (1 − ω)(ω − ω 2 )(ω − 1)u20 v0 + (1 − ω)(ω 2 − ω)(ω 2 − 1)u20 v0 + (1 − ω)(ω 2 − ω)(ω − 1)u0 v02 + (1 − ω 2 )(ω − ω 2 )(ω 2 − 1)u20 v0 + (1 − ω 2 )(ω − ω 2 )(ω − 1)u0 v02 + (1 − ω 2 )(ω 2 − ω)(ω 2 − 1)u0 v02 + (1 − ω 2 )(ω 2 − ω)(ω − 1)v03 . We calculate that the coefficient in front of u30 is √ (1 − ω)(ω − ω 2 )(ω 2 − 1) = 1 − ω − ω + ω 2 − ω + ω 2 + ω 2 − 1 = 3(ω 2 − ω) = −3 3i.

400

CHAPTER 7. FIELD EXTENSIONS

2 2 We √ calculate in a similar fashion that the coefficients for u0 v0 and for u0 v0 is 0, while the coefficient for v0 is 3 3i. Hence, √ d = −3 3i(u30 − v03 ) ! !! r r √ q ∆ q ∆ = −3 3i − + − − − + − 2 108 2 108 r √ ∆ = −3 3i · 2 − 108 √ = ∆. √ √ √ We have proved that ∆ ∈ K. The quantity ∆ is √ in F so either ∆ ∈√F or the degree of ∆ over √ F is 2. / F , then K = F (α1 , ∆) and Since [F (α1 ) : F ] = 3, then it is not possible to have ∆ ∈ F (α1 ). So if ∆ ∈ [K : F ] = 6.

Exercise: 10 Section 7.6 Question: Let p(x) ∈ F [x] be a polynomial of degree n and let K be the splitting field of p(x) over F . Prove that [K : F ] in fact divides n!. Solution: We prove this result by induction on n. More precisely, we prove that for all positive integers n, for all fields F and all polynomials p(x) ∈ F [x] of degree n, if K is the splitting field of p(x), then [K : F ] divides n!. This result is obviously true for n = 1. Indeed, if n = 1, for all field F and all polynomials of degree 1 in F [x], the splitting field is F itself so [F : F ] = 1 and therefore divides 1!. Suppose that for some n > 1, the result is true for all k with k < n. In particular, suppose that for all k with k < n, for every field F 0 and every polynomial p(x) ∈ F 0 [x] of degree k, the splitting field K 0 of p(x) satisfies [K 0 : F 0 ] divides k!. Now consider the integer n. Let F be a field and p(x) ∈ F [x] a polynomial of degree n. Suppose first that p(x) is irreducible. Let α be a root of p(x). Then [F (α) : F ] = n and p(x) = (x − α)q(x) where q(x) is a polynomial of degree n − 1 in F (α)[x]. The splitting field K of q(x) over F (α) is the splitting field of p(x) over F . Furthermore, by the induction hypothesis, [K : F (α)] divides (n − 1)!. Hence, since [K : F ] = [K : F (α)][F (α) : F ] = [K : F (α)]n then [K : F ] divides n!. Now suppose that p(x) is reducible with p(x) = a(x)b(x) and deg a(x) = m and deg b(x) = ` both greater than 0. Let E be a splitting field of a(x) over F . By the induction hypothesis, we know that [E : F ] divides m!. We now consider b(x) as a polynomial in E[x]. Let K be a splitting field of b(x) over E. Also by the induction hypothesis, [K : E] divides `!. Thus [K : F ] = [K : E][E : F ] divides m!`!, n n! which in turns divides n! since m!`! since ` = n − m. Again, [K : F ] divides n!. = m By induction, for fields F and all polynomials p(x) ∈ F [x] of degree n, the index of a splitting field of p(x) over F divides n!. Exercise: 11 Section 7.6 Question: Let p(x), q(x) ∈ F [x] be two polynomials with deg p(x) = m and deg q(x) = n. Notice that p(q(x)) is a polynomial of degree mn. Prove that the splitting field E of p(q(x)) has a degree that satisfies [E : F ] ≤ m!(n!)m . Prove also that for for m, n ≥ 2, this quantity strictly divides (mn)!. Solution: Let K be the splitting field of the polynomial p(x) and let α1 , α2 , . . . , αm be the roots of p(x) possibly listed with multiplicity. By Theorem 7.6.3, [K : F ] ≤ m!. The (possible) mn roots of the polynomial p(q(x)) are the solutions to q(x) = αi . This leads to n polynomials q(x) − αi in K[x]. Set K0 = K. Call Ki the splitting field of q(x) − αi over Ki−1 [x]. By Theorem 7.6.3, we have [Ki : Ki−1 ] ≤ m!. The splitting field of p(q(x)) is Kn and we have [Kn : F ] = [Kn : Kn−1 ] · · · [K2 : K1 ][K1 : K0 ][K0 : F ] ≤ (m!)n n!. The final part of the exercise is to remark that m!(n!)m divides (mn)!. We note that (mn)! (n!)m

7.6. SPLITTING FIELDS AND ALGEBRAIC CLOSURE

401

is the multibinomial coefficient, which is an integer. Indeed, for any integer k, kn(kn − 1)(kn − 2) · · · (kn − n + 1) kn = . n! (k − 1)n Thus

(mn)! = (n!)m

n 2n 3n mn ··· . n n n n

However, we also have k(kn − 1)(kn − 2) · · · (kn − n + 1) kn − 1 kn kn(kn − 1)(kn − 2) · · · (kn − n + 1) = =k . = n! (n − 1)! n−1 (k − 1)n Thus, for all positive integers, k |

(mn)! kn m divides (mn)!. (k−1)n . Thus, m! divides (n!)m , so we conclude that m!(n!)

Exercise: 12 Section 7.6 Question: Let P (x) be a polynomial in F [x]. Suppose that a dynatomic polynomial ΦP,n (x) has degree k. Prove that if the roots of a dynatomic polynomial are only primitive n-cycles, then k is divisible by n and the degree of the splitting field E of ΦP,n (x) has an index [E : F ] that is less than or equal to k(k − n)(k − 2n) · · · (2n) · n · 1.

Solution: Let F be a field and P (x) ∈ F [x] and suppose that for a positive integer n the dynatomic polynomial ΦP,n (x) has degree k. If the roots of the dynatomic polynomial consist only of primitive n-cycles, then for any given root α of ΦP,n (x), the elements α, P (α), P 2 (α), · · · , P n−1 (α) are distinct roots of ΦP,n (x). Since P (x) ∈ F [x], then P j (α) ∈ F (α). Thus, if α is a root of ΦP,n (x), then ΦP,n (x) = (x − α)(x − P (α)) · · · (x − P n−1 (α))Qα (x) where Qα (x) is a polynomial of degree k − n in F (α)[x]. If the roots of the dynatomic polynomial consist of distinct primitive n-cycles, then ΦP,n (x) factors in k/n polynomials of the form (x − α)(x − P (α)) · · · (x − P n−1 (α)) for certain roots α. We define the sequence αi of roots of ΦP,n (x) as follows. Let α1 be any root of ΦP,n (x). Let F1 = F (α1 ). Then ΦP,n (x) = (x − α1 )(x − P (α1 )) · · · (x − P n−1 (α1 ))Q1 (x) in F1 [x]. Now for i ≥ 2, let αi be another root of ΦP,n (x) that is not in Fi−1 . Set Fi = Fi−1 (αi ). Then αi is a root of Qi−1 (x) and we have Qi−1 (x) = (x − αi )(x − P (αi )) · · · (x − P n−1 (αi ))Qi (x) in Fi [x]. Note that deg Qi (x) = k − ni. This terminates say at i = m when Qi (x) is a constant or all the roots of Qi (x) are in Fi . By construction, setting F0 = F we have [Fi+1 : Fi ] = [Fi (αi+1 ) : Fi ] ≤ k − ni for all i ≥ 0. Thus, [E : F ] = [F1 : F ][F2 : F1 ][F3 : F2 ] · · · [E : Fm ] ≤ k(k − n)(k − 2n) · · · (k − (m − 1)n) ≤ k(k − n)(k − 2n) · · · n · 1

Exercise: 13 Section 7.6 Question: Let p(x) ∈ Q[x] be a palindromic polynomial of even degree 2n. Let K be the splitting field of p(x). Prove that [K : Q] ≤ 2n n!.

402

CHAPTER 7. FIELD EXTENSIONS

Solution: By previous results on palindromic polynomials, there exists a polynomial q(x) of degree n such that p(x) = xn q(x + 1/x). The constant term of a palindromic polynomial is automatically nonzero so none of the roots of p(x) are 0. The roots αi of p(x) would solve 1 q αi + =0 αi Therefore, every root α of p(x) solves α + α1 = β, where β is a root of q(x). Let K be the splitting field of q(x). Then [K : Q] ≤ n!. Let βi for i = 1, . . . , n be the roots of q(x), possibly with multiplicity. Consider the n quadratic polynomials 1 x + = βi =⇒ x2 − βi x + 1 = 0. x We denote by αi,± the roots αi,± =

βi ±

p

βi2 − 4 . 2

Note that αi,− ∈ K(αi,+ ) and K(αi,+ ) : K] = 1 or 2 so mi = degK αi,+ = 1 or 2. The splitting field L of p(x) is L = K(α1,+ , α2,+ , . . . , αn,+ ). Then [L : Q] = [L : K][K : Q] ≤ m1 m2 · · · mn [K : Q] ≤ 2n n!.

Exercise: 14 Section 7.6 Question: Prove that a field F is algebraically closed if and only if the only the irreducible polynomials in F [x] are precisely the polynomials of degree 1. Solution: Let F be algebraically closed. Then if p(x) ∈ F [x] is a polynomial of degree 2 or greater, we have p(x) = (x − α)q(x) in F [x], where α is a root of p(x) and q(x) is a nonconstant polynomial. Thus p(x) is reducible. Hence, if F is algebraically closed, the only irreducible polynomials have degree 1. Conversely, suppose that F is a field such that the only irreducible polynomials in F [x] have degree 1. Let p(x) ∈ F [x] be any polynomial. Since F [x] is a UFD, the polynomial p(x) factors into a product of irreducible polynomials, hence into a product of polynomials of degree 1, as p(x) = (a1 x + b1 )(a2 x + b2 ) · · · (an x + bn ). Then the element −a−1 1 b1 ∈ F [x] is a root of p(x). Thus, every polynomial in F [x] has a root in F . Therefore, F is algebraically closed.

Exercise: 15 Section 7.6 Question: Prove that a field F is algebraically closed if and only if it has no proper algebraic extension. Solution: Suppose that a field F is algebraically closed. Then every polynomial polynomial p(x) factors into p(x) = c(x − α1 )(x − α2 ) · · · (x − αn ) with c, α1 , α2 , . . . , αn ∈ F . Thus, every element that is algebraic over F is in F , so F has no proper algebraic extension. Conversely, if a field F has not proper algebraic extension, then for every irreducible p(x) ∈ F [x], we have F [x]/(p(x)) = F , which means that every root of p(x) is in F . In particular, p(x) has a root in F and therefore F is algebraically closed.

7.7. FINITE FIELDS

403

7.7 – Finite Fields Exercise: 1 Section 7.7 Question: Give the addition and multiplication tables for F9 . Solution: We first need to give an expression for F9 . We need to write F9 as F3 [x]/(a(x)), where a(x) is an irreducible quadratic polynomial in F3 [x]. It is easy to check (since it has no roots) that x2 + 1 is irreducible in F3 [x]. Hence, we can write F9 as F3 [x]/(x2 + 1) or as F3 (θ), where θ is a root of x2 + 1 = 0. The elements of F9 are 0, 1, 2, θ, θ + 1, θ + 2, 2θ, 2θ + 1, 2θ + 2. The addition table is + 0 1 2 θ θ+1 θ+2 2θ 2θ + 1 2θ + 2

0 0 1 2 θ θ+1 θ+2 2θ 2θ + 1 2θ + 2

1 1 2 0 θ+1 θ+2 θ 2θ + 1 2θ + 2 2θ

2 θ 2 θ 0 θ+1 1 θ+2 θ+2 2θ θ 2θ + 1 θ + 1 2θ + 2 2θ + 2 0 2θ 1 2θ + 1 2

θ+1 θ+1 θ+2 θ 2θ + 1 2θ + 2 2θ 1 2 0

θ+2 2θ θ+2 2θ θ 2θ + 1 θ + 1 2θ + 2 2θ + 2 0 2θ 1 2θ + 1 2 2 θ 0 θ+1 1 θ+2

2θ + 1 2θ + 1 2θ + 2 2θ 1 2 0 θ+1 θ+2 θ

2θ + 2 2θ + 2 2θ 2θ + 1 2 0 1 θ+2 θ θ+1

For the multiplication table, it is useful to remember that θ2 + 1 = 0 so θ2 = 2. The multiplication table is × 0 1 2 θ θ+1 θ+2 2θ 2θ + 1 2θ + 2

0 0 0 0 0 0 0 0 0 0

1 0 1 2 θ θ+1 θ+2 2θ 2θ + 1 2θ + 2

2 θ 0 0 2 θ 1 2θ 2θ 2 2θ + 2 θ + 2 2θ + 1 2θ + 2 θ 1 θ+2 θ+1 θ + 1 2θ + 1

θ+1 θ+2 0 0 θ+1 θ+2 2θ + 2 2θ + 1 θ + 2 2θ + 2 2θ 1 1 θ 2θ + 1 θ + 1 2 2θ θ 2

2θ 0 2θ θ 1 2θ + 1 θ+1 2 2θ + 2 θ+2

2θ + 1 2θ + 2 0 0 2θ + 1 2θ + 2 θ+2 θ+1 θ + 1 2θ + 1 2 θ 2θ 2 2θ + 2 θ + 2 θ 1 1 2θ

Exercise: 2 Section 7.7 Question: Give the multiplication table for F16 . Solution: First, we need to express F16 in a useful way. It is not hard to show that x4 + x + 1 is irreducible in F2 [x]. Hence, F16 ∼ = F2 [x]/(x4 + x + 1), or in other words, F16 = F2 (θ), where θ is a root of the equation 4 x + x + 1. Every element in F16 can then be written as aθ3 + bθ2 + cθ + d, where a, b, c, d ∈ F2 . To write this more briefly, we will use the code of abcd for the expression aθ3 + bθ2 + cθ + d. Also for brevity, we do not write the rows or columns for multiplication by 0000, since we know that multiplication by 0000 always gives 0000. Finally, since the table is symmetric about the main diagonal, we only show the multiplications above the main diagonal.

404

CHAPTER 7. FIELD EXTENSIONS The multiplication table is × 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

0001 0001

0010 0010 0100

0011 0011 0110 0101

0100 0100 1000 1100 0011

0101 0101 1010 1111 0111 0010

0110 0110 1100 1010 1011 1101 0111

0111 0111 1110 1001 1111 1000 0001 0110

1000 1000 0011 1011 0110 1110 0101 1101 1100

1001 1001 0001 1000 0010 1011 0011 1010 0100 1101

1010 1010 0111 1101 1110 0100 1001 0011 1111 0101 1000

1011 1011 0101 1110 1010 0001 1111 0100 0111 1100 0010 1001

1100 1100 1011 0111 0101 1001 1110 0010 1010 0110 0001 1101 1111

1101 1101 1001 0100 0001 1100 1000 0101 0010 1111 1011 0110 0011 1110

1110 1110 1111 0001 1101 0010 0010 1100 1001 0111 0110 1000 0100 1010 1011

1111 1111 1101 0010 1001 0110 0100 1011 0001 1110 0011 1100 1000 0111 0101 1010

Exercise: 3 Section 7.7 Question: Define F8 as F2 [x]/(x3 + x + 1). Let θ ∈ F8 be an element such that θ3 + θ + 1 = 0. a) Write down the addition and the multiplication tables for elements in this field. b) In F8 , solve the equation (θ2 + 1)(α − (θ + 1)) = θα + 1. c) Solve the following system of two linear equations in two variables in the field F8 : ( θ2 α + (θ + 1)β =θ 2 2 (θ + 1)α + θ β = 1 Solution: Define F8 as F2 [x]/(x3 + x + 1). Let θ ∈ F8 be an element such that θ3 + θ + 1 = 0. a) The addition and multiplication tables are + 0 1 θ θ+1 θ2 θ2 + 1 θ2 + θ θ2 + θ + 1

0 0 1 θ θ+1 θ2 θ2 + 1 θ2 + θ θ2 + θ + 1

1 1 0 θ+1 θ θ2 + 1 θ2 θ2 + θ + 1 θ2 + θ

θ θ θ+1 0 1 2 θ +θ 2 θ +θ+1 θ2 θ2 + 1

θ+1 θ+1 θ 1 0 2 θ +θ+1 θ2 + θ θ2 + 1 θ2

and × 0 1 θ θ+1 θ2 θ2 + 1 θ2 + θ θ2 + θ + 1

0 0 0 0 0 0 0 0 0

1 0 1 θ θ+1 θ2 θ2 + 1 θ2 + θ θ2 + θ + 1

θ 0 θ θ2 θ2 + θ θ+1 1 2 θ +θ+1 θ2 + 1

θ+1 0 θ+1 θ2 + θ θ2 + 1 θ2 + θ + 1 θ2 1 θ

θ2 θ2 θ2 + 1 θ2 + θ θ2 + θ + 1 0 1 θ θ+1

θ2 0 θ2 θ+1 θ2 + θ + 1 θ2 + θ θ θ2 + 1 1

θ2 + 1 θ2 + 1 θ2 θ2 + θ + 1 θ2 + θ 1 0 θ+1 θ

θ2 + 1 0 θ2 + 1 1 θ2 θ θ2 + θ + 1 θ+1 θ2 + θ

θ2 + θ θ2 + θ θ2 + θ + 1 θ2 θ2 + 1 θ θ+1 0 1

θ2 + θ 0 θ2 + θ θ2 + θ + 1 1 θ2 + 1 θ+1 θ θ2

b) We solve the equation (θ2 + 1)(α − (θ + 1)) = θα + 1 =⇒ (θ2 + 1)α + θ2 = θα + 1 =⇒ (θ2 + θ + 1)α = θ2 + 1 =⇒ α = θ where the last equality is obtained from looking at the multiplication table of F8 . c) We solve the system of equations ( θ2 α + (θ + 1)β =θ 2 2 (θ + 1)α + θ β = 1

θ2 + θ + 1 θ2 + θ + 1 θ2 + θ θ2 + 1 θ2 θ+1 θ 1 0

θ2 + θ + 1 0 θ2 + θ + 1 θ2 + 1 θ 1 2 θ +θ θ2 θ+1

7.7. FINITE FIELDS

405

by eliminating the variable α: (θ2 + 1)(θ2 α + (θ + 1)β) + θ2 ((θ2 + 1)α + θ2 β) = θ2 + 1 =⇒θβ = θ2 + 1 =⇒β = θ2 + θ + 1 Then we solve for α by θ2 α + (θ + 1)(θ2 + θ + 1) = θ =⇒ θ2 α + θ = θ =⇒ α = 0. So the solution to the equation is (α, β) = (0, θ2 + θ + 1). Exercise: 4 Section 7.7 Question: Write x9 − x as a product of irreducible polynomials in F3 [x]. Solution: Over F3 [x], the roots of the polynomial x9 − x = x9 + 2x consist of all the elements in F9 . In particular, x9 + 2x is divisible by x(x + 1)(x + 2) and then by three irreducible monic quadratic polynomials. We have x9 + 2x = x(x + 1)(x + 2)(x2 + 1)(x2 + x + 2)(x2 + 2x + 2).

Exercise: 5 Section 7.7 Question: Show that every element besides the identity 1 in F32 is a generator of U (F32 ). Solution: To prove this result directly with computations would be ridiculous. We simply need to notice that U (F32 ) is a group with 31 elements. Hence, U (F32 ) ∼ = Z31 = hz | z 31 = 1i. Again, since 31 is prime, for all a 0 < a < 31, we have |z | = 31/ gcd(a, 31) = 31. Thus, every nonidentity element of Z31 is a generator, so every nonidentity element in U (F32 ) is a generator. Exercise: 6 Section 7.7 Question: Let p be an odd prime. Prove that xp−1 − 1 =

Y

(x − α).

α∈U (Fp )

Deduce that (p − 1)! ≡ −1 (mod p). (This fact is called Wilson’s Theorem.) Prove also that if n is a positive composite integer greater than 4, then (n − 1)! ≡ 0 (mod n). Solution: By Fermat’s Little Theorem, every element in Fp solves xp − x = 0. There are p distinct roots of xp − x. By factoring out x, we deduce that every element of U (Fp ) is a root of xp−1 − 1. Thus in Fp [x], we have Y xp−1 − 1 = (x − α). α∈U (Fp )

There are even number of elements in U (Fp ) so, after expanding the above product and comparing constant terms, in Fp = Z/pZ we have 1 · 2 · · · p − 1 = −1. Hence in Z, −1 ≡ (p − 1)!

(mod p).

If n is a composite integer greater than 4, then n = ab with 1 < a, b < n. Suppose that a 6= b, then a and b are distinct factors in (n − 1)! so (n − 1)! ≡ 0 (mod n). If n = a2 , then, since n > 4, both a and 2a appear as factors in (n − 1)!. Thus, in this case as well, (n − 1)! ≡ 0 (mod n). Exercise: 7 Section 7.7 Question: Let a ∈ Fp − {0}. Prove that the polynomial xp − x + a ∈ Fp [x] is irreducible. Solution: We know that the p distinct roots of xp − x are precisely the p elements of Fp . Thus the roots of xp − x + a are in a nontrivial field extension of Fp . Let α be a root of xp − x + a, where a 6= 0. Since (a + b)p = ap + bp in any field of characteristic p, we observe that (α + 1)p − (α + 1) + a = αp + 1 − α − 1 + a = 0.

406

CHAPTER 7. FIELD EXTENSIONS

Hence, if α is a root of xp − x + a, then so is α + 1, and by extension so are α + b, for all b ∈ Fp . This gives us all the roots of xp − x + a in terms of one root α. Now suppose that f (x) ∈ Fp [x] is an irreducible factor of xp − x + a of degree d. Then f (x) will have d distinct roots. Furthermore, for all b ∈ F[ x], each polynomial f (x + b) is irreducible of degree d with d distinct roots. Then xp − x + a must be divisible by some f (x + b). Since f (x + b1 ) and f (x + b2 ) are either equal or relatively prime, we deduce that xp − x + a is a product of distinct polynomials of the form f (x + b). Thus, we deduce that d | p. Hence, either d = 1 or d = p. But we know that the roots of f are not in Fp [x], so d > 1. Hence, p and we deduce that xp − x + a is irreducible. Exercise: 8 Section 7.7 Question: Suppose that d | n. Prove that Fpd ⊆ Fpn and that [Fpn : Fpd ] = nd . k

Solution: Recall that the elements of Fpd consist of the roots of xp − x (which are distinct). Suppose that d | n with n = md, then pn − 1 = (pd )m − 1 = (pd − 1)(pd(m−1) + pd(m−2) + · · · + 1). Then in Fp [x], if we set A = (pd(m−1) + pd(m−2) + · · · + 1) we have A d n n d d d xp − x = x(xp −1 − 1) = x (xp −1 − 1 = x(xp −1 − 1) (xp −1 )A−1 + (xp −1 )A−2 + · · · + 1 . d

n

From this we see that xp − x divides xp − x. Hence, all the elements of Fpd are in Fpn . Thus, Fpn is a field extension of Fpd . 0 In particular, Fpn is a vector space over Fpd . If [Fpn : Fpd ] = m0 , then because we have finite sets, pn = (pd )m . Hence, p = dm0 , so m0 = m = n/d. Exercise: 9 Section 7.7 Question: Consider the polynomial p(x) = x4 + x + 1 ∈ F2 [x]. a) Show that p(x) is irreducible. b) Show that p(x) factors into two quadratics over F4 and exhibit these two quadratic polynomials. Solution: Consider the polynomial p(x) = x4 + x + 1 ∈ F2 [x]. a) It is easy to check that x4 + x + 1 has no roots in F2 . Hence, if x4 + x + 1 is reducible, then x4 + x + 1 is a product of two irreducible quadratic polynomials. The only irreducible quadratic polynomial in F2 [x] is x2 + x + 1. Since (x2 + x + 1)(x2 + x + 1) = x4 + x2 + 1, then x4 + x + 1 is irreducible. b) We can recognize F4 as F2 [x]/(x2 + x + 1). We write r as an element in F4 such that r2 + r + 1 = 0. We have in F4 [x] (x2 + x + r)(x2 + x + r + 1) = x4 + 2x3 + (r + r + 1 + 1)x2 + (r + r + 1)x + r2 + r = x4 + x + 1.

Exercise: 10 Section 7.7 Question: Prove that a polynomial f (x) over a field F of characteristic 0 is separable if and only if it is the product of irreducible polynomials that are not associates of each other. [Note: Consequently, separable polynomials over a field of characteristic 0 are polynomials that are square-free in F [x].] Solution: Suppose that a polynomial f (x) ∈ F [x] is separable. Then by definition it has no double roots in its splitting field E. To each root α ∈ E of f (x), there is associated a minimal polynomial mα,F (x). Furthermore, mα,F (x) divides f (x). We can define a relation R on the set of roots of f (x) in E by α1 R α2 to mean α2 is a root of mα1 ,F (x). It is easy to see that the relation R is an equivalence relation. This defines a partition A = {A1 , A2 , . . . , Ak } on the set of roots of f (x). Then fi (x) =

Y α∈Ai

(x − α) = mβ,F (x),

7.7. FINITE FIELDS

407

where β is any element in Ai . Then f (x) = af1 (x)f2 (x) · · · fk (x), where a is the leading coefficient of f (x) and each fi (x) is an irreducible polynomial. Furthermore, since none of the fi (x) share roots, they are not associates of each other. Suppose that a polynomial f (x) ∈ F [x] is such that f (x) = f1 (x)f2 (x) · · · fk (x) with each fi (x) irreducible and no two irreducible factors are associates of each other. Let α be a root of f (x). Then α is a root of some fi (x). As a consequence of Proposition 7.2.2, fi (x) is a constant multiple of mα,F (x). Since fi (x) are not associate of each other, then no root of fi (x) is also a root of fj (x) if i 6= j. However, by Proposition 7.7.6, every irreducible polynomial in a field of characteristic 0 is separable. Hence, none of the fi (x) have double roots and therefore, f (x) does not have double roots, which means that f (x) is separable. Exercise: 11 Section 7.7 Question: Find a generator of U (F27 ). Solution: We know that U (F27 ) is a cyclic group of order 26 = 2 × 3. Consequently, not every nonidentity element will serve as a generator for the group. Let us write F27 as F3 (θ), where θ solves the equation x3 +x+2. We simply need to find an element α ∈ F3 (θ) such that α2 6= 1 and α13 6= 1. We try θ itself first. Obviously, θ2 6= 1. We have θ3 = 2θ + 1

θ6 = (θ3 )2 = θ2 + θ + 1

θ12 = (θ6 )2 = 2θ2 + 1 θ13 = 2θ + 2. Since θ13 6= 1, then by a corollary to Lagrange’s Theorem, we know that the order of θ must be 26, and hence θ is a generator of U (F27 ). Exercise: 12 Section 7.7 Question: The polynomial p1 (x) = x2 + x + 1 ∈ F5 [x] is irreducible. Call θ an element in F25 = F5 [x]/(p1 (x)) that satisfies θ2 + θ + 1 = 0. a) Find all other irreducible monic quadratic polynomials in F5 [x]. b) For each of the 10 polynomials found in the previous part, write the two roots in F25 as aθ + b for a, b ∈ F5 . Solution: We work with F25 as F5 (θ), where θ is a root of the irreducible polynomial x2 + x + 1 ∈ F5 [x]. a) By Theorem 7.7.14, there are 21 (52 − 5) = 10 monic irreducible quadratic polynomials in F5 [x]. Given any a ∈ F5 , we know that a polynomial f (x) is irreducible if and only if f (x + a) is irreducible. Using this result and starting from x2 + x + 1, we get five irreducible monic polynomials: x2 + x + 1,

x2 + 3x + 3,

x2 + 2,

x2 + 2x + 3,

x2 + 4x + 1.

We can also easily check that x2 + x + 2 does not have any roots in F5 , so since it is a quadratic, it is irreducible. Then repeating the above trick, the 5 remaining irreducible polynomials: x2 + x + 2,

x2 + 3x + 4,

x2 + 3,

x2 + 2x + 4,

x2 + 4x + 2.

Exercise: 13 Section 7.7 Question: Let q = pn . Prove that the Frobenius automorphism ϕ = σp : Fq → Fq is a Fp -linear transformation. Prove also that ϕn is the identity transformation. Solution: Recall that Fq is a vector space of dimension n over Fp . To prove that ϕ is a linear transformation, we point out that for all α, β ∈ Fq and all c ∈ Fp , we have ϕ(α + β) = (α + β)p =

p X p i=0

because p |

p i

i

αp−i β i = αp + β p = ϕ(α) + ϕ(β),

for all i with 1 ≤ i ≤ p − 1. Furthermore, ϕ(cα) = cp αp = cαp = cϕ(α),

because cp = c for all c ∈ Fp .

408

CHAPTER 7. FIELD EXTENSIONS For all α ∈ Fq , we have 2

n

ϕn (α) = ϕn−1 (αp ) = ϕn−2 ((αp )p ) = ϕn−2 (αp ) = · · · = αp = αq . However, we know from the proof of Theorem 7.7.12 that Fq consists of all the solutions of the equation xq − x. Hence, ϕn (α) = α for all α ∈ Fq . Thus, ϕn is the identity linear transformation.. Exercise: 14 Section 7.7 Question: Consider the Frobenius map ϕ from the previous exercise. Determine the eigenvalues and all corresponding eigenspaces for ϕ. Solution: (This question is too complicated and too much variety in its answer to be asked as a regular exercise. This could be explored as a project. The Frobenius automorphism always has the eigenvalue of 1 with Fp as a one-dimensional subspace of eigenspace 1. In general, the expression of the eigenspaces will depend on the choice of irreducible polynomial used to define a generator of Fq ,) (The following exercise shows a particular example with this question. The eigenvalues are 1 and 2.) Exercise: 15 Section 7.7 Question: Consider the Frobenius automorphism σ3 : F9 → F9 . Show how σ3 maps the elements of F9 . [Hint: Use the identification F9 = F3 [x]/(x2 + x + 2).] Solution: We view F9 as F3 (θ), where θ solves the equation x2 + x + 2 = 0. An ordered basis of F9 is then (1, θ). We have σ(1) = 1 σ(θ) = θ3 = θ(1 + 2θ) = θ + 2θ2 = 2θ + 2 Hence, a matrix for σ is 1 0

2 . 2

in other words, with a, b ∈ F3 , σ(a + bθ) = a + b(2 + 2θ) = (a + 2b) + 2bθ.

Exercise: 16 Section 7.7 Question: Prove that (1 + xp )n = (1 + x)pn in Fp [x]. Deduce that

pn pk p

≡

n k

(mod p).

Solution: In the finite field of characteristic p, we note that (1 + x) = 1 + xp because p divides 1 ≤ k ≤ p − 1. Then, n (1 + x)pn = ((1 + x)p ) = (1 + xp )n .

p k

= 0 for

We consider the binomial formula for both sides of this identity and get pn n X pn j X n pk x = x . j k j=0 k=0

By identifying coefficients of polynomials, we see that ( if p - j pn ∼ 0 (mod p) = n j (mod p) if j = pk. k

Exercise: 17 Section 7.7 Question: Let Fq be a finite field and let f (x) be an irreducible polynomial of degree n in Fq [x]. Suppose that α is one of the roots of f (x) in the field Fqn . Prove that 2

α, αq , αq , . . . , αq

n−1

7.7. FINITE FIELDS

409

are the n distinct roots of f (x) in Fqn . Solution: Suppose that q = pk . By Exercise 13 of this section, we point out that the iterated Frobenius automorphisms σpk is the identity on Fq . If f (x) = an xn + · · · + a1 x + a0 , then an αn + · · · + a1 α + a0 = 0. Applying σpk to this equality and then using the automorphism properties, we have σpk (an )σpk (α)n + σpk (an−1 )σpk (α)n−1 + · · · + σpk (a0 ) = 0 ⇐⇒

an (αq )n + an−1 (αq )n−1 + · · · + a0 = 0.

m

Thus, αq is also a root of f (x). By induction, αq is a root of f (x) for all m ≥ 0. n kn m Now Fq [x]/(f (x)) ∼ = Fpkn ∼ = Fqn . So αq = αp = α, again by Exercise 13. At most n of the elements αq m are distinct. But if αq = α for some m < n, then α is an element of Fpkm = Fqm so α would be the root of m an irreducible polynomial of degree less that n. This is a contradiction, so αq , with m = 0, 1, . . . n − 1 are all different from α. Since α was an arbitrary root of f (x), then all elements 2

α, αq , αq , . . . , αq

n−1

are distinct and roots of f (x). Exercise: 18 Section 7.7 Question: Prove that a polynomial of degree m = 2k over F2 [x] is irreducible if and only if it divides 2k

(x2

2k−1

+ x)/(x2

+ x). m

m

Solution: By the proof of Theorem 7.7.12, the elements of F2m are precisely the roots of x2 − x = x2 + x. If m = 2k , then the roots of an irreducible polynomial f (x) of degree m are elements in F2m but not in any 2k

subfield. The subfields of F22k are F22d for any 0 ≤ d < k. So roots of f (x) are roots of x2 of x

22

+ x but not roots

k−1

+ x. Hence, roots of f (x) are roots of 2k

(x2

2k−1

+ x)/(x2

+ x).

Thus, a polynomial of degree m = 2k is irreducible if and only if it divides the above polynomial.

8 | Group Actions 8.1 – Introduction to Group Actions Exercise: 1 Section 8.1 Question: Consider the natural action of D7 on labeled vertices of a regular heptagon. Consider the induced permutation representation ρ : D7 → S7 . Determine ρ(r) and ρ(s) with respect to your labeling of the vertices and choice of reflection for s. Solution: We assume that the heptagon is centered at (0, 0) and symmetric about the x-axis. We label the vertices in a counterclockwise orientation around the heptagon starting with 1 at the vertex that is on the positive x-axis. Let ρ : D7 → S7 be the permutation representation induced by the action of D7 on {1, 2, 3, 4, 5, 6, 7}. Then ρ(r) = (1 2 3 4 5 6 7) and ρ(s) = (2 7)(3 6)(4 5).

Exercise: 2 Section 8.1 Question: Let n be a positive integer and consider the group GLn (R). Prove that the pairing GLn (R)×R → R defined by A · x = det(A)x is a group action. Prove also that it is not faithful. Solution: We consider the pairing GLn (R) × R → R defined by A · x = det(A)x. Let A, B ∈ GLn (R) and let x ∈ R. Then A · (B · x) = A · (det(B)x) = det(A) det(B)x = det(AB)x = (AB) · x. This is the first criterion of a group action. Furthermore, I · x = det(I)x = x. This verifies both criteria of a group action. The kernel of this group action is the subgroup of GLn (R) of matrices A such that det(A)x = x for all x ∈ R. This means that det(A) = 1, so the kernel is SLn (R). Since this is nontrivial for n ≥ 2, we see that the action is not faithful as soon as n ≥ 2. Exercise: 3 Section 8.1 Question: Let G = D6 and let H = hr2 i. Since H E G, conjugation of G on H is an action of G on H. Label the H elements 1, r2 , and r4 as 1, 2, and 3 respectively and consider the induced permutation representation ρ : D6 → S3 . a) Exhibit the images under ρ of all elements in D6 . b) State the kernel of ρ and show that the action is not faithful. Solution: We consider the induced permutation representation ρ : D6 → S3 . a) We know that g1g −1 = 1 for all g ∈ D6 . Hence, for all g ∈ D6 , ( 1 if gr2 g −1 = r2 ρ(g) = (2 3) if gr2 g −1 = r4 . It is easy to see that if g = rk , then gr2 g −1 = r2 , whereas if g = srk , then gr2 g −1 = srk r2 r−k s = sr2 s = r−2 s2 = r4 . Thus ρ(rk ) = 1 for all k and ρ(srk ) = (2 3). b) The kernel of the action if ker ρ = hri. Since this is not the trivial subgroup, the action is not faithful. Exercise: 4 Section 8.1 Question: Let G be a group acting on a set X. Show that defining def

g · S = {g · s | s ∈ S} for all g ∈ G and all S ⊆ X induces an action of G on P(X). 411

412

CHAPTER 8. GROUP ACTIONS def

Solution: Let G be a group acting on a set X. Define g · S = {g · s | s ∈ S} for all g ∈ G and all S ⊆ X. This is a pairing G × P(X) → P(X). For all g1 , g2 ∈ G and for all S ∈ P(X), we have g1 · (g2 · S) = g1 · {g2 · s | s ∈ S} = {g1 · (g2 · s) | s ∈ S} = {(g1 g2 ) · s | s ∈ S}

since G acts on X

= (g1 g2 ) · S. Furthermore, for all S ∈ P(X), we have 1 · S = {1 · s | s ∈ S} = {s | s ∈ S} = S. These two prove the two criteria of a group action. Exercise: 5 Section 8.1 Question: Let F be a field, let G = GLn (F ), and let X = Mn×n (F ) be the set of n × n matrices with entries in F . Discuss how the relation of similarity on square matrices in Mn×n (F ) is related to a group action of G on X. Solution: Two matrices A, B in X = Mn×n (F ) are similar if and only if there exists an invertible matrix P ∈ GLn (F ) such that B = P AP −1 . Define pairing GLn (F ) × Mn×n (F ) → Mn×n (F ) by P · A = P AP −1 . Then for P, Q ∈ GLn (F ), P · (Q · A) = P · (QAQ−1 ) = P QAQ−1 P −1 = (P Q)A(P Q)−1 = (P Q) · A. Furthermore, I · A = IAI −1 = A. These two statements affirm the criteria of a group action. Hence, the similarity P AP −1 is the result of a group action operation on A. Exercise: 6 Section 8.1 Question: Let G = D6 and let H = hr2 i. Since H E G, conjugation of G on H is an action of G on H. Label the H elements 1, r2 , and r4 as 1, 2, and 3 respectively and consider the induce permutation representation ρ : D6 → S3 . a) Exhibit the images under ρ of all elements in D6 . b) State the kernel of ρ and show that the action is not faithful. Solution: a) We will represent any generic element of H as r2a for some integer a. It is easy to see that for any power of r and any element of H we have, rb r2a r−b = r2a . So that any power of r does not permute the elements of H under conjugation. So ρ(rb ) = ( 1 ) for all b. Now if we have an element of the form srb we can see that srb (1)srb = 1 srb (r2 )srb = srb sr4 rb = ssr−b r4 rb = r4 srb (r4 )srb = srb sr2 rb = ssr−b r2 rb = r2 So that r2 goes to r4 and r4 is sent to r2 . This shows that ρ(srb ) = ( 2 3 ) for any b. b) From the part a we can see that Ker(ρ) = {1, r, r2 , r3 , r4 , r5 } = 6 {1}. So the action is not faithful. Exercise: 7 Section 8.1 Question: Consider the action defined in Exercise 8.1.6 where n = 4 and k = 2. The induced permutation representation is a homomorphism ρ : S4 → S6 . Label the elements in P2 ({1, 2, 3, 4}) according to the following chart. 1 2 3 4 5 6 label subset {1, 2} {1, 3} {1, 4} {2, 3} {2, 4} {3, 4} Give ρ(σ) as a permutation in S6 for the permutations σ = 1, (1 2), (1 2 3), (1 2)(3 4), and (1 2 3 4). Solution: The images of ρ of the desired elements are ρ(1) = 1 ρ((1 2)) = (2 4)(3 5) ρ((1 2 3)) = (1 4 2)(3 5 6) ρ((1 2)(3 4)) = (2 5)(3 4) ρ((1 2 3 4)) = (1 4 6 3)(2 5).

8.1. INTRODUCTION TO GROUP ACTIONS

413

Exercise: 8 Section 8.1 Question: Let Pn be the of polynomials in R[x] that have degree n or less (including the 0) polynomial. Consider σ ∈ Sn+1 as a permutation on {0, 1, 2, · · · , n} and define σ · (an xn + · · · + a1 x + a0 ) = aσ−1 (n) xn + aσ−1 (n−1) xn−1 + · · · + aσ−1 (1) x + aσ−1 (0) . a) Prove that this defines an action of Sn+1 on Pn . b) Decide with proof whether σ · (a(x) + b(x)) = σ · a(x) + σ · b(x). c) Decide with proof whether σ · (a(x)b(x)) = (σ · a(x))(σ · b(x)). Solution: We consider the action of Sn+1 on the set of polynomials of degree n or less by σ · (an xn + · · · + a1 x + a0 ) = aσ−1 (n) xn + aσ−1 (n−1) xn−1 + · · · + aσ−1 (1) x + aσ−1 (0) . a) Let σ, τ ∈ Sn+1 and let p(x) = an xn + · · · + a1 x + a0 . Then τ · (σ · p(x)) = τ · (aσ−1 (n) xn + aσ−1 (n−1) xn−1 + · · · + aσ−1 (1) x + aσ−1 (0) ) = aσ−1 (τ −1 (n)) xn + aσ−1 (τ −1 (n−1)) xn−1 + · · · + aσ−1 (τ −1 (1)) x + aσ−1 (τ −1 (0)) = a(τ σ)−1 (n) xn + a(τ σ)−1 (n−1) xn−1 + · · · + a(τ σ)−1 (1) x + a(τ σ)−1 (0) = (τ σ) · p(x). This is the first criterion of a group action. (In the above calculation, it is important to consider the coefficients aσ−1 (i) as bi so that when τ acts on σ · p(x), the τ is applied in the proper manner as done.) Also, it is obvious that 1 · p(x) = p(x). These show that the defined pairing is a group action. b) Let a(x) and b(x) be polynomials in Pn . If the degree of a(x) or b(x) are less than n, we assign a value of 0 to the coefficients ai or bi as necessary. Write a(x) = an xn + · · · + a1 x + a0 , b(x) = bn xn + · · · + b1 x + b0 , and c(x) = a(x) + b(x). Then for all σ ∈ Sn+1 , σ · (c(x)) = cσ−1 (n) xn + cσ−1 (n−1) + · · · + cσ−1 (1) x + cσ−1 (0) = (aσ−1 (n) + bσ−1 (n) )xn + · · · + (aσ−1 (1) + bσ−1 (1) )x + (aσ−1 (0) + bσ−1 (0) ) = σ · a(x) + σ · b(x). c) Now suppose that a(x), b(x) ∈ Pn are such that a(x)b(x) ∈ Pn (x). (This means that the sum of the degrees of a(x) and b(x) is less than or equal to n.) Call c(x) = a(x)b(x). Then the coefficients of c(x) are ck =

k X

ai bk−i .

i=0

Now σ −1 (k)

cσ−1 (k) =

X

ai bσ−1 (k)−i

i=0

Let n = 2 and let a(x) = x + 2 and b(x) = 3x + 4. Then a(x)b(x) = 3x2 + 10x + 8. Let σ = (0 1 2). Then σ · a(x) = x2 + 2x σ · b(x) = 3x2 + 4x σ(a(x)b(x)) = 10x2 + 8x + 3 (σ · a(x))(σ · b(x)) = 3x4 + 10x3 + 8x2 . It is clear that σ(a(x)b(x)) is generally not equal to (σ · a(x))(σ · b(x)). In fact, (σ · a(x))(σ · b(x)) might no longer be in Pn .

414

CHAPTER 8. GROUP ACTIONS

Exercise: 9 Section 8.1 Question: Suppose that H is a group acting on a set X and let ϕ : G → H be a group homomorphism. Prove that the pairing G × X → X defined by g · x = ϕ(g) · x, where the action symbol on the right is the action of H on X, defines an action of G on X. Solution: Let g1 , g2 ∈ G be arbitrary and let x ∈ X. Then, (g1 g2 ) · x = ϕ(g1 g2 ) · x

by definition

= (ϕ(g1 )ϕ(g2 )) · x

since ϕ is a homomorphism

= ϕ(g1 ) · (ϕ(g2 ) · x)

by the action of H on X

= g1 · (g2 · x)

by definition of the action of G on X.

Also, for all x ∈ X, we have 1G · x = ϕ(1G ) · x = 1H · x = x. We have shown the two criteria for a group action. Exercise: 10 Section 8.1 Question: Let G be a group acting on a set X and let ρ : G → SX be the induced permutation representation. Prove that the mapping (G/ Ker ρ) × X → X defined by (g Ker ρ) · x = g · x is an action of G/ Ker ρ on X. Prove also that this action is faithful. Solution: Let G be a group acting on a set X and let ρ : G → SX be the induced permutation representation. Since we define the mapping (G/ Ker ρ) × X → X by (g Ker ρ) · x = g · x namely based on an equivalence class representative, we first need to check that this is actually a function. Suppose that g1 Ker ρ = g2 Ker ρ. Then g2−1 g1 ∈ Ker ρ. Then for all x ∈ X, we have (g2−1 g1 ) · x = x. Hence, by group action properties g2 · x = g2 · ((g2−1 g1 ) · x) = (g2 g2−1 g1 ) · x = g1 · x. This shows that for any representative g of the coset g Ker ρ, the pairing g · x is the same. Hence, this pairing does define a function. Now let g, h ∈ G and x ∈ X. Then (gh Ker ρ) · x = (gh) · x = g · (h · x) = g · ((h Ker ρ) · x) = (g Ker ρ) · ((h Ker ρ) · x). Also, 1G/ Ker ρ · x = (Ker ρ) · x = x by definition of the kernel of a group action. We have shown the two criteria for a group action. The kernel of the action of G/ Ker ρ on x consists of all cosets g Ker ρ such that g · x = x for all x. But this implies that g ∈ Ker ρ, so g Ker ρ = Ker ρ. Hence, the action of G/ Ker ρ on X is faithful. Exercise: 11 Section 8.1 Question: In this exercise, we explore the actions described in Example 8.1.12. Consider the set of functions X = Fun({0, 1, 2}, {0, 1, 2}) be the set of functions from {0, 1, 2} to itself. The set X has cardinality 27. We label the functions in X by fk with 0 ≤ k ≤ 26 as the function fk (i) = ai where k = (a2 a1 a0 )3 as an integer expressed in base 3. Hence, f14 satisfies f14 (0) = 2,

f14 (1) = 1,

f14 (2) = 1

because in base 3, the integer 14 is 14 = (112)3 . Consider elements in S3 as permutations on {0, 1, 2}. Using the described labeling of functions in X, a) Give ρ1 (σ) as a permutation on {0, 1, 2, . . . , 26} for the permutation representation ρ1 induced from the action of S3 on X defined by (σ · f )(x) = σ · f (x). b) Give ρ2 (σ) as a permutation on {0, 1, 2, . . . , 26} for the permutation representation ρ2 induced from the action of S3 on X defined by (σ · f )(x) = f (σ −1 · x). Solution: We consider the set of functions X = Fun({0, 1, 2}, {0, 1, 2}) with the notation fi as described in the exercise. Since we are considering elements of S3 as bijections on {0, 1, 2}, we will write the permutations as (0 1) for example for the transposition switching 0 and 1. We also denote permutations on {0, 1, 2, . . . , 26} using usual cycle notation but with integers taken from {0, 1, . . . , 26}.

8.1. INTRODUCTION TO GROUP ACTIONS

415

a) Obviously, id ·fi = fi for all fi ∈ X, so as expected, ρ1 (id) = id. For the following calculations, (σ · fk )(i) = σ(fk (i)) = σ(ai ), where k = (a2 a1 a0 )3 as an integer expressed in base 3. Thus, σ · fk = σ · f(a2 a1 a0 )3 = f(σ(a2 )σ(a1 )σ(a0 ))3 . For example, (0 1) · f14 = (0 1) · f(112)3 = f(002)3 = f2 . The following calculations are tedious, but we get ρ1 ((0 1)) = (0 13)(1 12)(2 14)(3 10)(4 9)(5 11)(6 16)(7 15)(8 17)(18 22)(19 21)(20 23)(24 25) ρ1 ((0 2)) = (0 26)(1 25)(2 24)(3 23)(4 22)(5 21)(6 20)(7 19)(8 18)(9 17)(10 16)(11 15)(12 14) ρ1 ((1 2)) = (1 2)(3 6)(4 8)(5 7)(9 18)(10 20)(11 19)(12 24)(13 26)(14 25)(15 21)(16 23)(17 22) ρ1 ((0 1 2)) = (0 13 26)(1 14 24)(2 12 25)(3 16 20)(4 17 18)(5 15 19)(7 11 21)(8 9 22) ρ1 ((0 2 1)) = (0 16 13)(1 24 14)(2 25 12)(3 20 16)(4 18 17)(5 19 15)(7 21 11)(8 22 9).

b) Again, we have id ·fi = fi for all fi ∈ X, so as expected, ρ2 (id) = id. For this second action described on X, we define (σ · fk )(i) = fk (σ −1 · i) = aσ−1 (i) when k = (a2 a1 a0 )3 in ternary. Thus, (σ · fk ) = fk0 , where k 0 is expressed in ternary as (aσ−1 (2) aσ−1 (1) aσ−1 (0) )3 . We get ρ2 ((0 1)) = (1 3)(2 6)(5 7)(10 12)(11 15)(14 16)(19 21)(20 24)(23 25) ρ2 ((0 2)) = (1 9)(2 18)(4 12)(5 21)(7 15)(8 24)(11 19)(14 22)(17 23) ρ2 ((1 2)) = (3 9)(4 10)(5 11)(6 18)(7 19)(8 20)(15 21)(16 22)(17 23) ρ2 ((0 1 2)) = (1 3 9)(2 6 18)(4 12 10)(5 15 19)(7 21 11)(8 24 20)(14 16 22)(17 25 23) ρ2 ((0 2 1)) = (1 9 3)(2 18 6)(4 10 12)(5 19 15)(7 11 21)(8 20 24)(14 22 16)(17 23 25).

Exercise: 12 Section 8.1 Question: Suppose that a group G acts on a set X and let Y = Fun(X, X) be the set of all functions from X to X. a) Prove that the pairing ? : G × Y → Y defined by (g ? f )(x) = g · f (g −1 · x) is an action of G on Fun(X, X). b) Prove also that the the subset of bijections from X to X is invariant under this action. Solution: Suppose that a group G acts on a set X. We define the action of G on Y = Fun(X, X) by (g ? f )(x) = g · f (g −1 · x) for all x ∈ X. a) Let g1 , g2 ∈ G and let f : X → X be any function. Then for all x ∈ X, we have ((g1 g2 ) ? f )(x) = (g1 g2 ) · f ((g1 g2 )−1 · x) = (g1 g2 ) · f ((g2−1 g1−1 ) · x) = g1 · g2 · f (g2−1 · (g1−1 · x)) = g1 · (g2 ? f )(g1−1 · x) = (g1 ? (g2 ? f ))(x). Furthermore, for the identity 1G ∈ G we have (1G ? f )(x) = 1G · f (1G · x) = f (x) for all x, so 1G acts trivially on the function f . We have established both criteria of a group action, so G acts on Fun(F, F ). b) Consider the set SX of bijections in Fun(X, X). We wish to prove that if g ∈ G and σ ∈ SX , then g ? σ ∈ SX . Let σ −1 be the inverse function of σ. Then for all x ∈ X we have ((g ? σ) ◦ (g ? σ −1 ))(x) = (g ? σ) (g ? σ −1 )(x) = (g ? σ)(g · σ−1(g −1 · x)) = g · σ g −1 · (g · σ−1(g −1 · x)) = g · σ(1G · σ −1 (g −1 · x)) = g · σ(σ = g · (g = x.

−1

−1

(g

−1

· x)

· x))

by the group action property

416

CHAPTER 8. GROUP ACTIONS By replacing σ with σ −1 and vice versa in the above calculation, we also show that ((g?σ −1 )◦(g?σ))(x) = x. Thus, g ? σ is a bijection with inverse function g · σ −1 . Consequently, the action of g on Fun(X, X) maps bijections to bijections, so SX is invariant under the action. [Note: this is not synonymous with “fixed under the action.”]

Exercise: 13 Section 8.1 Question: Let G be a group. Let φ : X → Y and ψ : Y → Z be G-set homomorphisms between G-sets. Prove that the composition ψ ◦ φ is a G-set homomorphism. Solution: We use the following symbols to describe the action of G on X, Y and on Z: If x ∈ X, we denote g · x for the group action pairing; if y ∈ Y , we denote g ∗ y for the group action pairing on Y ; and if z ∈ Z, we denote g ? z for the group action pairing. Let φ : X → Y and ψ : Y → Z be G-set homomorphisms between G-sets. Thus, for all x ∈ X, and all g ∈ G, we have ψ(φ(g · x)) = ψ(g ∗ φ(x))

because φ is a G-set homomorphism

= g ? (ψ(φ(x)))

because ψ is a G-set homomorphism.

Thus, the composition ψ ◦ φ is a G-set homomorphism. Exercise: 14 Section 8.1 Question: Let f : X → Y be a G-set homomorphism between G-sets. a) Prove that if S is a G-subset of X, then f (S) is a G-subset of Y . b) Prove that if T is a G-subset of Y , then f −1 (T ) is a G-subset of X. Solution: Let f : X → Y be a G-set homomorphism between G-sets. Let us denote by g · x the pairing of the action of G on X and by g ? y the pairing of the action of G on Y . a) Suppose that S is a G-subset of X. We consider f (S) as a subset of Y . Let w ∈ f (S). Then there exists s ∈ S such that w = f (s). Then g ? f (s) = f (g · s), which is in f (S), since g · s ∈ S. Thus, f (S) is a G-subset of Y . b) Suppose that T is a G-subset of Y . Let u ∈ f −1 (T ). Then f (u) ∈ T , or in other words, f (u) = t for some t ∈ T . Thus f (g · u) = g ? f (u) = g ? t, which is in T . Hence, g · u is in f −1 (T ) and we conclude that f −1 (T ) is closed under the action of G, i.e., that f −1 (T ) is a G-subset of X. Exercise: 15 Section 8.1 Question: Let V be the set of vertices of a cube and let D be the long diagonals. Define the function f : V → D so that f (v) is the unique long diagonal that contains the vertex v. Let G be the group of rigid motions of the cube and consider the natural actions of G on V and on D. Prove that f is a G-set homomorphism. Solution: A cube has eight vertices v1 , v2 , . . . , v8 and four long diagonals. We can view each long diagonal as an unordered pair {vi , vj }. Note that each vertex arises in a single long diagonal. The function f : V → D is defined as f (v) = {v, v 0 }, where {v, v 0 } is the unique long diagonal containing v. Let g be a rigid motion of the cube. Since the group G of rigid motions of the cube satisfies g · {vi , vj } = {g · vi , g · vj }, then for all v ∈ V , we have g · f (v) = g · {v, v 0 } = {g · v, g · v 0 } = f (g · v). Thus, f is a G-set homomorphism. We also see that g · v 0 = (g · v)0 in this exercises notation. Exercise: 16 Section 8.1 Question: Let X be a set and let G and H be subgroups of SX . Prove that the actions of G on X and H on X are permutation isomorphic if and only if G and H are conjugate subgroups in SX . Solution: Let X be a set and let G and H be subgroups of SX . Suppose that the actions of G on X and H on X are permutation isomorphic. This means that there exists an isomorphism ϕ : G → H and a bijection ω : X → X such that for all g ∈ G and all x ∈ X, ω(g · x) = ϕ(g) · ω(x). Suppose that we set x = ω −1 (x0 ), then we get ω(g · ω −1 (x0 )) = ϕ(g) · x0 .

8.2. ORBITS AND STABILIZERS

417

Now x is arbitary so x0 is arbitrary in X. Furthermore, the actions of G, H, and SX on X are all the natural permutation action. Furthermore, an element in SX is uniquely determine by how it acts on X. Hence, we have shown that given the hypothesis, there exists a bijection ϕ : G → H such that ϕ(g) = ωgω −1 , for some ω ∈ SX . Hence, H = ωGω −1 , so G and H are conjugate subgroups. Conversely, suppose that G and H are conjugate subgroups. Then there exists ω ∈ SX such that H = ωGω −1 . Hence, there exists a bijection ϕ : G → H of the form ϕ(g) = ωgω −1 . It is obvious that such a bijection is also a homomorphism, so it is an isomorphism between G and H. Then for all x0 ∈ X we have ω ◦ g ◦ ω −1 (x0 ) = ϕ(g)(x0 ). Replacing x0 with x0 = ωx, we get (ω ◦ g)(x) = ϕ(g)(ω(x)), which we can write with action notation as ω(g · x) = ϕ(g) · ω(x). Since x was arbitrary, we deduce that the actions of G on X and of H on X are permutation isomorphic.

8.2 – Orbits and Stabilizers Exercise: 1 Section 8.2 Question: Let X = {1, 2, 3}3 = {(i, j, k) | 1 ≤ i, j, k ≤ 3}. a) Consider the action of S3 on X by σ · (a1 , a2 , a3 ) = (σ(a1 ), σ(a2 ), σ(a3 )). Explicitly list all the orbits of this action. b) Consider the action of S3 on X by σ · (a1 , a2 , a3 ) = (aσ−1 (1) , aσ−1 (2) , aσ−1 (3) ). Explicitly list all the orbits of this action. Solution: a) O1 = {(1, 1, 1), (2, 2, 2), (3, 3, 3)} O2 = {(1, 2, 3), (2, 3, 1), (3, 1, 2), (1, 3, 2), (2, 1, 3), (3, 2, 1)} O3 = {(1, 1, 2), (2, 2, 1), (1, 1, 3), (3, 3, 1), (2, 2, 3), (3, 3, 2)} O4 = {(1, 2, 1), (2, 1, 2), (1, 3, 1), (3, 1, 3), (2, 3, 2), (3, 2, 3)} O5 = {(1, 2, 2), (2, 1, 1), (1, 3, 3), (3, 1, 1), (2, 3, 3), (3, 2, 2)}. b) This time we can just permute the specific elements in our tuple. O1 = {(1, 1, 1)}, O2 = {(2, 2, 2)}, O3 = {(3, 3, 3)} O4 = {(1, 1, 2), (1, 2, 1), (2, 1, 1)}, O5 = {(1, 1, 3), (1, 3, 1), (3, 1, 1)} O6 = {(2, 2, 1), (2, 1, 2), (1, 2, 2)}, O7 = {(2, 2, 3), (2, 3, 2), (3, 2, 2)} O8 = {(3, 3, 1), (3, 1, 3), (1, 3, 3)}, O9 = {(3, 3, 2), (3, 2, 3), (2, 3, 3)} O10 = {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}

Exercise: 2 Section 8.2 Question: Let X be the set of bit strings of length 6, namely X = {0, 1}6 . a) Consider the action of S6 on X by σ · (b1 , b2 , . . . , b6 ) = (bσ−1 (1) , bσ−1 (2) , . . . , bσ−1 (6) ). Give a unique representative for each orbit and determine how many elements are in each orbit. b) View Z6 as a subgroup of S6 via Z6 ∼ = h(1 2 3 4 5 6)i. Determine all the orbits of order 1, 2, 3, and 6. Solution: Let X be the set of bit strings of length 6, namely X = {0, 1}6 . a) The exercise does not ask to list the elements in the orbits so we avoid an explicit list. When the group acting on X is S6 , the orbits are determined by the number of 1s in the bit string. For example, 010100 is in the orbit of 110000; we simply need a permutation σ such that σ(1) = 2 and σ(2) = 4. For example, (1 2 4) · 110000 = 010100.

418

CHAPTER 8. GROUP ACTIONS 1. 000000 gives an orbit of one element. 2. 100000 leads to an orbit with 6 elements (consisting of all bit strings with a single 1) 3. 110000 leads to an orbit with 62 = 15 elements. 4. 111000 leads to an orbit with 63 = 20 elements. 5. 111100 leads to an orbit with 64 = 15 elements. 6. 111110 leads to an orbit with 65 = 6 elements. 7. 111111 leads to an orbit with 66 = 1 element.

b) When we consider the action of Z6 on bit strings, we will have more orbits, though each orbit with this action will be a subset of some orbit in the S6 action. Furthermore, by the Orbit-Stabilizer Theorem, the size of each orbit is a divisor of 6. In the following list, all orbits are distinct. 1. 000000 gives an orbit of one element. 2. 100000 leads to an orbit with 6 elements. 3. 110000 leads to an orbit with 6 elements. 4. 101000 leads to an orbit with 6 elements. 5. 100100 leads to an orbit with 3 elements. (Note that the union of this and the previous two orbits gives the 15 bit strings with two 1s.) 6. 111000 leads to an orbit with 6 elements. 7. 110100 leads to an orbit with 6 elements. 8. 110010 leads to an orbit with 6 elements. 9. 101010 leads to an orbit with 2 elements. (Note that the union of this and the previous three orbits gives the 20 bit strings with three 1s.) 10. 111100 leads to an orbit with 6 elements. 11. 111010 leads to an orbit with 6 elements. 12. 110110 leads to an orbit with 3 elements. 13. 111110 leads to an orbit with 6 elements. 14. 111111 leads to an orbit with 1 element.

Exercise: 3 Section 8.2 Question: Consider the group G of rigid motions of a cube and consider the action of G on the set of points C that make up the surface of the rotated cube. (See Exercise 8.1.11) For each point of C, describe its orbits. In particular, for all divisors d of |G| = 24, determine which points of C have orbits of size d. Solution: The divisors of 24 are {1, 2, 3, 4, 6, 8, 12, 24}. There is no orbit of size 1, 2, 3, or 4. The orbit of size 6 is the set of center points of any face of the cube. The orbit of size 8 is the set of vertices of the cube. The orbit of size 12 is the set of midpoints of the edges. Any point which is on the interior of a face and is not a centerpoint is in an orbit of size 24. Additionally any edge point that is not a vertex or a midpoint of an edge is in an orbit of size 24. Exercise: 4 Section 8.2 Question: Let G be a group. Prove that the action of G on itself by left multiplication is a regular group action. Solution: Let G be a group, let X = G, and consider the group action defined by g · x = gx, where this is the usual group operation in G. Let g, h, x ∈ G and suppose that gx = hx. Then by multiplying on the right by x−1 , we deduce that g = h. Thus, this action of G on itself by left multiplication is free. Now let x, y ∈ G be arbitrary. Set g = yx−1 . Then g · x = gx = yx−1 x = y. Thus, the action of G on itself by left multiplication is transitive. These two statements together affirm that this action is regular.

8.2. ORBITS AND STABILIZERS

419

Exercise: 5 Section 8.2 Question: Consider the group G of invertible affine transformations of the plane acting on R2 . Recall that in coordinates, an invertible affine transformation on R2 has the form x x e 7−→ A + , y y f where A ∈ GL2 (R) and e, f ∈ R. Show that the natural action of G on R2 is 2-transitive. Solution: Let ~u1 , ~u2 , ~v1 , and ~v2 be four arbitrary elements in R2 with ~u1 6= ~u2 and ~v1 6= ~v2 . Let F ∈ G and let us write this affine transformation as F (~x) = A~x + ~b. We wish to have F (~u1 ) = ~v1 and F (~u2 ) = ~v2 . If this is true, then we must have ~v2 − ~v1 = A~u2 − A~u1 = A(~u2 − ~u1 ). With the conditions on ~u1 , ~u2 , ~v1 , and ~v2 , we know that ~u2 − ~u1 6= ~0 and ~v2 − ~v1 6= ~0. We now prove that under these arbitrary conditions, there exists an invertible matrix A that multiplies ~u2 − ~u1 to ~v2 − ~v1 . Let M be a matrix whose first column is ~u2 − ~u1 and whose second column is not a multiple of the first column. Then M is invertible and M −1 M = I, so in particular, M −1 (~u2 − ~u1 ) = 10 . Now let N be a matrix whose first column is ~v2 −~v1 and whose second column is not a multiple of the first column. Then N is invertible and N 10 = ~v2 − ~v1 . Thus, setting A = N M −1 , we see that A(~u2 − ~u1 ) = ~v2 − ~v1 . Given the above invertible matrix A, we define the affine transformation as F (~x) = A~x + (~v1 − A~u1 ). Then we find that F (~u1 ) = A~u1 + (~v1 − A~u1 ) = ~v1 F (~u2 ) = A~u2 + ~v1 − A~u1 = A(~u2 − ~u1 ) + ~v1 = ~v2 − ~v1 + ~v1 = ~v2 . Hence, the affine transformation F we constructed maps the pair (~u1 , ~u2 ) to the pair (~v1 , ~v2 ). We conclude that the action of G on R2 is 2-transitive. Exercise: 6 Section 8.2 Question: Let G act on a nonempty set X. Suppose that x, y ∈ X and that y = g · x for some g ∈ G. Prove that Gy = gGx g −1 . Deduce that if the action is transitive then the kernel of the action is \ gGx g −1 . g∈G

Solution: Let G act on a nonempty set X and let x, y ∈ X such that y = g · x for some g ∈ G. Let h ∈ gGx g −1 with h = gαg −1 where α ∈ Gx . Thus h · y = h · (g · x) = (gαg −1 g) · x = (gα) · x = g · (α · x) =g·x

because α ∈ Gx

= y. Thus h ∈ Gy . This shows that gGx g −1 ⊆ Gy . Conversely, suppose that k ∈ Gx . Then k · x = x. Hence, (gk) · x = g · x = y. Also, since x = g −1 · y, then we have (gk) · (g −1 · y) = (gkg −1 ) · y = y. Thus, gkg −1 ∈ Gy , so we deduce that gGx g −1 ⊆ Gy . These two points observations, show that gGx g −1 = Gy . Now suppose that the action of G on X is transitive. This means that for any fixed x ∈ X, every stabilizer group is of the form gGx g −1 for some g ∈ G. The kernel of a group action is precisely the intersection of all the stabilizers, the kernel of this action is \ gGx g −1 . g∈G

Exercise: 7 Section 8.2 Question: Let G be a group acting on a set X. Prove that a subset S of X is a G-invariant subset of X if and only if S is a union of orbits. Solution: Let G be a group acting on a set X and let S be a subset of X. Suppose that S is a G-invariant subset. Thus, for all s ∈ S and all g ∈ G, we have g · s ∈ S. In particular, the orbit of s is in S. Since the orbit of

420

CHAPTER 8. GROUP ACTIONS

every element in S is a subset of S, then S is a union of orbits. Conversely, suppose that S = O1 ∪ O2 ∪ · · · ∪ On , where Oi are orbits of the group action. Let s ∈ S and g ∈ G. Then s ∈ Oi for some i. Thus, g · s ∈ Oi and in particular g · s ∈ S. Thus, S is G-invariant. Exercise: 8 Section 8.2 Question: Suppose that a group G acts on a set X and also on a set Y . Prove that a G-set homomorphism f : X → Y maps orbits of G in X to orbits of G in Y . Solution: Let f : X → Y be a G-set homomorphism of G-sets X and Y . We write the action of G on X as g · x and the action of G on Y as g ∗ y. Consider the orbit G · x. For all g · x ∈ G · x, we have f (g · x) = g · f (x). Thus, f (G · x) ⊆ G · f (x). Now let y ∈ Y . Suppose first that y ∈ G · f (x) for some x ∈ X. Then y = g · f (x) = f (g · x), so y ∈ f (G · x). In other words, the orbit of f (x) in Y is the image of the orbit G · x. If y ∈ / G · f (x) for any x ∈ X, then in particular y ∈ / f (X) and the orbit of y with not be in any f (G · x). Thus, the image of any orbit under f is an orbit in Y , though not all orbits in Y are obtained as images of orbits in X. Exercise: 9 Section 8.2 Question: Show that a group of order 55 acting on a set of size 34 must have a fixed point. Solution: Let G act on a set X with |G| = 55 and |X| = 34. By the Orbit-Stabilizer Theorem, |G·x| = |G : Gx |. Assume that the action does not have a fixed point, then Gx 6= G for all x ∈ X. Hence, the sizes of orbits are |G : Gx |, where this can take on values of 5 or 11. (It cannot take the value of 55 because since |X| = 34, no orbit has 55 elements.) Thus, there must exist nonnegative integers n, m such that 5m + 11n = 34. If n = 0 then we must have m = 34/5; if n = 1 then we must have m = 23/5; if n = 2, we must have m = 12/5; if n = 3, we must have m = 1/5; and if n ≥ 4, the condition implies that m < 0. None of these values for m are positive integers, so there is a contradiction. Hence, the action must have at least one fixed point. Exercise: 10 Section 8.2 Question: Suppose G is a group of order 21. Determine with proof the positive integers n such that an action of G on a set X of size n must have a fixed point. Solution: Let G be a group of order 21. By the Orbit-Stabilizer Theorem, the sizes of the orbits of the action must be divisors of 21. Furthermore, an action has a fixed point if and only if there exists an orbit of size 1. To answer the question, we must find the positive integers n that cannot by written as 3a + 7b + 21c for some nonnegative integers a, b, c. Since 21 is a multiple of both 3 and 7, this numerical problem amounts to determining the integers n that cannot be written as 3a + 7b for some nonnegative integers a and b. We first prove that every number greater than or equal to 14 can be written as 3a + 7b with a, b ≥ 0. Suppose that n ≥ 14. Let b be the remainder of n when divided by 3. Thus, n = 3q + b with b = 0, 1, or 2. Then n − 7b = 3q − 6b, which is divisible by 3 with n − 7b = 3a where a = q − 2b. Furthermore, since b ≤ 2, then 7b ≤ 14 and since n ≥ 14, we deduce that n − 7b ≥ 0 and that a = 31 (n − 7b) ≥ 0. We explicitly check the integers between 1 and 13 and find that 1, 2, 4, 5, 8, 11 cannot be written as 3a + 7b with a, b ∈ N. Hence, sets X of these sizes must have a fixed point. Exercise: 11 Section 8.2 Question: Let X = R[x1 , x2 , x3 , x4 ]. Consider the action of G = S4 on X by σ · p(x1 , x2 , x3 , x4 ) = p(xσ(1) , xσ(2) , xσ(3) , xσ(4) ). a) Find the stabilizer of the polynomial x1 + x2 and give its isomorphism type. b) Find a polynomial q(x1 , x2 , x3 , x4 ) whose stabilizer is isomorphic to D4 . c) Explicitly list the elements in the orbit of x1 x2 + 5x3 . d) Explicitly list the elements in the orbit of x1 x22 x33 . Solution: We consider that action of G = S4 acting on X = R[x1 , x2 , x3 , x4 ] by permuting the variables. a) Obviously, the element (1 2) stabilizes x1 + x2 . The transposition (3 4) also stabilizes x1 + x2 . In the stabilizer, any permutation σ maps σ(1) ∈ {1, 2} and 2 to {1, 2} as well. With these conditions, all permutations are possible. Thus, the stabilizer is h(1 2), (3 4)i and this is isomorphic to Z2 ⊕ Z2 .

8.2. ORBITS AND STABILIZERS

421

b) The subgroup H = h(1 2 3 4), (2 4)i ≤ S4 is isomorphic to D4 . The polynomial x1 + x2 + x3 + x4 will not do since its stabilizer is all of S4 . Consider the polynomial q(x1 , x2 , x3 , x4 ) = x1 x2 + x2 x3 + x3 x4 + x4 x1 is stabilized by both (1 2 3 4) and (2 4).This observation shows that H stabilizes q but we must verify that no larger subgroup of S4 stabilizes it. We notice that |H| = 8 and that |S4 | = 24. Since 24/8 = 3 is a prime number, the only subgroup of S4 that strictly contains H is S4 . Hence, if the stabilizer of q is a strictly larger than H, it must be S4 . However, (1 2) · q(x1 , x2 , x3 , x4 ) = x1 x2 + x1 x3 + x3 x4 + x4 x2 6= q(x1 , x2 , x3 , x4 ). Thus, (1 2) does not stabilizer q so the stabilizer of q is H. c) The elements in the orbit of x1 x2 + 5x3 are x1 x2 + 5x3 from 1 and (1 2) x2 x3 + 5x1 from (1 3) and (1 3)(1 2) = (1 2 3) x2 x4 + 5x3 from (1 4) and (1 4)(1 2) = (1 2 4) x1 x3 + 5x2 from (2 3) and (2 3)(1 2) = (1 3 2) x1 x4 + 5x3 from (2 4) and (2 4)(1 2) = (1 4 2) x1 x2 + 5x4 from (3 4) and (3 4)(1 2) = (1 2)(3 4) x2 x3 + 5x4 from (1 3 4) and (1 3 4)(1 2) = (1 2 3 4) x2 x4 + 5x1 from (1 4 3) and (1 4 3)(1 2) = (1 2 4 3) x2 x3 + 5x4 from (2 3 4) and (2 3 4)(1 2) = (1 3 4 2) x3 x4 + 5x1 from (2 4 3) and (2 4 3)(1 2) = (1 4 3 2) x2 x3 + 5x1 from (1 3)(2 4) and (1 3)(2 4)(1 2) = (1 4 2 3) x3 x4 + 5x2 from (1 4)(2 3) and (1 4)(2 3)(1 2) = (1 3 2 4). d) The elements in the orbit of x1 x22 x33 are the 24 elements of the form xi1 xj2 xk3 x`4 , where (i, j, k, `) are quadruples such that i, j, k, ` ∈ {0, 1, 2, 3} and are distinct. Exercise: 12 Section 8.2 Question: Let G be a group acting on a set X. Let H be a subgroup of G. It acts on X with the action of G restricted to H. Let O be an orbit of H is X. a) Prove that for all g ∈ G, the set gO is an orbit of the conjugate subgroup gHg −1 . b) Deduce that if G is transitive on X and if H E G, then all the orbits of H are of the form gO. Solution: Let G be a group acting on a set X. A subgroup H ≤ G acts on X with the action of G restricted to H. a) Let g ∈ G and consider the set gO ⊆ X. For all ghg −1 ∈ gHg −1 and for all g · x ∈ gO, we have (ghg −1 ) · g · x = (gh) · x = g · (h · x), which is again in gO because h · x ∈ O. Hence, the orbit (gHg −1 ) · (g · x) is a subset of gO. Conversely, consider two distinct elements in gO, namely g · x and g · y with x, y ∈ O with y = h · x for some h ∈ H. Then (ghg −1 ) · (g · x) = (gh) · x = g · y. Thus, the action of gHg −1 on gO is transitive so gO is an orbit of gHg −1 . b) Suppose now that G acts transitively on X and the H E G. Let O be one orbit of H acting on X. By the previous part, for all g ∈ G, the set gO is an orbit of gHg −1 . But gHg −1 = H, since H E G. Since the action of G on X is transitive, [ X= gO. g∈G

So all orbits of H acting on X are of the form gO. Exercise: 13 Section 8.2 Question: Let A = {1, 2, . . . , n} and consider the set X = {(a, S) ∈ A × P(A) | a ∈ S}. Consider the action of

422

CHAPTER 8. GROUP ACTIONS

Sn on X by σ · (a, S) = (σ(a), σ · S), where σ · S is the power set action. (See Exercise 8.1.4.) Prove that the Orbit Equation for this action gives the combinatorial formula n X n k = n2n−1 . k

k=0

Solution: The set X = {(a, S) ∈ A × P(A) | a ∈ S} is not all of A × P(A). To determine the cardinality of X, we think first of the options for the element a in the pair (a, S). For each of the n options of a in the pair, we already know that a is in the set S so S = {a} ∪ T , where T is an arbitrary set in P(A − {a}), which has size 2n−1 . Hence, |X| = n2n−1 . We now consider the orbits of Sn acting on X. Let (a1 , {a1 , a2 , . . . , ak }) be any element in X. For any {b1 , . . . , bk } ⊂ A, since the ai are all distinct and the bi are all distinct from each other, then there exists a permutation σ ∈ Sn such that σ(ai ) = bi . This σ satisfies σ · (a1 , {a1 , a2 , . . . , ak }) = (b1 , {b1 , b2 , . . . , bk }). Furthermore, it is obvious from the power set action that σ·(a1 , {a1 , a2 , . . . , ak }) must be of the form (b1 , {b1 , b2 , . . . , b` }) with k = `. Hence, the orbits of Sn on X are Ok = {(a S) ∈ | |S| = k} with k = 1, . . . , n. We can determine the cardinality of Ok by first counting the number of possibilities for the sets S for pairs (a, S) in Ok , and for each of those S, there are k choices for the element a. Hence, |X| =

n X

n−1

|Ok | =⇒ n2

k=1

n X n = k . k k=1

Exercise: 14 Section 8.2 Question: Let A be a finite set of size k and consider the action of Sn on X = An via σ · (a1 , a2 , . . . , an ) = (aσ−1 (1) , aσ−1 (2) , . . . , aσ−1 (n) ). (See Example 8.1.13.) Prove that the Orbit Equation for this action is kn =

n! , s !s ! · · · sk ! 1 2 s +···+s =n X

1

k

where the summation is taken over all (s1 , s2 , . . . , sk ) ∈ Nk that add up to n. Solution: If A is a set of size k, then |An | = k n . We now need to determine the orbits of the action of Sn on Ak (which is by permuting the order of the entries in the n-tuple) and determine the size of each. For each n-tuple (a1 , a2 , . . . , an ) in X = An we associate the k-tuple (s1 , s2 , . . . , sk ) such that sk is the number of entries in (a1 , a2 , . . . , an ) that are equal to k. Write F (a1 , a2 , . . . , an ) = (s1 , s2 , . . . , sk ). We claim that orbits are in bijection with the k-tuples (s1 , s2 , . . . , sk ) such that 0 ≤ si ≤ n and s1 + s2 + · · · + sk = n. Let us denote O(s1 ,s2 ,...,sk ) = {(a1 , a2 , . . . , an ) ∈ X | F (a1 , a2 , . . . , an ) = a1 , a2 , . . . , an )}. Given (a1 , a2 , . . . , an ) ∈ An and σ ∈ Sn , the permuted element σ · (a1 , a2 , . . . , an ) contains the same number of any given type of element in A. Thus, O(s1 ,s2 ,...,sk ) is closed under the action of Sn on it. Furthermore, given (a1 , a2 , . . . , an ), (b1 , b2 , . . . , bn ) ∈ O(s1 ,s2 ,...,sk ) , since they involve the same elements of A in the same number, there is a permutation σ ∈ Sn of the indices such that bi = aσ−1 (i) , which means that (b1 , b2 , . . . , bn ) = σ · (a1 , a2 , . . . , an ). Thus, O(s1 ,s2 ,...,sk ) is an orbit. Furthermore, it is obvious that since we are working with n-tuples, s1 + s2 + · · · + sk = n. In order to apply the Orbit Equation, we need to determine |O(s1 ,s2 ,...,sk ) |. By the Orbit-Stabilizer Theorem, |O(s1 ,s2 ,...,sk ) | = |Sn : H| where H is the stabilizer of a single element in the orbit. Consider the element in O(s1 ,s2 ,...,sk ) that consists of s1 entries of 1s, followed by s2 entries of 2s, and so forth, finished with sk entries

8.2. ORBITS AND STABILIZERS

423

of ks. The stabilizer H of this element consists of all permutations σ ∈ Sn such that 1 ≤ i ≤ s1 implies 1 ≤ iσ(i) ≤ s1 and for 2 ≤ j ≤ k, s1 + s2 + · · · + sj−1 + 1 ≤ i ≤ s1 + s2 + · · · + sj =⇒ s1 + s2 + · · · + sj−1 + 1 ≤ σ(i) ≤ s1 + s2 + · · · + sj . It is clear that H ∼ = Ss1 ⊕ Ss2 ⊕ · · · ⊕ Ssk . Thus |H| = s1 !s2 ! · · · sk ! so |O(s1 ,s2 ,...,sk ) | = n!/(s1 !s2 ! · · · sk !). The Orbit Equation consists of adding the cardinalities of all the orbits, and we get kn =

n! , s !s ! · · · sk ! 1 2 s +···+s =n X

1

k

where the summation is taken over all (s1 , s2 , . . . , sk ) ∈ Nk that add up to n. Exercise: 15 Section 8.2 Question: Let A and B be finite disjoint sets of cardinality a and b, respectively. Let X = Pk (A ∪ B) be the set of subsets of A ∪ B of cardinality k. Let SA and SB be the groups of permutations on A and B, respectively. Consider the action of G = SA ⊕ SB on A ∪ B by ( σ(c) if c ∈ A (σ, τ ) · c = τ (c) if c ∈ B. Define the action of G = SA ⊕ SB on Pk (A ∪ B) as the standard set of subsets action. Prove that the Orbit Equation of G on X gives the Vandermonde Identity X a b a+b = . k i j i+j=k

Solution: Since A and B are disjoint, then |A ∪ B| = a + b and |X| = |Pk (A ∪ B)| = a+b k . Let C ∈ Pk (A ∪ B). Since A and B are disjoint, C ∩ A and C ∩ B are disjoint but since C ⊆ A ∪ B, then C = (C ∩ A) ∪ (C ∩ B). Then for any (σ, τ ) ∈ SA ⊕ SB and for any C ∈ Pk (A ∪ B), the action (σ, τ ) · C = (σ · (C ∩ A)) ∪ (τ · (C ∩ B)), and this union is disjoint. We know that for any finite set Y , the permutation group SY acts transitively on Pk (Y ), where k is any nonnegative integer. Thus, depending on σ the action σ · (C ∩ A) can be any subset of A of size |C ∩ A| (namely ai possibilities and depending on τ the action τ · (C ∩ B) can be any subset of B of size |C ∩ B| (a total of jb possibilities). Thus, if we call |C ∩ A| = i, then |C ∩ B| = j with i + j = k. There are (at b most) k + 1 orbits parametrized by i with 0 ≤ i ≤ k. The orbit corresponding to i has size ai k−i . The Orbit Equation for this action is the formula X a b a+b = k i j i+j=k

where the sum is taken over all pairs (i, j) ∈ N2 such that i + j = k. Exercise: 16 Section 8.2 Question: Let X be the set of functions from {1, 2, . . . , n} into itself and let G = Sn ⊕ Sn . Define the pairing G × X → X by ((σ, τ ) · f )(a) = σ · f (τ −1 · a) for all a ∈ {1, 2, . . . , n}. a) Prove that this pairing is an action of G on X. b) Prove that the set of orbits is in bijection with the partitions of n, where for any partition λ = (λ1 , λ2 , . . . , λs ), the corresponding orbit Oλ consists of functions in which the orders of the distinct fibers are λ1 , λ2 , . . . , λs . c) Determine the number of functions in the orbit Oλ . d) Supposing the n = 5, find the stabilizer of the function f such that f (1) = f (2) = 1, f (3) = f (4) = 2 and f (5) = 3.

424

CHAPTER 8. GROUP ACTIONS

Solution: Let X be the set of functions from {1, 2, . . . , n} into itself and let G = Sn ⊕ Sn . Define the pairing G × X → X by ((σ, τ ) · f )(a) = σ · f (τ −1 · a) for all a ∈ {1, 2, . . . , n}. a) Let (σ1 , τ1 ), (σ2 , τ2 ) ∈ G and let f be a function in X. For all a ∈ {1, 2, . . . , n} ((σ1 , τ1 ) · ((σ2 , τ2 ) · f ))(a) = σ1 (σ2 (f (τ −2 (τ −1 (a))))) = σ1 σ2 (f ((τ1 τ2 )−1 (a))) = ((σ1 σ2 , τ1 τ2 ) · f )(a). Furthermore, (id, id) · f = f for all functions f ∈ X. These show that the pairing is a group action. b) (Recall that a fiber of a function f is a subset of the form f −1 (a). Hence, the fibers of functions f : {1, 2, . . . , n} → {1, 2, . . . , n} are sets of the form f −1 (a) = {i ∈ {1 2 . . . , n} | f (i) = a}.) To any function f : {1, 2, . . . , n} → {1, 2, . . . , n} we associate the partition λ of n as follows. Consider the n-tuple of cardinalities of fibers (|f −1 (1)|, |f −1 (2)|, . . . , |f −1 (n)|), order this n-tuple in nonincreasing order, and drop the terminating 0s; the resulting s-tuple is the associated partition λ. We call G the function from X to the set of partitions of n as G(f ) = λ based on what we just described. We show that two functions of f1 , f2 ∈ X are in the same orbit if and only if L(f1 ) = L(f2 ). First, suppose that f1 , f2 ∈ X are in the same orbit, i.e., that there exists (σ, τ ) ∈ G such that f2 = (σ, τ ) · f1 = σ ◦ f1 ◦ τ −1 . We note that f2−1 (j) = {i ∈ {1, 2, . . . , n} | f2 (i) = j} = {i ∈ {1, 2, . . . , n} | σ(f1 (τ −1 (i))) = j} = {τ (i0 ) | i0 ∈ {1, 2, . . . , n} with σ(f1 (i0 )) = j} = {τ (i0 ) | i0 ∈ (σ ◦ f1 )−1 (j)}. In particular, |f2−1 (j)| = |(σ ◦ f2 )−1 (j)|. Then, the n-tuple of cardinalities of fibers is (|f2−1 (1)|, |f2−1 (2)|, . . . , |f2−1 (n)|) = (|(σ ◦ f1 )−1 (1)|, |(σ ◦ f1 )−1 (2)|, . . . , |(σ ◦ f1 )−1 (n)|) = (|f1−1 (σ −1 (1))|, |f1−1 (σ −1 (2))|, . . . , |f1−1 (σ −1 (n))|). This latter is simply a reordering of the n-tuple of cardinalities of fibers of f1 . Since G puts these n-tuples in nonincreasing order, then L(f2 ) = L(f1 ). Conversely, suppose that L(f1 ) = L(f2 ). Using the reasoning in the previous paragraph, since passing from the n-tuple of cardinalities of fibers reorganizes the fibers, there exists a permutation σ ∈ Sn that maps the n-tuple of cardinalities of fibers of f1 into the n-tuple of cardinalities of f2 . This permutation σ acts on n-tuples as (|f2−1 (1)|, |f2−1 (2)|, . . . , |f2−1 (n)|) = σ · (|f1−1 (1)|, |f1−1 (2)|, . . . , |f1−1 (n)|) = (|f1−1 (σ −1 (1))|, |f1−1 (σ −1 (2))|, . . . , |f1−1 (σ −1 (n))|). Then there exists a permutation τ ∈ Sn that, for each i, maps the specific elements of f2−1 (i) into the specific elements of f1−1 (σ −1 (i)). Then f2 = σ ◦ f1 ◦ τ −1 . This shows that the orbits of G acting on X are in bijective correspondence with partition of n. We denote the corresponding orbit by Oλ . c) We can use the Orbit-Stabilizer Theorem to calculate Oλ as |G|/|Gf | where Gf is the stabilizer of some element f ∈ Oλ . For a given partition, λ = (λ1 , λ2 , . . . , λs ) we select the representative function fλ ∈ Oλ defined by: f (j) = 1 for 1 ≤ j ≤ λ1 and, for all 1 ≤ i ≤ s, set f (j) = i for all j with λ1 + · · · + λi−1 + 1 ≤ j ≤ λ1 + · · · + λi . An element (id, τ ) ∈ G = §n ⊕ Sn that stabilizes fλ allows for permutation of input integers j whenever 1 ≤ j ≤ λ1 or λ1 + · · · + λi−1 + 1 ≤ j ≤ λ1 + · · · + λi for some i. There are λ1 !λ2 ! · · · λs ! such elements. Let us extend the partition λ from (λ1 , λ2 , . . . , λs ) to (λ1 , λ2 , . . . , λn ) setting some of the tailing parts to 0 if necessary. Suppose that λi = λi0 ; then we can map bijectively all the j with λ1 + · · · + λi−1 + 1 ≤ j ≤ λ1 + · · · + λi into {λ1 + · · · + λi0 −1 + 1, . . . , ≤ λ1 + · · · + λi0 } and vice versa and then apply a permutation σ that interchanges i and i0 and still recover the same function. Hence, suppose that in (λ1 , λ2 , . . . , λn ) there are µt parts (entries in the nonincreasing n-tuple λ) that are equal

8.2. ORBITS AND STABILIZERS

425

to t, where t is any integer between 0 and n. Then, for each of the λ1 !λ2 ! · · · λs ! elements in the stabilizer of the form (id, τ ), there are µ0 !µ1 ! · · · µn !. Hence, |Gλ | = (λ1 !λ2 ! · · · λs !)(µ0 !µ1 ! · · · µn !), so |Oλ | =

(n!)2 . (λ1 !λ2 ! · · · λs !)(µ0 !µ1 ! · · · µn !)

For example with n = 10 and λ = (3, 2, 2, 2, 1), we fill out λ so that it is a 10-tuple as λ = (3, 2, 2, 2, 1, 0, 0, 0, 0, 0) and then we have µ = (5, 1, 3, 1, 0, 0, 0, 0, 0, 0, 0) so |Oλ | =

(10!)2 3!(2!)3 1! × 5!1!3!1!

.

d) The function f such that f (1) = f (2) = 1, f (3) = f (4) = 2 and f (5) = 3 has the associated partition λ = L(f ) = (2, 2, 1). The stabilizer Gf is generated by the following pairs Gf = h(id, (1 2)), (id, (3 4)), ((1 2), (1 3)(2 4)), ((4 5), id)i.

Exercise: 17 Section 8.2 Question: Consider the action of G = S3 ⊕ S3 on X = {(i, j, k) | 1 ≤ i, j, k ≤ 3} by (σ, τ ) · (a1 , a2 , a3 ) = (τ (aσ−1 (1) ), τ (aσ−1 (3) ), τ (aσ−1 (3) )). This action has the effect of rearranging the elements in the triple according to σ and then permuting the outcome by τ . Explicitly calculate X g for all g ∈ G and verify the Cauchy-Frobenius Lemma. Solution: (Note that this exercise fits into the frame of Exercise 16 but with n = 3. The triples of elements in {1, 2, 3} can be understood as functions {1, 2, 3} → {1, 2, 3}. However, the roles of σ and τ in this exercise is the opposite from that in the previous exercise.) We have |G| = (3!)2 = 36 and |X| = 33 . We have X (id,id) = X X (id,(1 2)) = {(3, 3, 3)} X (id,(1 3)) = {(2, 2, 2)} X (id,(2 3)) = {(1, 1, 1)} X (id,(1 2 3)) = ∅ X (id,(1 3 2)) = ∅ X ((1 2),id) = {(a, a, x) | a, x ∈ {1, 2, 3}} X ((1 2),(1 2)) = {(1, 2, 3), (2, 1, 3), (3, 3, 3)} X ((1 2),(1 3)) = {(1, 3, 2), (3, 1, 2), (2, 2, 2)} X ((1 2),(2 3)) = {(2, 3, 1), (3, 2, 1), (1, 1, 1)} X ((1 2),(1 2 3)) = ∅ X ((1 2),(1 3 2)) = ∅ X ((1 3),id) = {(a, x, a) | a, x ∈ {1, 2, 3}} X ((1 3),(1 2)) = {(1, 3, 2), (2, 3, 1), (3, 3, 3)} X ((1 3),(1 3)) = {(1, 2, 3), (3, 2, 1), (2, 2, 2)} X ((1 3),(2 3)) = {(2, 1, 3), (3, 1, 2), (1, 1, 1)} X ((1 3),(1 2 3)) = ∅ X ((1 3),(1 3 2)) = ∅

426

CHAPTER 8. GROUP ACTIONS X ((2 3),id) = {(x, a, a) | a, x ∈ {1, 2, 3}} X ((2 3),(1 2)) = {(3, 1, 2), (3, 2, 1), (3, 3, 3)} X ((2 3),(1 3)) = {(2, 1, 3), (2, 3, 1), (2, 2, 2)} X ((2 3),(2 3)) = {(1, 2, 3), (1, 3, 2), (1, 1, 1)} X ((2 3),(1 2 3)) = ∅ X ((2 3),(1 3 2)) = ∅ X ((1 2 3),id) = {(1, 1, 1), (2, 2, 2), (3, 3, 3)} X ((1 2 3),(1 2)) = {(3, 3, 3)} X ((1 2 3),(1 3)) = {(2, 2, 2)} X ((1 2 3),(2 3)) = {(1, 1, 1)} X ((1 2 3),(1 2 3)) = {(1, 2, 3), (2, 3, 1)(3, 1, 2)} X ((1 2 3),(1 3 2)) = {(1, 3, 2)(2, 1, 3)(1, 3, 2)} X ((1 3 2),id) = {(1, 1, 1), (2, 2, 2), (3, 3, 3)} X ((1 3 2),(1 2)) = {(3, 3, 3)} X ((1 3 2),(1 3)) = {(2, 2, 2)} X ((1 3 2),(2 3)) = {(1, 1, 1)} X ((1 3 2),(1 2 3)) = {(1, 3, 2), (2, 1, 3)(3, 2, 1)} X ((1 3 2),(1 3 2)) = {(1, 2, 3)(2, 3, 1)(3, 1, 2)}.

We now verify the Cauchy Frobenius formula. According to the previous exercise, there are m = 3 orbits because that is the number of partitions of 3 (namely 3, 2 + 1, 1 + 1 + 1). The left hand side of the formula is m|G| = 3 · 36 = 108. The summation on the right hand side of the formula has (27 + 1 + 1 + 1 + 0 + 0) + 3(9 + 3 + 3 + 3 + 0 + 0) + 2(3 + 1 + 1 + 1 + 3 + 3) = 30 + 3 × 18 + 2 × 12 = 30 + 54 + 24 = 108. This verifies the formula. Exercise: 18 Section 8.2 Question: Recall that the group of rigid motions of a tetrahedron is A4 . Suppose that we color the faces of a tetrahedron with colors red, green, or blue. We consider two colorings equivalent if one coloring can be obtained from another by rotating the tetrahedron. How many inequivalent such colorings are there? Solution: Let X be the set of colorings of the tetrahedron. There are three possible colors for all four faces so |X| = 34 . The group of rigid motions A4 acts on elements in X. The inequivalent colorings of the tetrahedron consist of the number m of different orbits in the action of A4 on X. We use the Cauchy-Frobenius Lemma to find m, X |A4 |m = |X g |. g∈A4

We know that |A4 | = 12. • The identity obviously fixes all of X so |X 1 | = 34 = 81. • Let σ be a non-identity rotation by 2π/3 or 4π/3 about an axis through a vertex and the centroid of the opposite face. In order for σ to fix a coloring, the faces that touch the vertex must all be the same color and the face that contains the centroid of the axis of rotation can be any color. Thus |X σ | = 32 . Note there are 8 such rotations. • The remaining rigid motions are rotations τ by π about an axis that goes the the center C1 of an edge E and the center C2 of the one other edge that is not incident with E. In order for τ to fix a coloring, the faces that touch C1 must all be the same color and the faces that touch C2 must be the same. Thus |X σ | = 32 . Note there are 3 such rotations.

8.2. ORBITS AND STABILIZERS

427

The Cauchy-Frobenius Lemma affirms that 12m = 34 + 8 · 32 + 3 · 32 = 180. Hence m = 180/12 = 15. Exercise: 19 Section 8.2 Question: We consider colorings of the vertices of a square as equivalent if one coloring can be obtained from the other by any D4 action on the square. How many different colorings are there with (a) 3 colors; (b) 4 colors; (c) 5 colors? Solution: We use the Cauchy-Frobenius Lemma to answer this question. Let X be the set of colorings of the vertices of the square. If there are n colors, then |X| = n4 . Note that |D4 | = 8. • All colorings are fixed by the identity. • The only colorings fixed by a rotation of angle π/2 or 3π/2 are those in which all the vertices have the same color. Hence, for all 3 of those rotations there are n colorings fixed. • The colorings fixed by a rotation of angle π are those in which pairs of interchanged points have the same color. There are n2 such colorings. • Now consider a reflection s through two diagonally opposite vertices. The colorings that are fixed by s can have any color on either of the vertices on the line of reflection and the colors on the points that are interchanged must be the same. Thus, there are n3 such colorings. • If σ is a reflection through centers of opposite sides, then the colorings that are fixed by σ must be such that for each pair of points interchanged, the points in the pair must have the same color. There are n2 such colorings. The number of inequivalent colorings is the number m of orbits. By the Cauchy-Frobenius Lemma we find that n(n3 + 2n2 + 3n + 2) 1 4 (n + 2n + n2 + 2n3 + 2n2 ) = . 8 8 If n = 3, then m = 21. If n = 4, then m = 55. If n = 5, then m = 120. m=

Exercise: 20 Section 8.2 Question: We consider colorings of the vertices of an equilateral triangle as equivalent if one coloring can be obtained from the other by any D3 action on the triangle. How many different colorings are there using p colors? Solution: We use the Cauchy-Frobenius Lemma that says the number of inequivalent colorings of the vertices X g of the triangle is m, where 6m = |X |, where X is the set of all colorings of the vertices. g∈D3

• The identity fixes all elements of X so |X id | = |X| = p3 . • In order for a coloring to be fixed by either of the two rotations of 2π/3 and 4π/3, it must have all the same colors. Hence, there are p such colorings. • In order for a coloring to be fixed by any of the three reflections, the points reflected into each other must have the same color and the point on the axis of reflection can be any color. Hence, there are p2 colorings fixed by these reflections. = p+2 By the Cauchy-Frobenius Lemma, m = 16 (p3 + 2p + 3p2 ) = p(p+1)(p+2) 6 3 . Exercise: 21 Section 8.2 Question: Repeat Example 8.2.18 but consider bracelet colorings equivalent if one is obtained from another under the action of some D8 element. (Note that the bracelet in Figure 8.5 is only fixed by 1 under the Z8 action but by {1, sr} in D8 .) Solution: The set of colorings X of the bracelet has |X| = 38 . He have |D8 | = 16. The number of inequivalent colorings is 1 8 (3 + 3 + 32 + 3 + 34 + 3 + 32 + 3 + 4 × 35 + 4 × 34 ) 16 = 498.

m=

428

CHAPTER 8. GROUP ACTIONS

8.3 – Transitive Group Actions Exercise: 1 Section 8.3 Question: Let G be a group acting transitively on a set X with |X| prime. Prove that the action is primitive. Solution: By Proposition 8.3.5, the cardinality |B| of a block B in a transitive group action divides |X|. Thus is |X| = p is prime, then for any block B of the action, |B| = 1 or p. Thus all the blocks are trivial and the action is primitive. Exercise: 2 Section 8.3 Question: Show that if a finite group G acts transitively on a set X, then there exists some g ∈ G such that X g = ∅. Solution: This exercise assumes that |G| is finite. If G is finite and acts transitively on X, then X is also a finite set since given any x ∈ X, we must have X = G · x. The we can use the Cauchy-Frobenius Lemma. Since the group action is transitive, the number of orbits is 1 so X |G| = |X g |. g∈G

Assume that X g 6= ∅ for all g ∈ G. Then |X g | ≥ 1. We also know that X 1 = X so we find that |G| ≥ |X| + (|G| − 1) · 1 = |X| + |G| − 1 ⇐⇒ 1 ≤ |X|. So it is possible for X g 6= ∅ for all g ∈ G but only if |X| = 1. Otherwise, we have a contradiction. Thus if |X| ≥ 2, then there must exist some g ∈ G such that X g = ∅. Exercise: 3 Section 8.3 Question: We showed that the group G of rigid motions of a cube acting on its set of vertices has the long diagonals for blocks. We claim that this group action has more nontrivial blocks. Find all of them and prove you have found them all. Solution: From Proposition 8.3.5, the size |B| of a block B in a transitive group action must divide |X|, which is equal to 8 in this case, since X is the set of vertices of a cube. Hence, the sizes of blocks must be the trivial ones (of size 1 or 8) or of size 2 or 4. For |B| = 2, consider any pair of vertices {v1 , v2 } of the cube. It is not too hard to check visually from a diagram that if {v1 , v2 } is not a long diagonal, then v1 and v2 share a face F1 of the cube. Let g be a rotation of 2π/3 about the long diagonal through v1 . The g · v1 = v1 , whereas g · v2 6= v2 . Hence, {v1 , v2 } ∩ {g · v1 , g · v2 } = {v1 }, which is neither {v1 , v2 } nor the ∅. Hence, the only blocks of size 2 are the pairs of vertices on long diagonals. For |B| = 4, suppose we label the vertices of the cube as in Figure 8.2. A face cannot be a block because a (nontrivial) rotation about a long diagonal through any of the vertices of the face would map that face to another face of the cube that shares exactly an edge with the original face. Hence, the intersection of the original face with the rotated face would be neither empty nor the original face. For the same reason, but being more judicious concerning about with vertex to consider rotation, there can be no block that involves three vertices from the same face. Consequently, any block of size four must be such that it at most 2 vertices in a block can share a face. Suppose that two vertices in a block B that share a face also share an edge; then B would consist of parallel edges, for example {1, 5, 3, 7} using Figure 8.2. However, the rotation by 2π/3 around a long diagonal (with our example {1, 7}) produces the new set {1, 7, 2, 8} and {1, 5, 3, 7} ∩ {1, 7, 2, 8} is neither {1, 5, 3, 7} nor ∅. Hence, for blocks of size 4, the block must be such that no more that two vertices share a face and no two share an edge. There is only one configuration that satisfy these conditions, namely the two blocks {1, 3, 6, 8} and {2, 4, 5, 7}. These are vertices that arise at opposite corners of faces to each other. Exercise: 4 Section 8.3 Question: Find the largest subgroup of S6 that acts transitively on {1, 2, 3, 4, 5, 6} and has {1, 2, 3} as a block. Solution: The subgroup H = {σ ∈ S6 | σ(i) = i for i = 4, 5, 6} preserves {1, 2, 3} but this subgroup does not act transitively on {1, 2, 3, 4, 5, 6}. Suppose G is a subgroup of S6 that acts transitively on {1, 2, 3, 4, 5, 6} and has

8.3. TRANSITIVE GROUP ACTIONS

429

{1, 2, 3} as a block. Then because of transitivity, there must be an element σ ∈ G such that σ(1) = 4. But then since {1, 2, 3} is a block for the action of G, then σ(2), σ(3) ∈ {5, 6}. This inspires us to consider the subgroup G = h(1 2 3), (1 2), (1 4)(2 5)(3 6)i. We prove this is the largest subgroup of S6 that has {1, 2, 3} as a block. Note that {4, 5, 6} is also a block in this group action so that Σ = {{1, 2, 3}, {4, 5, 6}} is the corresponding system of blocks. Note that G contains the subgroup H1 = h(1 2 3), (1 2)i, which is the full symmetric group on {1, 2, 3}. G also contains the conjugate subgroup H2 = (1 4)(2 5)(3 6)h(1 2 3), (1 2)i(1 4)(2 5)(3 6) = h(4 5 6), (4 5)i which is the full symmetric group on {4, 5, 6}. Also G acts transitively since (1 4)(2 5)(3 6) · 1 = 4 and then composition with other elements of H2 will map 1 to 5 or 6. Thus, for all i ∈ {1, 2, 3, 4, 5, 6} there is a g ∈ G with g · 1 = i. This proves transitivity. It is not hard to check that all the generators preserve the block structure of Σ and hence all of G does. We show that G is the largest group with the desired property by a combinatorial count. We count all the permutations of {1, 2, 3, 4, 5, 6} that preserve the block {1, 2, 3}. There are six options for where to send 1. However, depending on whether σ maps 1 to {1, 2, 3} or {4, 5, 6}, the remaining elements of the block must be sent to the same block as σ(1). So we have 2 × 1 options for 2 and 3. The choice of σ(1) determines the block where 1 goes and also the block where 4 goes. But then there are 3 × 2 × 1 = 6 choices for where to send 4, 5, and 6. Hence, the number of desired permutations is 12 × 6 = 72. We know that G contains the group hH1 , H2 i ∼ = H1 ⊕ H2 , which has order 36. G also contains one more generator (1 4)(2 5)(3 6) so |G| is a multiple of 36 that is larger than 36. But |G| ≤ 72, so we deduce that G = 72. Thus, the presentation we have is the largest transitive group acting on {1, 2, 3, 4, 5, 6} that preserves the block {1, 2, 3}. Exercise: 5 Section 8.3 Question: Let K = h(1 2 3)(4 5 6)(7 8 9), (1 4 7)(2 5 8)(3 6 9)i be a subset of S9 and consider its natural action on X = {1, 2, . . . , 9}. Prove that K acts transitively and that it has the same system of blocks as the group H in Example 8.3.7. Solution: Let σ = (1 2 3)(4 5 6)(7 8 9) and τ = (1 4 7)(2 5 8)(3 6 9) be elements in S9 . We note that τ στ −1 = (4 5 6)(7 8 9)(1 2 3) = σ. Hence τ and σ commute. We can prove that K acts transitively on X = {1, 2, . . . , 9} by proving that the orbit of 1 is all of X. We note that σ(1) = 2 and σ 2 (1) = 3. Also τ (1) = 4, then στ (1) = 5 and σ 2 τ (1) = 6. Finally, τ 2 (1) = 7 so στ 2 (1) = 8 and σ 2 τ 2 (1) = 9. To prove that the action of K on X has the same system of blocks as the group H in Example 8.3.7, we only need to consider how σ and τ acts on the proposed blocks. Note that σ · {1, 2, 3} = {1, 2, 3}

σ · {4, 5, 6} = {4, 5, 6}

σ · {7, 8, 9} = {7, 8, 9}

τ · {1, 2, 3} = {4, 5, 6}

τ · {4, 5, 6} = {7, 8, 9}

τ · {7, 8, 9} = {1, 2, 3}.

Thus, for every block B in the system of blocks Σ, we do have σ cot B = B and τ · B ∩ B = ∅. Since K is generated by σ and τ , the whole group K has the same system of blocks as Example 8.3.7. Exercise: 6 Section 8.3 Question: The group described in Example 8.3.7 is not the largest subgroup of S9 that acts on X = {1, 2, . . . , 9} with {1, 2, 3} as a block. Find this largest subgroup. Solution: The largest subgroup of S9 that acts on X = {1, 2, . . . , 9} with {1, 2, 3} as a block, must involve every permutation within every block and must be able to permute between any block. A full permutation action group in {1, 2, 3} is h(1 2 3), (1 2)i. To permute between any block, we can use the subgroup, h(1, 4 7)(2 5 8)(3 6 9), (1, 4 7)(2 5 8)i. The desired group is the join of these two subgroups, which is H = h(1 2 3), (1 2), (1, 4 7)(2 5 8)(3 6 9), (1, 4 7)(2 5 8)i. Though it was not asked for, we can determine the order of this group as follows:

430

CHAPTER 8. GROUP ACTIONS • There are 9 options for where to send 1. (This choice also determines to which block {1, 2, 3} maps.) • Then there are only two choices for where to send 2. • With the above information, there are no more choices for where to send 3. • There are now 6 choices for where to send 4. (Any element in the two blocks where {1, 2, 3} does not go.) • There are 2 choices for where to send 5. • There is 1 choice for where to send 6. • For the choices of 7, 8, and 9, there is a total of 3! choices.

Hence, the total number of elements in H is 9 · 2 · 6 · 2 · 6 = 64 = 1296. Exercise: 7 Section 8.3 Question: Consider the group G of rigid motions on a cube. Show that G has a normal subgroup of order 4 and describe the associated system of blocks of vertices described in Proposition 8.3.11. Solution: We know that the group of rigid motions of a cube is isomorphic to S4 , corresponding to the permutations of the long diagonals. (We refer to Figure 8.2 and the relevant explanation.) Consider the subgroup N = {1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. It is a union of conjugacy classes (cycle types) of S4 so it must be a normal subgroup. The permutation • (1 2)(3 4) acts on the vertices as the rotation of 180◦ around the line through the center of face 1485 and center of the face 2376. • (1 3)(2 4) acts on the vertices as the rotation of 180◦ around the line through the center of face 1234 and center of the face 5678. • (1 4)(2 4) acts on the vertices as the rotation of 180◦ around the line through the center of face 1265 and center of the face 3487. The orbit of 1 when acted on by N is {1, 3, 6, 8}. The orbit of 2 when acted on by N is {2, 4, 5, 7}. The orbits correspond to two regular tetrahedra that occur as subsets of the cube. Exercise: 8 Section 8.3 Question: Show that no subgroup G of S5 acts transitively on X = {1, 2, 3, 4, 5} in such a way that Gx is an elementary abelian 2-group for any x ∈ X. Solution: Suppose that G acts transitively on X. Then for all x ∈ X, the orbit G · x is all of X. In particular |G · x| = 5. By the Orbit-Stabilizer Theorem, for all x ∈ X, we have 5 = |G : Gx |. Thus |G| = 5|Gx |. We now consider elementary abelian 2-groups (i.e., a subgroup isomorphic to Z2n ) in S5 . Every nonidentity element in an abelian 2-group has order 2. The only elements of order 2 in S5 are of the form (a b) and (a b)(c d). Suppose that a subgroup of S5 contains to nondisjoint transpositions (a b) and (b c). Then it also contains the 3cycle (a b)(b c) = (a b c). Thus an elementary abelian 2-group cannot contain nondisjoint transpositions. Suppose that a subgroup H of S5 contains an element of the form (a b)(c d). If H also contains (a e), then H contains (a b e)(c d) which is of order 6. If H contains an element of the form (b c), then H contains (a b)(c d)(b c) = (a b d c). Hence an elementary abelian 2-subgroup of S5 cannot contain (a b)(c d) and any transposition different from (a b) or (c d). So the only elementary abelian 2-groups in S5 are of the form h(a b)i, h(a b)(c d)i, h(a b), (c d)i or h(a b)(c d), (a c)(b d)i. In particular, every elementary abelian 2-group is isomorphic to Z2 or Z2 ⊕ Z2 . Assume that there is a subgroup G ≤ S5 that acts transitively on X such that for some x ∈ X the stabilizer Gx is an elementary abelian 2-group. It must have one of the 4 above forms, where x ∈ / {a, b, c, d}. Furthermore, this implies that |G| = 10 or 20. All the elements in G must have order 1, 2, 4 or 5. They cannot have order 10 or 20 since the highest order of any element in S5 is order 6. Case Gx = h(a b), (c d)i: We now consider elements that do not fix x. Then G cannot contain a transposition τ with x, since, composed with either (a b) or (c d) would produce a 3-cycle, which is a contradiction. Assume

8.3. TRANSITIVE GROUP ACTIONS

431

that G contains a 4-cycle with x, say (a b c x), then (a b)(a b c x) = (b c x) which is impossible since G contains no 3-cycles. Assume that G contains a 4-cycle with x, say (a c b x); then (c d)(a c b x) = (a d c b x). But by Exercise 3.5.42, then G = S5 , which leads to a contradiction. Assume that G contains a 5-cycle, then again by Exercise 3.5.42, G = S5 , which is another contradiction. Hence, this case does not occur. Case: G = h(a b)(c d), (a c)(b d)i: We now consider elements that do not fix x. Assume that G contains a transposition (a x), then it also contains (a x)(a b)(c d) = (a b x)(c d), which has order 6. This is a contradiction. Assume that G contains a 4-cycle. Without loss of generality, we can assume this 4-cycle has the form (a b c x). Then G also contains (a c)(b d)(a b c x) = (a d b)(c x) which is an element of order 6, a contradiction. Assume that G contains a 5-cycle, which, without loss of generality, is (a b c d x). Then (a b)(c d)(a b c d x) = (b d x), which is a 3-cycle so a contradiction. Case: G = h(a b)(c d)i: The same possibilities and reasoning occur as the above example and we find that all cases lead to a contradiction. Case Gx = h(a b)i: We now consider elements that do not fix x. Then G cannot contain (a x) or (b x), since otherwise G has an element of order 3. If G contains (c x), then we can use a fifth element d as the fixed element return to one of the above cases, all of which lead to contradictions. Assume that G contains a 4-cycle, then it either contains a 3-cycle or a 5-cycle, both of which lead to contradictions by Exercise 3.5.42. And of course G cannot contain a 5-cycle, without leading to a contradiction. Hence, we have proved that all possibilities of Gx lead to contradictions. Hence, there is no subgroup G of S5 that acts transitively on {1, 2, 3, 4, 5} such that there exists some x with Gx ∼ = Z2 or Z2 ⊕ Z2 . [Note: This exercise would be much easier if we could use Cauchy’s Theorem but it is not introduced until section 8.4. By Cauchy’s Theorem, we conclude immediately that G contains a 5-cycle and quickly show that G either contains a 3-cycle (a contrdiction) or is all of S5 , another contradiction.] Exercise: 9 Section 8.3 Question: Let B be a nontrivial block of a transitive action of G on a set X. Prove that the pointwise stabilizer G(B) and the setwise stabilizer G{B} are subgroups of G and that G(B) E G{B} . Solution: Let g ∈ G{B} be an element that stabilizes the set B, let n ∈ G(B) , and let b ∈ B. Then (gng −1 ) · b = (gn) · (g −1 · b)

and g −1 · b ∈ B since g −1 ∈ G{B}

= g · (n · (g −1 · b)) = g · (g −1 · b)) = (gg

−1

since n ∈ G(B)

)·b

= b. This establishes that gng −1 ∈ G(B) . Consequently, G(B) E G{B} . Exercise: 10 Section 8.3 Question: Let G be a group acting transitively on a set X and let α ∈ X. Let B be the set of all blocks B in the group action such that α ∈ B, and let S = {H ∈ Sub(G) | Gα ≤ H}. The function Ψ : B → S defined by Ψ(B) = G{B} is a poset isomorphism between the posets (B, ⊆) and (S, ≤). Solution: We first prove that Ψ is a function, i.e., that for all blocks B containing α, that Ψ(B) = G{G} contain Gα . For any g ∈ Gα , we have B ∩ gB 6= ∅ since α ∈ B ∩ gB. Since B is a block, we deduce that gB = B so g ∈ G{G} . Thus Gα ⊆ G{G} and thus Ψ(B) ∈ S. We now show that Ψ is a bijection by presenting its inverse. Let H ∈ S and consider the orbit H·α. This subset of X contains α. We claim that the cosets gH · α form a system of blocks. Suppose that (g1 H · α) ∩ (g2 H · α) 6= ∅. Then there exists h1 , h2 ∈ H such that −1 −1 −1 g1 h1 · α = g2 h2 · α =⇒ h−1 2 g2 g1 h2 · α = α =⇒ h2 g2 g1 h2 ∈ Gα ,

432

CHAPTER 8. GROUP ACTIONS

which in turn implies that g2−1 g1 ∈ H. Thus g1 H = g2 H and g1 H · α = g2 H · α. This proves the claim. Hence, we define the function Φ : S → B defined by Φ(H) = H · α, since this is a block that contains α. Let B be a block in B and consider the orbit G{B} · α. Obviously, G{B} · α ⊆ B. If β is any element in the block B, then since the action is transitive, there exists some g ∈ G with β = g · α. Since B is a block, then g · B = B as β ∈ g · B ∩ B, so g ∈ G{B} . Thus, B ⊆ G{B} · α, which shows that B = G{B} · α. We can restate this as Φ(Ψ(B)) = B. Conversely, let H ∈ S and consider G{H·α} . First, for all h ∈ H we have h · (H · α) = H · α, so H ⊆ G{H·α} . Second, if g · (H · α) = H · α, then there exists h1 , H2 ∈ H such that (gh1 ) · α = h2 · α. Hence −1 h−1 2 gh1 · α = α, so h2 gh1 ∈ H, by which we deduce that g ∈ H. Thus G{H·α} ⊆ H and therefore G{H·α} = H. We have proven that Ψ(Φ(H)) = H. In summary Φ and Ψ are inverse functions of each other, which establishes a bijection between B and S. To prove that Ψ preserves the order given on the posets, suppose that B, B 0 ∈ B with B ⊆ B 0 . Then for all g ∈ G{B} we have gB 0 ∩ B 0 6= ∅ since g · b ∈ B ⊆ B 0 for all b ∈ B. Since B 0 is a block, we deduce that g · B 0 = B 0 so G{B} ≤ G{B 0 } . Thus, Ψ(B) ≤ Ψ(B 0 ). To prove that Φ preserves the order, consider subgroups H and H 0 with Gα ≤ H ≤ H 0 ≤ G. It is obvious that H · α ⊆ H 0 · α. Hence, Φ(H) ⊆ Φ(H 0 ). Exercise: 11 Section 8.3 Question: Let G act transitively on a set X and let x ∈ X. Prove that the fixed set X Gx is a block of X. Deduce that if G acts primitively, then X Gx = {x} or else Gx = {1} and X is a finite set with |X| prime. Solution: The subset X Gx consists of all elements in X fixed by the subgroup Gx . Note that for all x ∈ X, we know that x ∈ X Gx . Consider the set of subsets of the form {X Ga | a ∈ X}. We prove that this forms a system of blocks for this action. Let g ∈ G and let a = gx. Suppose that y ∈ X Gx . We want to show that gy ∈ X Ga . Recall that Ga = gGx g −1 . Hence, any element in Ga has the form ghg −1 where h ∈ Gx . We then have (ghg −1 ) · (gy) = gh · y =g·y

because y is fixed by Gx .

Thus gy is fixed by any element in Ga so gy ∈ X Ga . If G acts primitively, then by definition the blocks are the singletons or all of X. If the blocks are singletons, then X Gx = {x}, which means that the only elements stabilized by a stabilizer Gx is x itself. If the block obtained as X Gx is all of X, then every element is fixed by Gx , so Gx is the kernel Ker ρ of the group action. By Corollary 8.3.9, Gx is maximal so by the Fourth Isomorphism Theorem G/ Ker ρ contains no nontrivial subgroups. In particular, G/ Ker ρ is cyclic, finite, and of prime order. By the Orbit Stabilizer Theorem, |X| = |G : Gx | so |X| is finite and of prime order. Exercise: 12 Section 8.3 Question: Prove that if G acts r-transitively on a set X with |X| = n, then |G| is divisible by n!/(n − r)!. Solution: Define Pr (X) = {(x1 , x2 , . . . , xr ) ∈ X r | xi 6= xj for i 6= j}. By definition, G is r-transitive on X if and only if G is transitive on Pr (X). By usual counting methods, we see that |Pr (X)| = n!/(n − r)!. Suppose that G acts r-transitively on X and let x̄ = (x1 , x2 , . . . , xr ) ∈ Pr (X). By the Orbit-Stabilizer Theorem |Pr (X)| = |G : Gx̄ |, so (supposing G is finite), |Gx̄ ||Pr (x)| = |G| so |Pr (X)| = n!/(n − r)! divides |G|. Exercise: 13 Section 8.3 Question: Prove that a 2-transitive group action is primitive. Solution: Suppose that a group G acts on a set X 2-transitively. The singleton sets are always blocks in any transitive group action. Let B be a block that contains more than one element, say x 6= y. Let z be any other element in X. Since the action is 2-transitive, there exists g ∈ G such that g · (x, y) = (g · x, g · y) = (x, z). Then x ∈ (g · B) ∩ B, so (g · B) ∩ B 6= ∅ and since B is a block we have g · B = B. Furthermore, z ∈ g · B = B. Hence, we have shown that X ⊆ B, so X = B. Consequently, the only blocks of a 2-transitive group action are trivial and hence, the action is primitive. Exercise: 14 Section 8.3 Question: Suppose that (G, X, ρ) is an r-transitive group action. Prove that if N E G then N is (r − 1)transitive.

8.3. TRANSITIVE GROUP ACTIONS

433

Solution: Suppose that G acts r-transitively on a set X and suppose that N E G and N acts nontrivially. The problem supposes that r ≥ 2 for the concept of (r − 1)-transitivity to make sense. Using Exercise 8.3.13, we point out that since the group action is 2-transitive, it is also primitive. And by Proposition 8.3.11, this means that N acts transitively on X, since it does not act trivially. Consequently, previous results establish this exercise for r = 2, so we may assume that r ≥ 3. For a set X, denote by X (r) = {(x1 , x2 , . . . , xr ∈ X r | xi 6= xj for i 6= j} the set of distinct r-tuples. Given a group action (G, X, ρ), there is a natural group action (G, ρ(r) , X (r) ), i.e., a homomorphism ρ(r) : G → SX (r) defined by g · (x1 , x2 , . . . , xr ) = (g · x1 , g · x2 , . . . , g · xr ). Saying that the action of G on X is r-transitive means that G acts transitively on the set X (r) via ρ(r) . Suppose that G acts r-transitively on X and we study the blocks of the action of G on X (r−1) , which is obviously transitive. Let (x1 , x2 , . . . , xr−1 ) ∈ X (r−1) . Since G acts transitively on X (r) , for any y and z different from the xi , there exists g ∈ G such that g · (x1 , x2 , . . . , xr−1 , y) = (x1 , x2 , . . . , xr−1 , z). Consequently, by considering all but the last coordinate or all but the second to last coordinate, we see that if (x1 , x2 , . . . , xr−1 ) and (x1 , . . . , xr−2 , y) are in the same block B, then (x1 , x2 , . . . , xr−1 ) and (x1 , . . . , xr−2 , z) are in the same block gB. However, since (x1 , x2 , . . . , xr−1 ) ∈ B ∩ gB, then gB = B so (x1 , . . . , xr−2 , z) ∈ B. Since y and z were arbitrary, we see that not only does G act (r −1)-transitively on X but that for all (x1 , x2 , . . . , xr−2 ) ∈ X (r−2) , the stabilizer G(x1 ,...,xr−2 ) acts primitively on X − {x1 , x2 , . . . , xr−2 }. Consequently, the action of G on X (r−1) is primitive. By Proposition 8.3.11, a normal subgroup N E G that does not act trivially on X (r−1) and therefore X is transitive. We deduce that if a group action G is r-transitive and N a nontrivially acting normal subgroup, then N acts (r − 1)-transitively. Exercise: 15 Section 8.3 Question: Let G act transitively on a finite set X. Suppose that for some element x ∈ X, the stabilizer Gx has s orbits on X. Prove that X |X g |2 = s|G|. g∈G

Deduce that G acts 2-transitively if and only if X

|X g |2 = 2|G|.

g∈G

Solution: Let G act in the natural way on X 2 . Then (X 2 )g = {(x, y) ∈ X | g · x = x and g · y = y}. We note that |(X 2 )g | = |X g |2 . Let m(x) be the number of orbits of Gx acting on X. Since G acts transitively on G, then for all z ∈ X, there exists g ∈ G with z = g · x. We know that Gz = gGx g −1 . Since X = g · X, we see that m(x) = m(z). Now consider the set S = {(g, x, y) ∈ G × X 2 | g · (x, y) = (x, y)}. We can count |S| is two different ways. First, as we consider all g ∈ G, then (g, x, y) ∈ S if and only if x, y ∈ X g . Thus |S| =

X

|(X 2 )g | =

g∈G

X

|X g |2 .

g∈G

We can also count the cardinality of S in reference to the variable x. We get X X X X |S| = |X g | = m(x)|Gx | = s |Gx |, x∈X g∈Gx

x∈X

x∈X

where the second equality holds by the Cauchy-Frobenius Lemma. Since G is transitive, then G · x = X so |X| = |G|/|Gx | so |Gx | = |G|/|X|. Hence, |S| = s

X |G| x∈X

|X|

= s|G|

|X| = s|G|. |X|

434

CHAPTER 8. GROUP ACTIONS

For the second part of the exercise, suppose that G acts 2-transitively on X. Then obviously, {x} is one orbit of Gx . Otherwise, for all y, z different from x, there exists some element g ∈ G such that g · (x, y) = g · (x, z). Note that this g is in Gx so G − {x} is another orbit of Gx . Hence, Gx has two orbits so by the first part X |X g |2 = 2|G|. g∈G

Conversely, suppose that X

|X g |2 = 2|G|.

g∈G

Then there exists some element x such that the number of orbits of Gx is 2. However, the stabilizer group always has {x} as one of the orbits. If Gx only has two orbits, then the second orbit must be G − {x}. Now let (x, y) and (z, w) be in X × X with x 6= y and z 6= w. Since G is transitive, there is some element g ∈ G such that z = g · x. Suppose that w = g · w0 . Since z 6= w, we know that w0 6= x. Let h ∈ Gx such that h · w0 = y. Then hg −1 · (z, w) = hg −1 · (g · x, g · w0 ) = h · (x, w0 ) = (x, y). In particular, gh−1 · (x, y) = (z, w). Thus, G is 2-transitive. Exercise: 16 Section 8.3 Question: Let F be a finite field of order pm for some prime p. Let G be the set of functions in Fun(F, F ) of the form f (x) = αx + β such that α ∈ F − {0} and β ∈ F . a) Prove that G is a nonabelian group of order pm (pm − 1). b) Prove that the action of G on F via f · α = f (α) is faithful and transitive. c) Prove that G acts 2-transitively on F . d) Prove that G contains a normal subgroup of order pm that is abelian. e) Determine all the maximal subgroups of G. Solution: Let F be a finite field of order pm for some prime p. Let G be the set of functions in Fun(F, F ) of the form f (x) = αx + β such that α ∈ F − {0} and β ∈ F . a) Let f1 , f2 ∈ G with f1 (x) = α1 x + β1 and f2 (x) = α2 x + β2 . Then (f1 ◦ f2 )(x) = α1 α2 x + (α1 β2 + β1 ). Since α1 , α2 6= 0, then α1 α2 6= 0. Hence, function composition is a binary operation on G. Function composition is associative in every context. The function e(x) = 1x + 0 serves as the identity function. β Finally, if y = αx + β, then x = α1 y − α . Hence, for every f ∈ G, there is an inverse function f −1 in G. Thus G is a group. Furthermore, |G| = (|F | − 1)(|F |) = (pm − 1)pm . β b) Consider the action of G on F by evaluation. Suppose that α, β 6= 0. Then f (x) = α x satisfies f (α) = β. Suppose that α 6= 0. Then f (x) = x + α satisfies f (0) = α and if g(x) = x − α, then g(α) = 0. Finally, the identity function maps 0 to 0. Thus, the action of G on F is transitive. Suppose that a function f (x) = αx + β satisfies f (γ) = γ for all γ ∈ F , then with γ1 6= γ2 , we have f (γ2 ) − f (γ1 ) = γ2 − γ1 ⇐⇒ α(γ2 − γ1 ) = γ2 − γ1 =⇒ α = 1. But then this also forces β = 0. Hence, the kernel of this action is the identity function f (x) = x. Hence, the action is faithful. c) Let (α1 , α2 ) and (β1 , β2 ) be two elements in F × F with α1 6= α2 and β1 6= β2 . Since α1 6= α2 , then we consider the function obtained from the point slope formula: f (x) =

β2 − β1 β2 − β1 x + β1 + α1 . α2 − α1 α2 − α1

We easily see that f (α1 ) = β1 but also f (α2 ) =

β2 − β1 (α2 − α1 ) + β1 = β2 . α2 − α1

This shows that f maps (α1 , α2 ) to (β1 , β2 ) and we conclude that G is 2-transitive on F .

8.3. TRANSITIVE GROUP ACTIONS

435

d) Consider the subgroup H in G of functions of the form h(x) = x + γ, where γ ∈ F . Then for all f ∈ G with f (x) = αx + β, we have 1 β 1 β (f ◦ h ◦ f −1 )(x) = (f ◦ h) x− =f x − + γ = x + αγ. α α α α Hence, f ◦ h ◦ f −1 ∈ H, so H E G and H is abelian of order pm . Furthermore, the quotient group G/H is isomorphic to U (F ), the group of units on F . e) By Exercise 8.3.13, since G acts 2-transitively on F , the action is primitive. By Corollary 8.3.8, Gx is a maximal subgroup all x. Let γ ∈ F . If f (x) = αx + β with f (γ) = γ, we must have β = (α − 1)γ. So elements in Gγ have the form fα (x) = α(x − γ) + γ. Let fα and fα0 be two elements in Gγ . Then (fα ◦ fα0 )(x) = α(α0 (x − γ) + γ − γ) + γ = αα0 (x − γ) + γ = fαα0 (x). This shows that for all γ ∈ F , the function Ψ : U (F ) → Gγ defined by Ψ(α) = fα is an isomorphism (where U (F ) means the group of units of F , equipped with multiplication). Note that the stabilizer groups Gγ are conjugates of the stabilizer subgroup G0 = {x 7→ αx | α ∈ U (F )}. All these groups are maximal. We find all the other maximal subgroups of G as follows. Note that if a maximal subgroup of G contains an element of the form f (x) = x + γ, then it must contain all of H since H ≤ G. We notice that G/H ∼ = U (F ). By Proposition 7.5.2, U (F ) is cyclic with |U (F )| = pm − 1, in particular U (F ) is not of prime order. Let M̄ be an maximal subgroup of U (F ). Then by the Fourth Isomorphism Theorem, there exists M ∈ G such that M̄ = M/H and this group M is maximal in G. The above two constructions give all the maximal subgroups of G. Exercise: 17 Section 8.3 Question: Consider the elements in X = {1, 2, . . . , 7} and the diagram shown in Figure 8.7. The diagram is called the Fano plane and arises in the study of finite geometries. Consider the subset L (called lines) of P(X) whose elements are the subsets of size 3 depicted in the Fano plane diagram either as a straight line or the circle. So for example {1, 2, 5} and {5, 6, 7} are in L. Let G be the largest subgroup of S7 that maps lines to lines, i.e., acts on the set L. a) Prove that the action of G on X is 2-transitive. b) Prove that |G| = 168. c) Prove that G is simple. Solution: Consider the Fano plane as depicted in Figure 8.7. a) Note that every point in X lies on exactly 3 lines in L and that every line lies on exactly 3 points. Also there are 7 lines in L, just as there are 7 points in X. We also observe that for every pair of points, there is a unique line that contains that pair. Let σ ∈ G. Consider two pairs (a, b) and (c, d) with a 6= b and c 6= d of elements in X. If σ(a) = c and ← → σ(b) = d, then we immediately know that σ maps the third point on the line ab to the remaining point on ← → ← → the line cd . For the other two lines on a besides ab , the function σ must map them to either of the other ← → two lines on c besides cd . And similarly for lines on b mapped to lines on d. The action is 2-transitive. b) We determine |G| by determining the options for where σ ∈ G may send successive elements. We refer to Figure 8.7. • 3: Since G is transitive, there are 7 options for σ(3). • 5: Since every two points lie on a common line, the requirements for the action of G do not limit the options for σ(5) except that σ(5) 6= σ(3). Hence, there are 6 options for σ(5). ←−−−−→ • 4: The image σ(4) must be the third point on the line σ(3)σ(5) so there is 1 option for σ(4). ←−−−−→ • 7: The image σ(7) may be any point not on the line σ(3)σ(5). There are 4 options for σ(7). ← → ← → ← → • All three of the remaining points correspond to the third point on 3 7, 5 7 and 4 7 and hence are ←−−−−→ ←−−−−→ ←−−−→ mapped respectively to the third point on σ(3)σ(7), σ(5)σ(7) and σ(4)(7). We point out that every element σ of G is determined completely and uniquely by where σ maps 3 reference noncollinear points. Furthermore, any mapping of 3 noncollinear points into any other 3 noncollinear points

436

CHAPTER 8. GROUP ACTIONS

is possible. With the above labeling, we used where 3, 5, and 7 as the three noncollinear reference points. Note that since points 1, 2, and 3 are noncollinear, we can also use them as reference points to uniquely describe the action of a group element σ ∈ G. c) By Exercise 8.3.13, the action of G on {1, 2, . . . , 7} is primitive because it is 2-transitive. It is also easy to see that the action is faithful s owe can view G as a subgroup of S7 . Referring to Corollary 8.3.14 we attempt to show that the only normal transitive subgroup is all of G. The elements in G have cycle types of the form (a b)(c d),

(a b c d)(e f ),

(a b c)(d e f ),

(a b c d e f g),

though not all elements of these cycle types occur in G. The elements of the form (a b)(c d) correspond to interchanging two pairs of lines. It is impossible to interchange a single pair of lines. (We note in passing that G ≤ A7 .) Now just as Sn is generated by the set of transpositions, G can be generated by the set of elements of the form (a b)(c d). By conjugacy in Sn (see Example 4.2.13), the element gσg −1 behaves the same as σ after the vertices {1, 2, . . . , 7} have been relabeled according to g. We discuss some properties of conjugacy classes in G by cycle type: Consider elements of the form (a b)(c d). Since we can uniquely determine elements in G by how they act on the set {1, 2, 3} and since G is 2-transitive, every element in G of the form is conjugate to one of the form (1 2)(a b). But since it is uniquely determined by how it acts on the element 3, we can check that the only possibility is (1 2)(3 4) and (1 2)(6 7), which occurs when 3 is fixed. Note that these correspond ← → ← → ← → to situations in which the lines 12 and cd intersect at 5, the third point on the line 12 . These two are conjugate to each other via (1 2)(6 7) = (3 6)(4 7)(1 2)(3 4)(3 6)(4 7). Hence, all the elements in G of the form (a b)(c d) form a single conjugacy class. From now on, as we consider elements that are not of the above cycle type, Since G is 2-transitive, every element is conjugate to one that sends 1 to 2 and 2 to 3. We check subsequent possibilities given this condition: • if 3 goes to 1, the only possibility is (1 2 3)(5 7 6). • 3 cannot go to 2 or 3. • if 3 goes to 4, the only possibility is (1 2 3 4)(5 7). • if 3 goes to 5, the only possibility is (1 2 3 5 7 4 6). • if 3 goes to 6, the only possibility is (1 2 3 6 4 5 7). • 3 cannot go to 7. Now, let H be a transitive subgroup of G (i.e., a subgroup of G that acts transitively on X) that is a normal subgroup. By our above discussion, if H contains a single element of the form (a b)(c d), then it contains all such elements and hence H = G. So now we only need to suppose that H contains one of the four types of elements listed above and we will show in all cases that H = G. • H contains (1 2 3)(5 7 6). Then H contains (1 2)(3 4)(1 2 3)(5 7 6)(1 2)(3 4) · (1 3 2)(5 6 7) = (1 3)(2 4) and hence H = G. • H contains (1 2 3 4)(5 7). Then H contains the square of this element, which is (1 3)(2 4) so again H = G. • H contains (1 2 3 5 7 4 6). Then H contains (1 2)(3 4)(1 2 3 5 7 4 6)(1 2)(3 4) · (1 6 4 7 5 3 2) = (1 2 4 3)(5 6), and so also contains the square of this, which is (1 4)(2 3). Thus again H = G. • H contains (1 2 3 6 4 5 7) is identical in approach to the previous item. By Corollary 8.3.14, we conclude that G is a simple group.

8.4. GROUPS ACTING ON THEMSELVES

437

8.4 – Groups Acting on Themselves Exercise: 1 Section 8.4 Question: Consider the action of Q8 on itself by left multiplication. Consider the induced permutation representation ρ : Q8 → S8 . Determine ρ(−1), ρ(i), ρ(j), and ρ(k). Solution: We first need to label the elements as Q8 with integers. Let 1, −1, i, −i, j, −j, k, −k correspond to 1, 2, 3, 4, 5, 6, 7, 8. Then the action of Q8 on itself by left multiplication induces the following representations: ρ(−1) = (1 2)(3 4)(5 6)(7 8) ρ(i) = (1 3 2 4)(5 7 6 8) ρ(j) = (1 5 2 6)(3 8 4 7) ρ(k) = (1 7 2 8)(3 5 4 6)

Exercise: 2 Section 8.4 Question: Consider the action of D4 on itself by conjugation. After labeling the elements in D4 , define the induced permutation representation ρ : D4 → S8 , and write down ρ(g) for all g ∈ D4 . Solution: We label the elements 1, r, r2 , r3 , s, sr, sr2 , sr3 of D4 by g1 , g2 , . . . , g8 . We defined the induced representation ρ : D4 → S8 is the homomorphism such that that ρ(g) = σ, where σ(i) = j when ggi = gj . With these labels, we have ρ(1) = 1 ρ(r) = (1 2 3 4)(5 8 7 6) ρ(r2 ) = (1 3)(2 4)(5 7)(6 8) ρ(r3 ) = (1 4 3 2)(5 6 7 8) ρ(s) = (1 5)(2 6)(3 7)(4 8) ρ(sr) = (1 6)(2 7)(3 8)(4 5) ρ(sr2 ) = (1 7)(2 8)(3 5)(4 6) ρ(sr3 ) = (1 8)(2 5)(3 6)(4 7).

Exercise: 3 Section 8.4 Question: Label the S3 elements 1, (1 2), (1 3), (2 3), (1 2 3), and (1 3 2) with the integers 1, 2, 3, 4, 5, and 6 respectively. Consider the action of S3 on itself by left multiplication and write the image of each element under the induced permutation representation as an element in S6 . Solution: With the given labels of elements, the images of the induced representation are ρ(1) = 1 ρ((1 2)) = (1 2)(3 6)(4 5) ρ((1 3)) = (1 3)(2 5)(4 6) ρ((2 3)) = (1 4)(2 6)(3 5) ρ((1 2 3)) = (1 5 6)(2 3 4) ρ((1 3 2)) = (1 6 5)(2 4 3).

Exercise: 4 Section 8.4 Question: Let G be a group and let X = G. Show that the pairing G × X → X given by (g, x) 7→ xg is not generally a (left) group action. Show, however, that the pairing G × X → X given by (g, x) 7→ xg −1 does give a group action of G on itself.

438

CHAPTER 8. GROUP ACTIONS

Solution: With the pairing G × X → X given by (g, x) 7→ xg, we have g1 · (g2 · x) = g1 · xg2 = xg2 g1 whereas (g1 g2 ) · x = xg1 g2 . Hence, this axiom is satisfied if and only if G is abelian. Hence, this pairing does not generally define a group action. With the pairing G × X → X given by (g, x) 7→ xg −1 , we have g1 · (g2 · x) = g1 · xg2−1 = xg2−1 g1−1 = x(g1 g2 )−1 = (g1 g2 ) · x. In addition, 1 · x = x1−1 = x, so this pairing does define a group action. Exercise: 5 Section 8.4 Question: Prove that the action of a group on itself by left multiplication is primitive if and only if G ∼ = Zp , where p is prime. Solution: Suppose that G = Zp . Let B be a block containing two elements x1 and x2 . Then the element x2 x−1 is a nonidentity element of G and thus it is a generator of G = Zp . The block B also contains all the 1 k elements (x2 x−1 1 ) x1 , namely the orbit of x1 , which is all of G since the action of G on itself is transitive. Since the blocks of the action are either singletons or all of G, the action is primitive. Conversely, suppose that the action of a group on itself by left multiplication is primitive. Let B be a block that contains the identity 1. Let b1 , b2 ∈ B. Then with g = b2 b−1 1 we have b2 ∈ gB ∩ B, so gB = B since B is a block. Thus g · 1 = g ∈ B. Thus b2 b−1 ∈ B. By the One-Step Subgroup Criterion, B ≤ G. Furthermore, for 1 any subgroup H, the cosets gH form a system of blocks of the left multiplication of G on itself. So if this action is primitive, then C contains only two subgroups, {1} and G. The only groups with this property are cyclic of prime order p. Exercise: 6 Section 8.4 Question: Prove that the action of a group on itself by conjugation is trivial if and only if G is abelian. Solution: The action of G on itself by conjugation is trivial if and only if gxg −1 = x for all g ∈ G and for all x ∈ G. This is equivalent to gx = xg for all g, x ∈ G. Hence, this is equivalent to G being abelian. Exercise: 7 Section 8.4 Question: Let ρ : G → SG be the homomorphism induced from G acting on itself by left multiplication. Suppose x ∈ G with |x| = n and that |G| = mn. Prove that ρ(x) is a product of m disjoint n-cycles. Solution: Consider the subgroup hxi in G. Let {g1 , g2 , . . . , gm } be a subset of G such that hxig1 , hxig2 , . . . , hxigm are the distinct cosets of hxi. Then x acts as an n-cycle on each right coset hxigi . Since the right cosets form a partition of G, any labeling of the elements of G is such that the imagine ρ(x) is a permutation of m disjoint n-cycle. Exercise: 8 Section 8.4 Question: Let G be a group and let H be a subgroup of G. def

a) Prove that the mapping g · (xH) = (gx)H defines an action of G on the set of left cosets of H. b) Denote by ρH the homomorphism of G onto the set of permutations on left cosets of H. Prove that the kernel of this action is \ Ker ρH = xHx−1 . x∈G

c) Prove that Ker ρH is the largest normal subgroup contained in H. Solution:

8.4. GROUPS ACTING ON THEMSELVES

439

a) We know that the mapping is well defined for any choice of representative for the left coset from Section 4.1. We will check if it is a group action. For some g1 , g2 ∈ G we have g1 · (g2 · xH)) = g1 · (g2 x)H = (g1 (g2 x))H = (g1 g2 )xH = (g1 g2 ) · xH. And for the identity, 1 · xH = (1x)H = xH. So the mapping defines an action on the set of left cosets of H. b) Let g ∈ G and g ∈ Ker ρH so that g · xH = gxH = xH for all x ∈ G. For any x ∈ G and any element h1 ∈ H there exists some h2 ∈ H where gxh1 = xh2 −1 ⇒ g = x(h2 h−1 ∈ xHx−1 . 1 )x T T Since this holds ∈ G, we have that g ∈ x∈G xHx−1 . This implies that Ker ρH ⊆ x∈G xHx−1 . T for any x Now let g ∈ x∈G xHx−1 . So for any element x we can write g = xh1 x−1 for some element h1 ∈ H. Consider, for any x ∈ G,

g · xH = (gx)H = (xh1 x−1 x)H = (xh1 )H = xH. Since this holds for every x ∈ G, this implies that ρH (g) = (1) or g ∈ Ker ρH . This shows the equality. c) It is clear that since Ker ρH is a kernel of a homomorphism that it is a normal subgroup. Since 1 ∈ G this implies that \ Ker ρH = xHx−1 x∈G

 =

 \

xHx−1  ∩ H

x∈G and x6=1

⇒ Ker ρH ⊆ H. So the kernel is a normal subgroup and is contained in H. Now we show that any normal subgroup contained in H is contained in the kernel. Let K ≤ H and K E G. Consider any element k1 ∈ K. Then for any x ∈ G we have k1 · xH = (k1 x)H = (xx−1 )k1 xH = (x(x−1 k1 x))H = (xk2 )H = xH. This show that k1 · xH = xH for every x ∈ G. Equivalently, k1 ∈ Ker ρH ⇒ K ⊆ Ker ρH . So any normal subgroup contained in H is subsequently contained in the kernel of ρH as well. This show that Ker ρH is the largest normal subgroup contained in H. Exercise: 9 Section 8.4 Question: Use Exercise 8.4.8 to prove the following theorem. Suppose that p is the smallest prime dividing the order of G. Then if H ≤ G with |G : H| = p, then H E G. Solution: We will set up the same homomorphism ρH : G → S{xH} from the group into the set of permutations on the left cosets of H. We realize that S{xH} = Sp which has size p!. Now, we know that | Im ρH | |G|. Additionally, | Im ρH | |Sp | = p!. This implies that | Im ρH | divides the gcd(|G|, p!) which is either p or 1 since p is the smallest prime that divides G and p only divides p! once. If it is p, then | Ker ρH | = |G|/p = |H| which implies that Ker ρH = H since we already know that the kernel is contained inside of H. If | Im ρH | = 1 this implies that | Ker ρH | = |G|/1 = |G|. Then we contradict the fact that Ker ρH is contained in H. This is because

440

CHAPTER 8. GROUP ACTIONS

H is strictly smaller then G since |G : H| = p which is a prime and cannot be one. So we must have that Ker ρH = H, which implies that H is a normal subgroup by the first isomorphism theorem. Exercise: 10 Section 8.4 Question: Let G be a group and let Sub(G) be the set of subgroups of G. For all H ∈ Sub(G) and all g ∈ G, define the pairing g · H = gHg −1 . Prove that this pairing defines a group action of G on Sub(G). Solution: Let H ∈ Sub(G) and let g1 , g2 ∈ G. Then g1 · (g2 · H) = g1 · (g2 Hg2−1 ) = g1 g2 Hg2−1 g1−1 = (g1 g2 )H(g1 g2 )−1 = (g1 g2 ) · H. Also, 1 · H = 1H1−1 = H. Thus, the action of conjugation on subgroups of G is a group action. Exercise: 11 Section 8.4 Question: Consider the action of G on itself by conjugation. Prove that the invariant subsets of G are unions of conjugacy classes. Conclude that a subgroup of G is invariant under the action if and only if it is a normal subgroup. Solution: Let S be a subset of G that is invariant under the action of G on itself by conjugation. Suppose that x ∈ S. Then gxg −1 ∈ S for all g ∈ G. Thus, the conjugacy class of x is in S. Consequently, S is the union of conjugacy classes (of all its elements). A subgroup H is invariant under conjugation if and only if gxg −1 ∈ H for all g ∈ G and x ∈ H, which is one of the equivalent conditions for a subgroup to be normal. Exercise: 12 Section 8.2 Question: Let G be a group. The automorphism group Aut(G) acts on G by ψ · g = ψ(g) for all ψ ∈ Aut(G) and all g ∈ G. a) Find the orbits of this action if G = Z12 . b) Find the orbits of this action if G = Z7 . c) Find the orbits of this action if G = D4 . [Hint: First determine Aut(D4 ).] Solution: a) If G = Z12 = hzi then the automorphism is completely determined by where we send the generator. Any power the is relatively prime to 12 can serve as a generator. So we can have z map to any of the following, {z, z 5 , z 7 , z 11 }. So the size of Aut(Z12 ) is 4 and the size of any orbit of the action is 1, 2, or 4. Our orbits are actually contain all elements of the same order. So we get the following orbits, O1 = {1}, O2 = {z 6 }, O3 = {z 3 , z 9 } O4 = {z 2 , z 10 }, O5 = {z 4 , z 8 }, O6 = {z, z 5 , z 7 , z 11 } b) If G = Z7 , then any non-identity element can serve as the generator. So we have the following orbits O1 = {1}, O2 = {z, z 2 , z 3 , z 4 , z 5 , z 6 }. c) If G = D4 , then our orbits are O1 = {1}, O2 = {r, r3 }, O3 = {r2 }, O4 = {s, sr2 , sr, sr3 }.

Exercise: 13 Section 8.4 Question: Let p < q be primes and let G be a group of order pq. Prove that G has a nonnormal subgroup of index q. Deduce that there exists an injective homomorphism of G into Sq . Solution: [This exercise is not correct as stated. For example, if G is abelian, then all subgroups are normal.] We will prove that for all nonabelian groups G with the stated property that there exists an injective homomorphism ϕ from G into Sq . By Cauchy’s Theorem, G contains an element x of order q. Hence, the index |G : hxi| = p. By Exercise 8.4.9, since p is the least prime dividing |G|, then hxi E G. Also by Cauchy’s Theorem, there exists an element y ∈ G of order p. Since hxi E G, then yxy −1 = xk , where 1 < k < q. (Note that k 6= 1 because G is nonabelian.) This relation on x and y shows that in any word involving x and y all the x generators can be moved to the left of the word, possibly changing powers on x as needed. Furthermore, it is easy to show

8.4. GROUPS ACTING ON THEMSELVES

441

that xa y b = xc y d if and only if a ≡ c (mod q) and b ≡ d (mod p). Hence, hx, yi has pq elements and therefore is all of G. n Note that if yxy −1 = xk , then y n xy −n = xk . So since y p = 1, this relation can only hold if k p ≡ 1 (mod q), and more precisely that the order of k̄ in U (q) is p. We begin to define the function ϕ as follows. First, set ϕ(x) = (1 2 3 · · · q). Then ϕ(xm ) = (1 2 3 · · · q)m . Since ϕ(x)k is another q cycle, there exists a permutation σ ∈ Sq such that σ(1 2 3 · · · q)σ −1 = (1 2 3 · · · q)k . Now set ϕ(y) = σ. Since the order of k̄ ∈ U (q) is p, then the order of σ ∈ Sq is p. Thus, by the Extension Theorem on Generators, ϕ defines a homomorphism G → Sq . Since |σ| = p and since σ and (1 2 3 · q) satisfy the same relation yxy −1 = xk , then the homomorphism is injective. Exercise: 14 Section 8.2 Question: Let p be a prime. Use the Class Equation to prove that every p-group, i.e., a group of order pk for some integer k, has a nontrivial center. Solution: The class equation is X |G| = |Z(G)| + |G : CG (x)|. x∈K

where K is a complete set of distinct representatives of conjugacy classes of size greater than or equal to 2. For any x, |G : CG (x)| is equivalent to the size of the conjugacy class of x. We know that the size of any conjugacy class of some element x must divide the order of the group G. Furthermore, since we also know that any x ∈ K has a conjugacy class greater than 1 so that we must have |G : CG (x)| = pi for some positive i. Then if we take the entire class equation mod p we see that 0 ≡ |Z(G)| + 0 (mod p). This implies that |Z(G)| ≡ 0 (mod p). Since 1 ∈ Z(G), we must have that |Z(G)| = pi for some postive i. This shows that the center is nontrivial. Exercise: 15 Section 8.4 Question: Let p be prime. Use Exercise 8.4.14 to prove that every group of order p2 is abelian. In particular, if |G| = p2 , then G is isomorphic to Zp2 or Zp ⊕ Zp . Solution: Let G be a group of order p2 where p is prime. By Exercise 8.4.14, the center of G is nontrivial. By Lagrange’s Theorem, Z(G) has order p or p2 . Obviously, if |Z(G)| = p2 , then Z(G) = G and G is abelian. By the Fundamental Theorem on Finitely Generated Abelian Groups, G ∼ = Zp2 or Zp ⊕ Zp . Assume that |Z(G)| = p. Then G/Z(G) has order p so G/Z(G) ∼ = Zp . By Exercise 4.3.21, since G/Z(G) is cyclic, then G is abelian. But then Z(G) = G. Hence, the case |Z(G)| = p leads to a contradiction. Exercise: 16 Section 8.4 Question: If G is a p-group and H is a proper subgroup, show that the normalizer NG (H) properly contains H. Solution: Let G be a p-group and H a proper subgroup. Note by Lagrange’s Theorem that every subgroup of G is itself a p-group. Then NG (H) acts on H by conjugation. Before setting up the proof, we give a few useful observations. 1) If G is abelian, then all subgroups are normal in G, in which case NG (H) = G and the result holds trivially since H is assumed to be proper in G. So to prove the exercise, we only need to suppose that G is nonabelian. Hence, |G| ≥ p2 and, in fact, by Exercise 8.4.15, we may suppose that |G| ≥ p3 . 2) We also observe that H normalizes itself and Z(G) normalizes H since elements in Z(G) commute with all elements in H. Hence, if Z(G) is not contained in H then H hZ(G), Hi ≤ NG (H) and so H is a proper subgroup of NG (H). We prove the Theorem of this exercise by (strong) induction on the size of G, more precisely on ordp |G|. As we observed, if ordp |G| ≤ 2, then G is abelian and the result holds. Suppose that the result holds for all p-groups G0 with ordp |G0 | < m and let G be a p-group of order pm . If Z(G) is not contained in H, then as observed the theorem holds (without reference to the induction hypothesis). So suppose that Z(G) ≤ H and consider the quotient group G/Z(G). By Exercise 8.4.14, ordp |G/Z(G)| < m and so by the induction hypothesis the statement of the theorem holds for G/Z(G). Thus, subgroups K/Z(G) in G/Z(G) by K̄, we see that H̄ is a proper subgroup of NḠ (H̄). By the Fourth Isomorphism Theorem, H is a proper subgroup of NG (H). By induction, the theorem holds for all p-groups.

442

CHAPTER 8. GROUP ACTIONS

Exercise: 17 Section 8.4 Question: Let G be a group. Show that the pairing (G ⊕ G) × G → G defined by (g, h) · x = gxh−1 is an action of G ⊕ G on G. Show that the action is transitive. Also determine the stabilizer of the identity 1. Solution: Let x, g1 , g2 , h1 , h2 ∈ G. Then the pairing (G ⊕ G) × G → G defined by (g, h) · x = gxh−1 satisfies −1 −1 −1 (g1 , h1 ) · ((g2 , h2 ) · x) = (g1 , h1 ) · g2 xh−1 = (g1 g2 , h1 , h2 ) · x. 2 = g1 g2 xh2 h1 = g1 g2 x(h1 h2 )

Also (1, 1) · x = 1x1−1 = x for all x ∈ G. Hence, the pairing defines an action of G ⊕ G on G. Simply by considering elements of the form (g, 1), since the action of left multiplication of G on itself is transitive, we see that this action is also transitive. The stabilizer of the identity is all elements (g, h) ∈ G ⊕ G such that g1h−1 = 1, namely all pairs where h = g. Hence, the stabilizer is the diagonal subgroup {(g, g) ∈ G ⊕ G | g ∈ G}. Exercise: 18 Section 8.4 Question: (Cauchy’s Theorem) The original proof to Cauchy’s Theorem did not use the group action described in the proof we gave, but it relied instead on the Class Equation. Let p be a prime that divides the order of a finite group G. a) Prove Cauchy’s Theorem for finite abelian groups. [Hint: Use induction on |G|.] b) Prove Cauchy’s Theorem for finite nonabelian groups by induction on |G| and using the Class Equation. Solution: Let G be a finite group and let p be a prime number such that p divides |G|. a) First suppose that G is an abelian group. Note that G = {1} is an abelian group and for every prime divisors p of |G| there exists an element in G of order p. (This is true trivially since there are no prime divisors of |G|. Now let G be an abelian group and suppose that Cauchy’s Theorem is true for all abelian groups of order strictly less than |G|. Let p be a prime dividing |G|. Since |G| > 1, then G contains a nonidentity element x. Case 1: p divides |x|, say with |x| = pm, then |xm | = p, so xm is an element of order p. Case 2: p does not divide |x|. Set n = |x|. Since G is abelian, hxi E G. Then |G/hxi| = |G|/|x| so |G/hxi| < |G| and p divides |G/hxi|. Hence, using the induction hypothesis, there exists ȳ ∈ G/hxi of order p. Let y ∈ G be a representative of the coset ȳ. Then y p ∈ hxi so y pn = 1. So the order of y n divides p, but since y n 6= 1, we deduce that |y n | = p. By induction, for all finite abelian groups G, if a prime p divides |G| then there exists an element of order p in G. (Note that this proof does not use the Fundamental Theorem of Finitely Generated Abelian groups, which would also have led to a proof of Cauchy’s Theorem for abelian groups.) b) We no longer suppose that G is abelian. We prove this by induction on |G|. Note first that as in the previous part, Cauchy’s Theorem is trivially satisfied for |G| = 1. Now let G be a finite group and suppose that the statement of Cauchy’s Theorem is true for all groups of order strictly less than |G|. Case 1: Suppose that p divides |Z(G)|. Then since Z(G) is abelian, by the previous part Z(G) and hence G contains an element of order p. Case 2: Suppose that p does not divide |Z(G)|. The Class Equation gives |G| = |Z(G)| +

s X

|G : CG (gi )|,

i=1

where {g1 , g2 , . . . , gs } is a complete set of distinct representatives of conjugacy classes of more than two elements. Since p does not divide |Z(G)|, then it is not possible for all the indices |G : CG (gi )| to be divisible by p. Let g 0 be such that p does not divide |G : CG (g 0 )| = |G|/|CG (g 0 )|. Then since p divides |G|, we deduce that p divides |CG (g 0 )|. But 1 < |G : CG (gi )| < |G| so 1 < |CG (g 0 )| < |G|. By the induction hypothesis, CG (g 0 ) contains an element of order p, which is an element of G. By induction, every finite group G such that a prime p divides |G| contains an element of order p. Exercise: 19 Section 8.4 Question: Suppose that G is a finite group with m conjugacy classes. Show that the number of ordered pairs (x, y) ∈ G × G such that yx = xy is equal to m|G|.

8.5. SYLOW’S THEOREM

443

Solution: Consider the action of G on itself by conjugation and apply the Cauchy-Frobenius Lemma. Note that for g ∈ G, the fixed set Gg = {x ∈ G | gxg −1 }. Thus, following the proof of the Lemma, the set S = {(x, y) ∈ G × G | xy = yx} = {(x, y) ∈ G × G | xyx−1 = y} is equal to the disjoint union [ S= Gg . g∈G

Since the union is disjoint, by the Cauchy-Frobenius Lemma we have |S| = m|G|, where m is the number of orbits of the action, i.e., the number of conjugacy classes in G.

8.5 – Sylow’s Theorem Exercise: 1 Section 8.5 Question: Determine the isomorphism type of the Sylow 2-subgroups of A6 . Solution: We know that |A6 | = 360 = 23 · 32 · 5 so the Sylow 2-subgroups of A6 have order 8. Consider the subgroup H = h(1 2 3 4)(5 6), (1 3)(5 6)i. If we set σ = (1 2 3 4)(5 6) and τ = (1 3)(5 6), it is easy to see that |σ| = 4 and |τ | = 2, and that τ στ −1 = σ −1 . Thus, H = {σ i τ j | i = 0, 1, 2, 3; j = 0, 1} so that |H| = 8. By the Extension Theorem on Generators, we deduce that H ∼ = D4 . Since all Sylow 2-subgroups are isomorphic, they are all isomorphic to H. Exercise: 2 Section 8.5 Question: Let p be a prime and let suppose that 2 ≤ k ≤ p − 1. Find a Sylow p-subgroup of Skp by expressing it using generators. Solution: The order of Skp is (kp)!. Since 2 ≤ k ≤ p − 1, there exactly k integers divisible by p, each having only one factor of p in the product (kp)!. Hence, ordp ((kp)!) = k. Thus, the Sylow p-subgroups of Skp have order pk . Define the permutations σ0 , σ1 , . . . , σk by as the p-cycle σj = (jp + 1 jp + 2 · · · jp + p − 1). Note that for i, j ∈ {0, 1, . . . , p − 1}, since σi and σj are disjoint p-cycles, they commute. Consider the subgroup H = hσ0 , σ1 , . . . , σp−1 i. Since all the generators commute and have order p, we see that H ∼ = Zpk . Hence, |H| = pk and thus is a Sylow p-subgroup of Skp . By Sylow’s Theorem, all Sylow p-subgroups of Skp are isomoprhic to Zpk . Exercise: 3 Section 8.6 Question: Let p be a prime. Find a Sylow p-subgroup of Sp2 by expressing it using generators. Show also that it is a nonabelian group of order pp+1 . Solution: Note that ordp (p2 !) = pp+1 . Hence, this is the order of the Sylow p-subgroups of Sp2 . As in the previous exercise, define the permutations σ0 , σ1 , . . . , σp−1 by as the p-cycles σj = (jp + 1 jp + 2 · · · jp + p − 1). Now also define the permutation τ defined by τ (k) = k + p mod p2 . So in cycle notation τ = (1 p + 1 2p + 1 · · · p2 − p + 1)(2 p + 2 2p + 2 · · · p2 − p + 2) · (p 2p 3p · · · p2 ). We observe that σi σj = σj σi since they are disjoint cycles. Furthermore, τ σi τ −1 = σi+1 for 0 ≤ i ≤ p − 2 and τ σp−1 τ −1 = σ0 . Thus, in every word in the σi and the element τ , the element τ can move to the right and every word can be written uniquely as kp−1 kp σ0k0 σ1k1 · · · σp−1 τ , where 0 ≤ ki ≤ p − 1. Hence, if H = hσ0 , σ1 , . . . , σp−1 , τ i, we see that |H| = pp+1 . This shows that H is a Sylow p-subgroup of Sp2 . Furthermore, it is nonabelian since τ σi τ −1 6= σi .

444

CHAPTER 8. GROUP ACTIONS

Exercise: 4 Section 8.5 Question: Exhibit all Sylow 2-subgroups of S4 . Solution: Since |S4 | = 24 = 23 · 3, the Sylow 2-subgroups have order 8. Note that h(1 2 3 4), (1 3)i is a subgroup of order 8, and it is isomorphic to D4 . Note that every Sylow 2-subgroup is isomorphic by conjugation to this group. There are 6 distinct 4-cycles in S4 and they occur in pairs by inverses in a given Sylow 2-subgroup. We get the following distinct subgroups h(1 2 3 4), (1 3)i

h(1 3 4 2), (2 3)ih(1 4 2 3), (3 4)i

Exercise: 5 Section 8.5 Question: Exhibit a Sylow 2-subgroup of SL2 (F3 ) by expressing it using generators. Solution: The group SL2 (F3 ) has 48/2 = 24 elements, so its Sylow 2-subgroups have order 8. Consider the element 1 1 A= . 1 0 We note that

1 1

2 2

A= A5 =

1 2 A2 = 0 1 2 1 A6 = 0 2

1 0 A3 = 1 2 2 0 A7 = 2 1

2 2 A4 = 1 0 1 1 A8 = 2 0

0 2 0 . 1

Thus, |hAi, which is isomorphic to Z8 , is a Sylow 2-subgroup of SL2 (F3 ). Exercise: 6 Section 8.5 Question: Suppose that G = GL2 (Fp ). Prove that np (G) = p + 1. Solution: By Exercise 3.2.32, the order of G = GL2 (Fp ) is (p2 − 1)(p2 − p) = p(p + 1)(p − 1)2 . Hence, Sylow p-subgroups of G have order p. By Sylow’s Theorem np (G) ≡ 1 (mod p) and np (G)|(p + 1)(p − 1)2 . From this information alone, the possible values of np (G) are 1, p + 1, (p − 1)2 , or (p + 1)(p − 1)2 . One particular Sylow p-subgroup is 1̄ 1̄ . 0̄ 1̄ 1̄ 1̄ By Sylow’s Theorem, all Sylow p-subgroups are conjugate to this one. Notice that the determinant of A = 0̄ 1̄ is 1̄. Then for all M ∈ GL2 (Fp ), the determinant det(M AM −1 ) = det(A) = 1̄. Hence, all the Sylow p-subgroups are in SL2 (Fp ), which has order p(p + 1)(p − 1). Now np (G) = np (SL2 (Fp )) and np (SL2 (Fp )) ≡ 1 (mod p) and also np (SL2 (Fp )) divides (p + 1)(p − 1). Thus np (G) = 1 or p + 1. However, 1̄ 0̄ 1̄ 1̄ is another Sylow p-subgroup so np (G) > 1. We conclude that np (G) = p + 1. Exercise: 7 Section 8.5 Question: Exhibit a Sylow 3-subgroup of GL2 (F17 ) by expressing it using generators. Solution: The order of GL2 (F17 ) is (172 − 1)(172 − 17) = 78336 = 29 · 32 · 17. Thus, the Sylow 3-subgroups have order 9. As we look for matrices with order 3 or 9, we point out that by Propostion 7.5.2, U (F17 ) = U (17) is a cyclic group of order 16. Furthermore, det : GL2 (F17 ) → U (17) is a homomorphism. If the order of a matrix A is 3 or 9, since the order of det(A) will divide the order of A and divide 16, we deduce that the order of det A is 1. Thus, any such matrix A must have determinant 1. It is also easy to check that A cannot be upper triangular or lower triangular. We try calculating various power of some matrices. As a first example, ¯ 8̄ ¯ 0̄ 2̄ 1̄ 5̄ 3̄ 13 16 2 3 9 A= =⇒ A = , A = , and A = ¯ . 1̄ 1̄ 3̄ 2̄ 8̄ 5̄ 0̄ 16

8.5. SYLOW’S THEOREM

445

From this, we notice that A18 = I. We can also check that A6 6= I so the order of A is 18. Thus, A2 has order 9 and so hA2 i is a Sylow 3-subgroup. In particular, the Sylow 3-subgroups of GL2 (F17 ) are cyclic. Exercise: 8 Section 8.5 Question: Let p be a prime number and consider the group S2p . 2 a) Show that np = 21 2p p ((p − 2)!) . 2 b) Use Sylow’s Theorem to conclude that 12 2p p ((p − 2)!) ≡ 1 (mod p). Solution: Let p be an odd prime and consider the group S2p . a) A Sylow p-subgroup of S2p is any subgroup conjugate to h(1 2 · · · p), (p + 1 p + 2 · · · 2p)i, which is isomorphic to Zp2 . Hence, Sylow 2-subgroups are the subgroups generated by two disjoint pcycles. We count up the number of subgroups we can make in this way. Note that all the integers in {1, 2, . . . , 2p} appear in one or the other of the two p-cycles generating the subgroup. Without loss of generality, we require that the first generating p-cycle (a1 a2 · · · ap ) begins with 1, i.e., a1 = 1. Then, the are (2p − 1)(2p − 2) · · · (p + 1) possible choices for the first generating p-cycle. Now all the integers in {1, 2, . . . , 2p} − {a1 , a2 , . . . , ap } will occur in the second generating p-cycle. To write the cycle in standard notation, the first entry of the second p-cycle begins with the least element in {1, 2, . . . , 2p}−{a1 , a2 , . . . , ap }. And then there are (p − 1)! options for the second generating p-cycle. However, any of the p − 1 powers of the first cycle and any of the p − 1 powers of the second cycle will generate the same subgroup. Thus, we have found that (2p)! 2p 1 2p (p!)2 (2p − 1)(2p − 2) · · · (p + 1)(p − 1)! np (S2p ) = = 2 = = ((p − 2)!)2 . (p − 1)2 2p (p − 1)2 p 2p2 (p − 1)2 2 p b) Sylow’s Theorem tells us that np (S2p ) ≡ 1 (mod p). Hence, in this particular case 1 2p ((p − 2)!)2 ≡ 1 (mod p). 2 p

Exercise: 9 Section 8.5 Question: Show that a group of order 418 has a normal subgroup of order 11 and a normal subgroup of order 19. Solution: Notice that 418 = 2 · 11 · 19. Then n11 ≡ 1 (mod 11) and n11 | 38. We only need to consider the possibilities of n11 = 1, 12, 23, 34, where the only possibility if n11 = 1. Hence, the Sylow 11-subgroup, which has order 11, is normal. Furthermore, n19 ≡ 1 (mod 19) and n19 | 22. We only need to consider the possibilities of n19 = 1, 20, where the only possibility if n11 = 1. Hence, the Sylow 19-subgroup, which has order 19, is normal.

Exercise: 10 Section 8.5 Question: Prove that there is no simple group of order 225. Solution: The prime factorizatin of 225 is 225 = 32 · 52 . By Sylow’s Theorem n5 ≡ 1 (mod 5) and n5 | 9. The options from the first condition with n5 < 9 give n5 = 1, 6 and show that the only situation in which n5 | 9 is n5 = 1. Hence, every group of order 225 has a normal Sylow 5-subgroup, which has order 25. Consequently, no group of order 225 is simple. Exercise: 11 Section 8.5 Question: Prove that there is no simple group of order 825. Solution: The prime factorization of 825 is 825 = 3 · 52 · 11. By Sylow’s Theorem, n11 must be a divisor of 75, (1, 3, 5, 15, 25, or 75), but also n11 ≡ 1 (mod 11) and so must be 1. If n11 = 1, then by Proposition 8.4.9 the Sylow 11-subgroup is normal and the group is not simple. Exercise: 12 Section 8.5 Question: Prove that there is no simple group of order 2907.

446

CHAPTER 8. GROUP ACTIONS

Solution: The prime factorization of 2907 is 2907 = 32 · 17 · 19. By Sylow’s Theorem n19 ≡ 1 (mod 19) and n19 must divide 32 · 17 = 153. The divisors of 153 are 1, 3, 9, 17, 51, 153. Obviously 3, 9, and 17 are not congruent to 1 modulo 19. Also 51 ≡ 13 (mod 19) but 153 ≡ 1 (mod 19). So it may be possible for n19 to be equal to 1 or 153. This does not yet confirm that a group of order 2907 must contain a normal subgroup. We consider the prime factor of 17. By Sylow’s Theorem n17 ≡ 1 (mod 17) and n17 must divide 32 · 19 = 171. The divisors of 171 are 1, 3, 9, 19, 57, 171. We note that of these 3, 9, 19, and 57 are not congruent to 1 modulo 17 but that 1 and 171 are. We prove the desired result by contradiction. Assume that a group of order 2907 is simple. Then n17 6= 1 and n19 6= 1. Thus, we must have n17 = 171 and n19 = 153. If there are 171 cycle subgroups of order 17, these subgroups can only intersect in the identity, since each nonidentity element in a cyclic group generates the whole group. Hence, 171 cyclic subgroups of 17 accounts for 16 × 171 = 2736 elements of order 17. (We used the factor of 16, because in each cyclic group of order 17, there are 16 elements of order 17.) By the same reasoning, there are 18 × 153 = 2754 elements of order 19 in the group. This already accounts for 5490 elements, which is more that 2907. Hence, it is not possible for n17 6= 1 and n19 6= 1. Thus, the group has a normal subgroup of order 17 or a normal subgroup of order 19. Consequently, no group of order 2907 is simple. Exercise: 13 Section 8.5 Question: Prove that there is no simple group of order 3124. Solution: The prime factorization of 3124 is 3124 = 22 · 11 · 71. We quickly notice that 4 · 11 = 44 is less than 71 so if n71 ≡ 1 (mod 71) and n71 | 44, then immediately we deduce that n71 = 1. Hence, every group of order 3124 has a normal subgroup of order 71, so there exists no simple group of order 3124. Exercise: 14 Section 8.5 Question: Prove that there is no simple group of order 4312. Solution: Let G be a group of order 4312. The prime factorization of 4312 is 4312 = 23 · 72 · 11. By Sylow’s Theorem, n11 ≡ 1 (mod 11) and also n11 | 23 · 72 = 392. The divisors of 392 are 1, 2, 4, 8, 7, 14, 28, 56, 49, 98, 196, and 392. The only divisors congruent to 1 modulo 11 are 1 and 56. These are possible values of n11 . By Sylow’s Theorem, n7 ≡ 1 (mod 7) and n7 | 88. Hence, n7 can be 1 or 8 or 22. We prove that G is not a simple group by contradiction. Assume that G is simple. Then we must have n11 = 56. Then for a Sylow 11-subgroup P of G, by Sylow’s Theorem |G : NG (P )| = 56 so |NG (P )| = 77. By Example 8.5.10, since 7 - 10, groups of order 77 are cyclic. Let H be a Sylow 7-subgroup of NG (P ), which has order 7. Again, by Sylow’s Theorem, there is a Sylow 7-subgroup Q of G that contains H. Since Q has order 49, a square of a prime, Q is abelian so Q centralizes H. Furthermore, since NG (P ) is abelian, NG (P ) also centralizes H. Consider the normalizer NG (H) of H in G. It contains both Q and NG (P ) so 77 and 49 divide |NG (H)|, so |NG (H)| is divisible by 72 · 11. Then the index |G : NG (H)| divides 8. If |G : NG (H)| = 1, then NG (H) = G, which implies that H E G contradicting the assumption that G is simple. If |G : NG (H)| = 2, 4, or 8, consider the action of G on the set of left cosets of NG (H). The induced permutation representation is a homomorphism ρ : G → S8 . However, this homomorphism cannot be injective because G contains elements of order 11, whereas S8 does not. Hence, every element of order 11 in G maps to the identity in S8 . This establishes that Ker ρ 6= {1}. We also know that Ker ρ 6= G since the described action is transitive. Consequently, Ker ρ is a nontrivial, proper normal subgroup of G. This again contradicts the assumption that G is simple. Exercise: 15 Section 8.5 Question: Prove that there is no simple group of order 132. Solution: The prime factorization of 132 is 132 = 22 · 3 · 11. By Sylow’s Theorem, the requirement that n11 ≡ 1 (mod 11) and n11 | 12 leads to the possibilities of n11 = 1 or 12. The requirement that n3 ≡ 1 (mod 3) and n3 | 44 leads to the possibilities of n3 = 1, 4 or 22. Assume that G is simple. Then n11 = 12 and n3 = 4 or 22. Since Sylow 11-subgroups of G have order 11, they would be cyclic and each contains 10 nonidentity elements, none of which occurs in another Sylow 11-subgroup (since in Zp with p prime, every nontidentity element is a generator). Hence G contains 12 · 10 = 120 elements of order 11. If n3 = 4, by the same reasoning, G contains 8 elements of order 3. So far, this accounts for 128 elements of order 11 or 3. (If n3 = 22, then we would already have more elements of order 3 and 11 than there are elements in the group. Hence, we cannot have both n1 1 = 12 and n3 = 22.) A Sylow 2-subgroup of G has 4 elements, 1 of which is the identity. Hence, this subgroup accounts for the exact number of remaining elements of

8.5. SYLOW’S THEOREM

447

G. Consequently, there is a unique Sylow 2-subgroup, which means that it is normal a normal subgroup. Hence, we have prove that n11 = 1, n3 = 1, or n2 =. Thus G is not simple. Exercise: 16 Section 8.5 Question: Prove that there is no simple group of order 351. Solution: The prime factorization of 351 is 351 = 33 · 13. Let G be a group of order 351. By Sylow’s Theorem, n13 ≡ 1 (mod 13) and n13 | 27 so n13 = 1 or 27. Furthermore, n3 ≡ 1 (mod 3) and n3 | 13 so n3 = 1 or 13. (The trick from the previous exercise does not easily work in this case because there are many possible structures for the Sylow 3-subgroup, and they may overlap.) Assume that G is simple. Then n13 = 27 and n3 = 13. Let P be a Sylow 13-subgroup and let Q be a Sylow 3-subgroup. Since P ∼ = Z13 , then any nonidentity element of P generates all of P . Hence, none of the distinct Sylow 13-subgroups gP g −1 intersect except at the origin so they contribut 27 · 12 = 324 distinct elements of order 13. Notice that 351 − 324 = 27, so all elements in G that are not order 13 must constitute exactly one Sylow 3-subgroup. Thus n3 = 1. Hence, we deduce that n13 = 1 or n3 = 1 and so G cannot be simple. Exercise: 17 Section 8.5 Question: Prove that a group of order 273 has a normal subgroup of order 91. Solution: Let G be a group of order 273. Note that 273 = 3 · 7 · 13. By Sylow’s Theorem, n13 ≡ 1 (mod 13) and n13 | 21 so n13 = 1. Furthermore, n7 ≡ 1 (mod 7) and n7 | 39. The divisors of 39 are 1, 3, 13, and 39. We easily see that n7 = 1. Thus, the Sylow 13-subgroup P is normal in G as is the unique Sylow 7-subgroup Q. Since, they are both normal subgroups, the set P Q is also a subgroup. Since |G : P Q| = 3 and 3 is the small prime factor of |G|, then by Exercise 8.4.9, P Q E G. Also |P Q| = |P ||Q|/|P ∩ Q| = 91. Exercise: 18 Section 8.5 Question: Prove that if |G| = 2015, then G contains a normal subgroup of order 31 and subgroup of order 13 in Z(G). Solution: Let G be a group of order 2015. Note that the prime factorization of 2015 is 2015 = 5 · 13 · 31. By Sylow’s Theorem n31 ≡ 1 (mod 31) and n31 | 5 · 13 = 65. It is easy to see that the only option is n31 = 1. Hence, G contains a normal subgroup of order 31. By Sylow’s Theorem n13 ≡ 1 (mod 13) and n13 | 5 · 31 = 155. The divisors of 155 are 1, 5, 31, and 155. The only divisor congruent to 1 modulo 13 is 1. Hence, n13 = 1. Let P be the unique (and hence normal) Sylow 13-subgroup of G. Consider the action of G on P by conjugation; it induces a homomorphism Ψ : G → Aut(P ) via Ψ(g) = ψg , where ψg (x) = gxg −1 . However, Aut(P ) ∼ = U (13), which is a cyclic group of order 12. Since gcd(|G|, | Aut(P )|) = gcd(2015, 12) = 1, then the only homomorphism Ψ is trivial, namely Ψ(g) = idP . Hence ψg (x) = gxg −1 = x for all g ∈ G and all x ∈ P . Thus, P ≤ Z(G). Exercise: 19 Section 8.5 Question: Prove that if |G| = 459, then G contains a Sylow 17-subgroup in Z(G). Solution: Let G be a group of order 459. The prime factorization of 459 is 459 = 33 · 17. By Sylow’s Theorem, n17 ≡ 1 (mod 17) and n17 |27. It is clear that n17 = 1. Let P be the unique and thus normal Sylow 17-subgroup of G. Consider the action of G on P by conjugation; it induces a homomorphism Ψ : G → Aut(P ) via Ψ(g) = ψg , where ψg (x) = gxg −1 . However, Aut(P ) ∼ = U (17), which is a cyclic group of order 16. Since gcd(|G|, | Aut(P )|) = gcd(459, 16) = 1, then the only homomorphism Ψ is trivial, namely Ψ(g) = idP . Hence, ψg (x) = gxg −1 = x for all g ∈ G and all x ∈ P . Thus, P ≤ Z(G). Exercise: 20 Section 8.5 Question: Prove that every group of order 1001 is abelian. Solution: Theorem,

Let G be a group of order 1001. The prime factorization of 1001 is 1001 = 7 · 11 · 13. By Sylow’s

• n13 ≡ 1 (mod 13) and n13 | 77, hence n13 = 1; • n11 ≡ 1 (mod 11) and n11 | 91, hence n11 = 1; and • n7 ≡ 1 (mod 7) and n7 | 143, hence n7 = 1.

448

CHAPTER 8. GROUP ACTIONS

This shows that all the Sylow subgroups of 1001 are normal. Call these P13 , P11 and P7 respectively. By the direct sum decomposition, we observe that G ∼ = P13 ⊕ P11 ⊕ P7 . Since Pi ∼ = Zi for i = 13, 11, 7, then ∼ G∼ Z ⊕ Z ⊕ Z Z . In particular, G is abelian. = 13 11 7 = 1001 Exercise: 21 Section 8.5 Question: Prove if |G| = 9163, then G has a Sylow 11-subgroup in Z(G). Solution: Note that 9163 = 72 · 11 · 17. By Sylow’s theorem, n7 ∈ {1, 11, 17, 187}, n11 ∈ {1, 7, 17, 49, 119}, and n17 ∈ {1, 7, 11, 49, 77, 539}. And by congruence to 1 mod p we get that n7 = n11 = n17 = 1. Then by Proposition 8.4.9, we have normal subgroups of order 49, 7, and 11. Now consider the mapping of Ψ : G → Aut(N ) by g → ψg where ψg is the automorphism defined by conjugating each element by g. This is a homomorphism by Exercise 3.7.38. So |Ψ(G)| divides | Aut(N )| by Lagrange’s theorem and also divides |G| = 72 · 11 · 17 since |Ψ(G)| = |G|/| Ker Ψ|. Since | Aut(N )| = 10, we have that |Ψ(G)| = 1 and every element of G acts trivially on N . Equivalently, ∀g ∈ G and ∀n ∈ N , gng −1 = n. This shows that N ∈ Z(G). Exercise: 22 Section 8.5 Question: How many elements of order 7 must exist in a simple group of order 168? Solution: Let G be a simple gorup of order 168. The prime factorization of 168 is 168 = 23 · 3 · 7. By Sylow’s Theorem, n7 ≡ 1 (mod 7) and n7 | 23 ·3 = 24. The divisors of 24 are 1, 2, 4, 8, 3, 6, 12, and 24. The only divisors that are congruent to 1 modulo 7 are 1 and 8. Assuming G is simple, then n7 = 8. Each Sylow 7-subgroup is isomorphic to Z7 , which contains 6 elements of order 7. Furthermore, each nonidentity element of a cyclic of prime order generates the whole group. Hence, none of the Sylow 7-subgroups intersect except in the identity. Thus, these 8 Sylow 7-subgroups account for exactly 8 · 6 = 48 elements of order 7. Exercise: 23 Section 8.5 Question: Prove that np (G) = 1 is equivalent to the property that all subgroups of G generated by elements of order p-power are p-subgroups. Solution: Let G be a group and let p be a prime number that divides the order of G. Suppose that np (G) = 1. This is equivalent to the Sylow p-subgroup Q being unique and normal. Let H be a subgroup generated by elements x1 , x2 , . . . , xn of orders pki for any power k1 , k2 , . . . , kn ∈ N∗ . Since hxi i are all subgroups of order pk , they are p-subgroups of G. By Sylow’s Theorem, they are in a Sylow p-subgroup, and hence in Q. Thus, H ≤ Q. By Lagrange’s Theorem, H is a p-subgroup of G. We now prove the converse, so we assume that the group G has the property every subgroup generated by elements of order p-power is a p-subgroup. Consider the subset S = {x ∈ G | |x| = pk with k ∈ N} and let H = hSi. By Lagrange’s, every element of a p-subgroup of G is in S so every p-subgroup is contained in the subgroup S. By the hypothesis, H is a p-subgroup. Since every Sylow p-subgroup is contained in H, then H is the unique Sylow p-subgroup and np (G) = 1. Exercise: 24 Section 8.5 Question: Let p be an odd prime. Show that every group of order 2p is isomorphic to Z2p or to Dp . Solution: Let G be a group of order 2p and let P be the Sylow p-subgroup of G. Since |G : P | = 2, then P E G. Call t a generator of the cyclic subgroup P . Since P is normal in G, then the action of G on P by conjugation defines a homomorphism Ψ : G → Aut(P ) ∼ = U (p) by Ψ(g) = ψg , where ψg (x) = gxg −1 for all x ∈ P . By Proposition 7.5.2, U (p) is cyclic so the automorphisms of P are of the form ϕa defined uniquely by ϕa (t) = ta , where a ∈ U (p). By Cauchy’s Theorem, G contains an element of order 2, call this s. Since hsi ∩ P = {1} by Lagrange’s Theorem, then hsiP = hs, ti has order 2p, so hs, ti = G. By homomorphism properties, the order of Ψ(s) divides |s| = 2. If Ψ(s) has order 1, then Ψ(s) = ϕ1 , which is the identity function on P , which tells us that sts−1 = t or in other words st = ts. This shows that hsi E G so by the Direct Sum Decomposition Theorem, G∼ = Z2p , since gcd(2, p) = 1. On the other hand, if |Ψ(s)| = 2, then Ψ(s) = ϕp−1 , since = hsi ⊕ P ∼ = Z2 ⊕ Zp ∼ p − 1 ≡ −1 (mod p) is the only element in the cyclic group of order 2. Thus, sts = t−1 so st = t−1 s. This shows that G∼ = hs, t | s2 = tp = 1, st = t−1 si = D2p . Thus G is isomorphic to either Z2p or D2p .

8.5. SYLOW’S THEOREM

449

Exercise: 25 Section 8.5 Question: Suppose that |G| = pm where p is a prime and p - m. Prove that gcd(p, m − 1) = gcd(p − 1, m) = 1 if and only if G has a normal Sylow p-subgroup in Z(G). Solution: This exercise question is not true. Let G be a group of order 3445 = 5·13·53. Though gcd(12, 5·53) = gcd(13, 264) = 1 the number of Sylow 13-subgroups n13 is not constrained to be 1 but can be 1 or 53. By Sylow’s Theorem, it is not hard to show that n5 = 1 and n53 = 1. Thus, G contains a normal Sylow 5-subgroup Q1 and a normal Sylow 53-subgroup, Q2 . Call H = Q1 Q2 , which is a subgroup of G since Q1 (and Q2 ) is normal. By Example 8.5.10, since 5 - 52, the normal subgroup H is a cyclic group with H ∼ = Z265 . Let x be a generator of H. Let y be an element of order 13 in G. The action of conjugation of G on H induces a homomorphism Ψ : G → Aut(H) with Ψ(g) = ψg , where ψg (x) = gxg −1 for x ∈ H. We know that Aut(H) ∼ = U (265). But U (265) is an abelian group of order φ(265) = 4 · 52 = 24 · 13. In particular, by Cauchy’s Theorem, U (265) contains an element of order 13. It takes a little checking but 16 is an element of order 13 in U (265). We can then define a group of order 3445 by hx, y | x265 = y 13 = 1, yxy −1 = x16 i. This group has order 3445 but the Sylow 13-subgroup hyi is not normal, and in particular y ∈ / Z(G) since y does not commute with x. Exercise: 26 Section 8.5 Question: Suppose that for every prime p dividing |G|, the Sylow p-subgroups are nonabelian. Prove that |G| is divisible by a cube. Solution: By Exercise 8.4.15, every group of order p2 is abelian. Hence, if all the Sylow p-subgroups (for all primes p dividing G) are nonabelian, then the order of every Sylow p-subgroup must be at least p3 . Hence, for every prime p dividing |G|, the power p3 must also divide |G|. Thus, |G| is divisible by the produce of cubes of primes dividing |G|. Exercise: 27 Section 8.5 Question: Suppose that |G| = p2 q 2 with p and q distinct primes. Prove that if p - (q 2 − 1) and q - (p2 − 1), then G is abelian. Solution: By Sylow’s theorem, we have that np ∈ {1, q, q 2 } and nq ∈ {1, p, p2 }. By observing that p - (q 2 − 1) we have that q 2 6≡ 1 mod p. Since p - q 2 − 1 we also have that p - q − 1 so that q 6≡ 1 mod p. Then np = 1. The reasoning above is symmetric for q so that nq = 1. Now let A be the normal Sylow p-subgroup and B the normal Sylow q-subgroup. Consider the product AB which has order |A||B|/|A ∩ B| = p2 q 2 /1 = p2 q 2 . So AB = G. Each element then looks like a1 b1 for some a1 ∈ A and b1 ∈ B. Then suppose that a1 b1 = b2 a2 . Then we have, a1 b1 a−1 2 = b2 −1 a1 b1 (a−1 1 a1 )a2 = b2 −1 (a1 b1 a−1 1 )a1 a2 = b2

b3 (a1 a−1 2 ) = b2 −1 a1 a−1 2 = b3 b2 .

Then this element exists in the intersection of the two Sylow subgroups and so must be the identity element. Then a1 = a2 . Now we can do the same procedure and show b1 = b2 by a1 b1 = b2 a2 b−1 2 a1 b1 = a2 −1 b2 b1 b−1 1 a1 b1 = a2 b−1 2 b1 a3 = a2 −1 b−1 2 b1 = a2 a3 .

Then b−1 2 b1 = 1 so that b2 = b1 . This shows that for all elements a ∈ A, b ∈ B, we have ab = ba. Since these elements serve as generators for the group and they are abelian, G is abelian.

450

CHAPTER 8. GROUP ACTIONS

Exercise: 28 Section 8.5 Question: Suppose that H is a subgroup of G such that gcd(| Aut(H)|, |G|) = 1. Prove that NG (H) = CG (H). Solution: Let H be a subgroup of G and consider the normalizer subgroup NG (H). Then NG (H) acts on H by conjugation. This action defines a homomorphism Ψ : NG (H) → Aut(H) via Ψ(g) = ψg , where ψg is the automorphism with ψg (h) = ghg −1 for all h ∈ H. However, since |NG (H)| divides |G| and gcd(| Aut(H)|, |G|) = 1, then we also know that gcd(| Aut(H)|, |NG (H)|) = 1. Hence, the homomorphism Ψ is trivial, so ψg is the identity automorphism on H for all g ∈ NG (H). In other words, ghg −1 = h for all g ∈ NG (H) and all h ∈ H. Thus NG (H) = CG (H). Exercise: 29 Section 8.5 Question: Prove that if N E G, then np (G/N ) ≤ np (G). Solution: Let π : G → G/N be the canonical projection. By the Fourth Isomorphism Theorem, there is a bijection between {H ∈ Sub(G) | N ≤ H} and Sub(G/N ) via H 7→ π(H) and H̄ = H/N 7→ π −1 (H̄). We note that |H| = |H̄| |N |. Suppose that ordp (|G|) = α and ordp (|N |) = β. Then Sylow p-subgroups of G have order pα and Sylow p-subgroups of G/N have order pα−β . Consider the set Sylp (G/N ) of Sylow p-subgroups of G/N . We have Sylp (G/N ) = {H̄ ≤ G/N | |H̄| = pα−β }. By the Fourth Isomorphism Theorem, Sylp (G/N ) is in bijection with {H ≤ G | N ≤ H, and |H| = pα }. Because of the condition N ≤ H, this set is a subset of Sylp (G) so np (G/N ) ≤ np (G). Exercise: 30 Section 8.5 Question: Let P be a normal Sylow p-subgroup of a group G and let H ≤ G. Prove that P ∩ H is the unique Sylow p-subgroup of H. Solution: By Exercise 4.2.14, P ∩ H E H. Furthermore, since P E G, the set P H is a subgroup of G. Furthermore, |P | divides |P H| so ordp (|P H|) = ordp (|P |) = ordp (|G|). By Proposition 4.1.16, |P H| =

|P ||H| |P ∩ H|

and by the logarithmic properties of the ordp : Q>0 → Z function (see Section 2.1.5), we have ordp (|P H|) = ordp (|P |) + ordp (|H|) − ordp (|P ∩ H|) ⇐⇒ ordp (|H|) = ordp (|P ∩ H|). This is equivalent to saying that P ∩ H is a Sylow p subgroup of H. However, since P ∩ H E H, this Sylow p-subgroup is unique. Exercise: 31 Section 8.5 Question: Let P ∈ Sylp (G) and let N E G. Prove that P ∩ N ∈ Sylp (N ). Prove also that P N/N is a Sylow p-subgroup of G/N . Solution: Since N E G, the set P N is a subgroup of G. Furthermore, |P | divides |P N | so ordp (|P N |) = ordp (|P |) = ordp (|G|). By Proposition 4.1.16, |P N | =

|P ||N | |P ∩ N |

and by the logarithmic properties of the ordp : Q>0 → Z function (see Section 2.1.5), we have ordp (|P N |) = ordp (|P |) + ordp (|N |) − ordp (|P ∩ N |) ⇐⇒ ordp (|N |) = ordp (|P ∩ N |). This is equivalent to saying that P ∩ N is a Sylow p subgroup of N . Thus P ∩ N ∈ Sylp (N ). Since P ∩ N ∈ Sylp (N ), we have ordp (|P ∩ N |) = ordp (|N |). Then applying ordp to the relation |P N |/|N | = |P |/|P ∩ N |, we get |P N | |P | = ordp = ordp (|P |)−ordp (|P ∩N |) = ordp (|G|)−ordp (|N |) = ordp (|G/N |). ordp (|P N/N |) = ordp |N | |P ∩ N | Hence, P N/N is a Sylow p-subgroup of G/N .

8.6. A BRIEF INTRODUCTION TO REPRESENTATIONS OF GROUPS

451

Exercise: 32 Section 8.5 Question: Let G1 and G2 be two groups, both of which have orders divisible by a prime p. Prove that all Sylow p-subgroups of G1 ⊕ G2 are of the form P1 ⊕ P2 , where P1 ∈ Sylp (G1 ) and P2 ∈ Sylp (G2 ). Solution: Let P ∈ Sylp (G1 ⊕ G2 ). Call G̃1 and G̃2 the subgroups of G1 ⊕ G2 defined by G̃1 = {(g, 1) ∈ G1 ⊕ G2 | g ∈ G1 } and similarly for G̃2 . Obviously, G̃1 ∼ = G1 and G̃2 ∼ = G2 and also G̃i E G1 ⊕ G2 . By Exercise 8.5.31, P ∩ G̃1 ∈ Sylp (G̃1 ) and P ∩ G̃2 ∈ Sylp (G̃2 ). Call P1 = P ∩ G̃1 and P2 = P ∩ G̃2 . By construction, every element in G̃1 commutes with every element in G̃2 . Thus P1 P2 = P2 P1 , so P1 P2 ≤ G. Since G̃1 ∩ G̃2 = {1}, we also deduce that P1 ∩ P2 = {1}. Thus |P1 P2 | =

|P1 | |P2 | = |P1 | |P2 | ¶ 1 ∩ P2 |

Suppose that ordp (|G1 |) = α1 and ordp (|G2 |) = α2 , then ordp (|G1 ⊕ G2 |) = α1 + α2 . But |P1 P2 | = pα1 pα2 = pα1 +α2 so P1 P2 is a Sylow p-subgroup of G contained in P , so P = P1 P2 . Now as a subset of P , we know that Pi = P ∩ G̃i E P since G̃i E G1 ⊕ G2 . By the Direct Sum Decomposition Theorem, we deduce that P ∼ = P1 ⊕ P2 , where P1 ∈ Sylp (G1 ) and P2 ∈ Sylp (G2 ). Exercise: 33 Section 8.5 Question: Let G be a finite group and let M be a subgroup such that NG (P ) ≤ M ≤ G for some Sylow p-subgroup P . Prove that |G : M | ≡ 1 (mod p). Solution: Suppose that M is a subgroup such that NG (P ) ≤ M ≤ G for some Sylow p-subgroup P . Since P ≤ NG (P ) we also have P ≤ M . Since M ≤ G, by Sylow’s Theorem, ordp (|M |) ≤ ordp (|G|) but since P ≤ M , we deduce that ordp (|M |) = ordp (|G|) and that P ∈ Sylp (M ). This leads to a chain of subgroups P ≤ NM (P ) ≤ NG (P ) ≤ M ≤ G. However, since all the elements in G that normalize P are in M , then NM (P ) = NG (P ). We then have np (G) = |G : NG (P )| = |G : M | |M : NG (P )| = |G : M | |M : NM (P )| = |G : M |np (M ). By Sylow’s Theorem, np (G) ≡ np (M ) ≡ 1 (mod p) so by elementary modular arithmetic, we deduce that |G : M | ≡ 1 (mod p). Exercise: 34 Section 8.5 Question: Let p be a prime dividing |G|. Prove that the intersection of all Sylow p-subgroups is the largest normal p-subgroup in G. Solution: We will denote the intersection of all Sylow p-subgroups as I = ∩P ∈Sylp (G) P . First we show that every normal p-subgroup in G is contained in I. Let N be a normal p-subgroup. Since N is a p-subgroup, by Theorem 8.4.6 part 1, N ≤ Q for some Sylow p-subgroup Q. Then for any g ∈ G we have that gN g −1 = N ≤ gQg −1 . Since each Sylow p-subgroup is conjugate, ∀P ∈ Sylp (G), N ≤ P . Then by the definition of intersection, N ≤ I. Now we show that I is itself a normal p-subgroup and so must be the largest. I is a p-subgroup since it is the intersection of other p-subgroups. Now, I ≤ P for all P ∈ Sylp (G) by the definition of intersection. Then for any g ∈ G, gIg −1 ≤ gP g −1 for all P ∈ Sylp (G). Since g acts transitively on the set of all Sylow p-subgroups, we can rewrite this as gIg −1 ≤ P for all P ∈ Sylp (G) and all g ∈ G which implies that gIg −1 ≤ I for all g ∈ G. This shows that I is a normal p-subgroup and therefore must be the largest.

8.6 – A Brief Introduction to Representations of Groups Exercise: 1 Section 8.6 Question: Consider the cyclic group Z6 = hz | z 6 = 1i. Prove that mapping 0 −1 z 7→ 1 1 defines a representation of Z6 on R2 .

452

CHAPTER 8. GROUP ACTIONS

Solution: Let us calculate the order of the matrix

0 1

−1 1

−1 1

1 −1

−1 0 1 0

A= in GL2 (R). We have 0 −1 1 1 0 1 A4 = −1 −1 A=

A2 = A5 =

A3 = A6 =

−1 0

1 0

0 −1 0 . 1

Thus, A6 = 1. By the Extension Theorem on Generators, the exists a unique homomorphism ϕ : Z6 → GL2 (R) with ϕ(z) = A. Consequently, this is a representation of Z6 on R2 . [Though the exercise did not ask to prove it, this representation is faithful since Ker ϕ = {1}.] Exercise: 2 Section 8.6 Question: Consider the dihedral group D4 . Prove that mapping i 0 1 r 7→ and s 7→ 0 −i 0

0 −1

does not define a representation on D4 in C2 .

0 1 Solution: It is not too hard to see that the order of in GL2 (C) is 4, and the order of −i 0 However, i 0 1 0 i 0 = 0 −i 0 −1 0 i while

1 0

0 −1

i 0

−1 0 1 = −i 0

i 0

0 −i 0 −i = −1 0 i 0

0 −1

is 2.

0 . −i

Since the matrices do not satisfy the relation corresponding to rs = sr−1 , the mapping cannot extend to a homomorphism from D4 to GL −2(C). Hence, these matrices do not define a representation of D4 on C2 . Exercise: 3 Section 8.6 Question: Find four nonisomorphic one-dimensional representations of D4 . Solution: The following one-dimensional representations are valid over any field F . Note that for any field F , the general linear group GL1 (F ) is the multiplicative group F × . Homomorphisms with D4 as domain are uniquely determined by how they map the generators r and s of D4 . Consider the following 4 options r s ϕ1 1 1 ϕ2 1 −1 ϕ3 −1 1 ϕ4 −1 −1 It is easy to see that for i = 1, 2, 3, 4, the identities ϕi (r)4 = 1, ϕi (s)2 = 1, and ϕi (r)ϕi (s) = ϕi (s)ϕi (r)−1 . Hence, by the Extension Theorem on Generators, all four of these functions on the sets of generators extends to homomorphisms ϕi : D4 → GL1 (R). To show that the four induced homomorphisms ϕi : D4 → GL1 (F ), we point out that a vector space isomorphism T : F → F is a function T (x) = ax, where a 6= 0. By Definition 8.6.13, T is an isomorphism of representations between the one for ϕi and the one for ϕj when, for all g ∈ D4 T (ϕi (g)v) = ϕj (g)T (v) for all v ∈ F . This is equivalent to aϕi (g)v = ϕj (g)av ⇐⇒ (ϕi (g) − ϕj (g))v = 0

8.6. A BRIEF INTRODUCTION TO REPRESENTATIONS OF GROUPS

453

Since this must hold for v 6= 0, then we deduce that ϕi and ϕj induce isomorphic representations if and only if ϕi = ϕj . Hence, all four homomorphisms induce nonisomorphic one-dimensional representations of D4 . Exercise: 4 Section 8.6 Question: Prove that the standard representation of Dn in R2 is an irreducible representation over the field R. Solution: The standard representation of Dn in R2 is described at the beginning of the section. We denote this representation by V and, as in the section, call ϕ : Dn → GL2 (R) the corresponding homomorphism. Assume that V is the direct sum of two one-dimensional subrepresentations of Dn . Call these subrepresentations V1 and V1 . Thus V1 = Span(~v1 ) and V2 = Span(~v2 ) and for all g ∈ Dn , we have ϕ(g)~v1 = λ1 (g)~v1 for some nonzero real λ1 (g) and ϕ(g)~v2 = λ2 (g)~v2 for some nonzero real λ2 (g). In particular, ϕ(g) must have two real eigenvalues for all g ∈ Dn . However, 2π − sin cos 2π n n ϕ(r) = cos 2π sin 2π n n has the characteristic equation 2

λ − 2 cos

2π n

λ + 1 = 0 ⇐⇒ λ = cos

2π n

s

±

cos2

2π n

− 1 = cos

2π n

± sin

2π n

.

With n > 2, these eigenvalues are complex. This leads to a contradiction. Hence, the standard representation V is irreducible. Exercise: 5 Section 8.6 Question: Prove that a function ϕ : Q8 → GL4 (R) such that    0 0 0 −1 0 0  0 0 1 0 0 0  and ϕ(j) =  ϕ(i) =  1 0 0 0 0 1 0 1 0 0 −1 0

−1 0 0 0

 0 −1  0 0

extends to homomorphism and thereby defines a representation of Q8 on R4 . Solution: Recall that a presentation of Q8 is hi, j | i4 = j 4 = 1, i2 = j 2 , ij = j 3 ii. We need to verify that the image matrices satisfy the relations corresponding to i and j. Call     0 −1 0 0 0 0 −1 0 1 0   0 0  and B = 0 0 0 −1 . A= 0 0 1 0 0 0 1 0 0 0 −1 0 0 1 0 0 We can easily check by direct calculation that A2 = B 2 = −I. So furthermore, we also get A4 = B 4 = I. We can also check that     0 0 0 1 0 0 0 1  0 0 −1 0  0 0 −1 0 3    AB =   0 1 0 0 and B A =  0 1 0 0 . −1 0 0 0 −1 0 0 0 Since the images A and B satisfy the relations corresponding to i and j, by the Extension Theorem on Generators, the function on the set of generators extends to a homomorphism ϕ : Q8 → GL4 (R) and thus a representation of Q8 on the vector space R4 . Exercise: 6 Section 8.6 Question: Prove that a function ϕ : Q8 → GL2 (C) such that i 0 0 ϕ(i) = and ϕ(j) = 0 −i 1

−1 0

454

CHAPTER 8. GROUP ACTIONS

extends to homomorphism and thereby defines a representation of Q8 on C2 . Solution: Recall that a presentation of Q8 is hi, j | i4 = j 4 = 1, i2 = j 2 , ij = j 3 ii. We need to verify that the image matrices satisfy the relations corresponding to i and j. Call i 0 0 −1 A= and B = . 0 −i 1 0 We can easily check by direct calculation that A2 = B 2 = −I. So furthermore, we also get A4 = B 4 = I. We also check that 0 −i 0 −i 3 AB = and B A = . −i 0 −i 0 Since the images A and B satisfy the relations corresponding to i and j, by the Extension Theorem on Generators, the function on the set of generators extends to a homomorphism ϕ : Q8 → GL2 (C) and thus a representation of Q8 on the vector space C2 . Exercise: 7 Section 8.6 Question: Let G = Z3 = hz | z 3 = 1i. Show that the mapping 7 −3 ρ(z) = 19 −8 defines a representation of Z3 of degree 2 over C. Also find a nontrivial subrepresentation. Solution: Consider the matrix 7 −3 A= . 19 −8 We easily calculate A2 =

−8 −19

3 , 7

A3 =

1 0

0 . 1

Thus, A has order 3 in GL2 (R). By the Extension Theorem on Generators, we see that ρ extends to a homomorphism Z3 → GL2 (C). Hence, ρ defines a representation of Z3 on C2 . We look for the eigenspaces of A. The eigenvalues come from det(λI − A) = 0 ⇐⇒ (λ − 7)(λ + 8) + 57 = 0 ⇐⇒ λ2 + λ + 1 = 0 ⇐⇒ λ = e2πi/3 or e−2πi/3 . To find an eigenvector for λ = e2πi/3 we solve the vector equation 3 7 − e2πi/3 −3 v1 v1 ~ = 0 ⇐⇒ =t for t ∈ C. v2 v2 7 − e2πi/3 19 −8 − e2πi/3 So if we call W , the subspace W = Span(

3 ), 7 − e2πi/3

then ρ(r)w ~ = e2πi/3 w ~ for all w ~ ∈ W . This is a one-dimensional subrepresentation. Exercise: 8 Section 8.6 Question: Prove that the determinant of the standard representation of Sn corresponds to the sign homomorphism. Solution: From Example 8.6.7, the action of a permutation σ ∈ Sn on vector ~v ∈ Rn , corresponds to permuting the entries by σ (so that the first coordinate goes to the σ(1) coordinate, etc.). For each permutation, this action leads to an invertible matrix σ · ~v = ϕ(σ)~v = Mσ ~v ,

8.6. A BRIEF INTRODUCTION TO REPRESENTATIONS OF GROUPS

455

where Mσ is the matrix whose entries are mij , where ( 1 if i = σ(j) mij = 0 otherwise. By Proposition 5.3.6, det M = (sign σ) det I = sign σ. Thus, the determinant of the standard representation of Sn corresponds to the homomorphism sign : Sn → U (F ). Exercise: 9 Section 8.6 Question: Let G be the group G = hx, y | x7 = y 3 = 1, yxy −1 = x2 i. Let F be the finite field F7 . Prove that the mapping 1 1 2 0 x 7→ and y 7→ 0 1 0 1 defines a degree-2 representation of G over the field F7 . Solution: Let A, B ∈ GL2 (F7 ) be A=

1 0

1 1

1 0

n 1

and B =

2 0

0 . 1

It is easy to tell that n

A =

n 2 and B = 0 n

0 . 1

In the field F7 , we have 7 = 0, which implies that |A| = 7 and 23 = 1, which implies that |B| = 3. We also have 2 0 1 1 4 0 BAB −1 = 0 1 0 1 0 1 2 2 4 0 = 0 1 0 1 1 2 = 0 1 = A2 . By the Extension Theorem on Generators, the function ϕ that maps x and y to A and B respectively extends to a homomorphism ϕ : G → GL2 (F7 ), i.e. a degree-2 representation of G over the field F7 . Exercise: 10 Section 8.6 Question: Let V be the n-dimensional standard representation of G = Sn over C. Show that W = {~x ∈ V | x1 + x2 + · · · + xn = 0} is a subrepresentation of V . Solution: Let w ~ ∈ W . By definition, if w ~ = (w1 , w2 , . . . , wn ) with respect to a basis, then w1 +w2 +· · ·+wn = 0. Then σ·w ~ = (wσ−1 (1) , wσ−1 (2) , . . . , wσ−1 (n) ). However, as the index i runs through 1, 2, . . . , n, the permuted indices σ −1 (i) also runs through the set {1, 2, . . . , n}. Thus wσ−1 (1) + wσ−1 (2) + · · · + wσ−1 (n) = w1 + w2 + · · · + wn = 0 so σ · w ~ ∈ W as well. The homomorphism properties so hold so restricting the action to W , defines a homomorphism §n → GL(W ). Exercise: 11 Section 8.6 Question: Let G = GLn (R) and let V = Mn×n (R) be the vector space of n × n matrices of real coefficients. a) Prove that the action of G on V defined by g · A = gAg −1 is a representation ρ of GLn (R).

456

CHAPTER 8. GROUP ACTIONS

b) Identifying M2×2 (R) with R4 via the standard basis on Mn×n (R), express a b ρ( ) c d for all invertible matrices in GL2 (R). Solution: Let G = GLn (R) and let V = Mn×n (R) be the vector space of n × n matrices of real coefficients. a) The space V is a real vector space of dimension n × n. For any g ∈ GLn (R), the transformation ρ(g) as the function A 7→ gAg −1 satisfies g(A + B)g −1 = g(Ag −1 + Bg −1 ) = gAg −1 + gBg −1 g(cA)g −1 = c(gAg −1 ) Thus, it is a linear transformation V → V . However, it is also a bijection with A 7→ g −1 Ag as its inverse. Hence, ρ(g) is an isomorphism on V . Thus, for all g ∈ G, we have ρ(g) ∈ GL(V ). In addition, for all A ∈ V , we have ρ(g1 g2 )(A) = (g1 g2 )A(g1 g2 )−1 = g1 (g2 Ag2−1 )g1−1 = (ρ(g1 ) ◦ ρ(g2 ))(A). Thus, ρ defines a homomorphism ρ : GLn (R) → GL(V ), i.e. V carries the structure of a representation of GLn (R). b) The standard basis on M2×2 (R) is (in this order) 1 0 0 1 0 0 0 0 E11 = , E12 = , E21 = , E22 = . 0 0 0 0 1 0 0 1 With g =

a c

b , we have d 1 1 ad −ab −ac a2 gE12 g −1 = ad − bc cd −cb ad − bc −c2 ac 1 1 bd −b2 −bc ab −1 gE21 g −1 = . gE g = 22 ad − bc d2 −bd ad − bc −cd ad gE11 g −1 =

Thus, with respect to this standard basis of M2×2 (R), the matrix of ρ(g) is   ad −ac bd −bc −ab a2 −b2 ab  1 .  ad − bc  cd −c2 d2 −cd −bc ac −bd ad

Exercise: 12 Section 8.6 Question: Let V be a representation of a group G. Prove that the subset V G = {v ∈ V | gv = v for all g ∈ G} is a subrepresentation of V . Solution: Let w ∈ V G and let g ∈ G. Then by definition gw = w ∈ V G . Thus, the set of fixed elements in the representation V is a subrepresentation. (It is a representation in its own right, i.e., a vector subspace that carries the structure of a G-representation.) Exercise: 13 Section 8.6 Question: Prove Proposition 8.6.15. Solution: Let V and W be two representations of a group G and let ϕ : V → W by a representation homomorphism.

8.6. A BRIEF INTRODUCTION TO REPRESENTATIONS OF GROUPS

457

a) Consider the set Ker ϕ in V . Let v ∈ Ker ϕ and let g ∈ G. Then ϕ(g · v) = g · ϕ(v) = g · 0 = 0. Hence, gv ∈ Ker ϕ. Thus, by the definition of subrepresentation, we see that Ker ϕ is a subrepresentation of V . b) Let w ∈ Im ϕ and let g ∈ G. Then there exists v ∈ V such that w = ϕ(v). But then g·w = g·ϕ(v) = ϕ(g·v). Thus, g · w ∈ Im ϕ. Thus Im ϕ is a subrepresentation of W . Exercise: 14 Section 8.6 Question: Let G be a finite group and let V be a finite-dimensional representation over C. Prove that g is diagonalizable and that all the eigenvalues of g are roots of unity. Solution: A finite dimensional representation V of G over C, induces by restriction a representation of the subgroup hgi on V , for every element g ∈ G. In order to prove the result, we can therefore assume that G = hgi is finite and cyclic. We also suppose that |g| = n. We prove this by induction on dim V . Suppose that V is 1-dimensional. A one-dimensional representation of G corresponds to a homomorphism φ : G → GL1 (C) = U (C). Such a homomorphism is determine by the image of the generator φ(g). But 1 = φ(1G ) = φ(g)n . Thus, the complex number φ(g) is an n-th root of unity (not necessarily primitive). Thus, the result holds for all representations of dimension 1. Now suppose that the result holds for all representations of G of dimension m. We prove the result holds for all representations of G of dimension m + 1. Let V be a representation of G with dim V = m + 1. Then φ(g) is an invertible linear transformation on V . Since C is an algebraically closed field, φ(g) has at least one eigenvalue λ ∈ C. The eigenspace Eλ , as a subspace of V , may have more than 1 dimension, but we only need to consider a 1-dimensional subspace W of Eλ . Then for all w ∈ W , we have gw = φ(g)(w) = λw. Furthermore, g n = 1G so w = g n w = λn w. Hence, assuming w is a nonzero vector, we deduce that λn = 1, i.e., that λ is an nth root of unity. By Maschke’s Theorem, there exists another subrepresentation of V such that V = W ⊕ U . Since W is 1-dimensional, the subspace U is m-dimensional and we can apply the induction hypothesis on U . By the induction hypothesis, there exists a basis {u1 , u2 , . . . , um } with respect to which φ(g) is diagonal with eigenvalues that are nth roots of unity. Then if w0 is a nonzero vector in W , then set {w0 , u1 , . . . , um } is a basis of V and with repsect to it φ(g) is diagonal with entries that are all nth roots of unity. By induction, for every element g ∈ G and every finite dimensional representation, the action of g on V , given by φ(g) is diagonalizable and has eigenvalues that are nth roots of unity, where n = |g|. Exercise: 15 Section 8.6 Question: Let V be a representation of G and let v ∈ V . Prove that Span(gv | g ∈ G) is a subrepresentation of V . Conclude that every representation V of a finite group G has a subrepresentation of a dimension less than or equal to |G|. Solution: We simply need to check that for all v ∈ V , the subspace W = Span(gv | g ∈ G) is closed under the action of G on it. Any element in w ∈ W is a finite sum of the form w=

n X

ci (gi v)

i=1

with ci in the base field of V and {g1 , g2 , . . . , gn } ⊆ G. Then for any g ∈ G, we have ! n n X X gw = g ci (gi v) = ci ((ggi )v), i=1

i=1

so gw ∈ W . Thus Span(gv | g ∈ G) is closed under the action of G. Hence, it is a subrepresentation of V . In particular, if G is finite, then for any fixed v ∈ V the set {gv | g ∈ G} is finite with at most |G| elements and so the span of that set is finite dimensional, of dimension less than or equal to |G|. Exercise: 16 Section 8.6 Question: Consider the abelian group G = (Z, +) and the function ρ : G → GL2 (C) given by 1 n ρ(n) = . 0 1

458

CHAPTER 8. GROUP ACTIONS

Show that ρ is a representation that is not completely reducible. (This gives a counterexample to Maschke’s Theorem when G is infinite.) Solution: Call V the two-dimensional representation described by the problem. We first note that W = Span( 10 ) is a subrepresentation. In fact, it is a trivial representation because ρ(n)w = w for all w ∈ W . Hence, the described representation is not irreducible. We prove the result by contradiction. Assume that V is completely reducible, then we could write V = V1 ⊕V2 , where V1 and V2 are subrepresentations of V . Because of our above comment, one of these cannot by all of V , so V1 and V2 must be one-dimensional subrepresentations. If V1 = Span(v1 ) and V2 = Span(v2 ), then for all n ∈ Z we would have ρ(n)v1 ∈ V1 and ρ(n)v2 ∈ V2 . In other words, V1 and V2 are eigenspaces for ρ(n) for all n ∈ Z and hence ρ(n) is diagonalizable. This is a contradiction because ρ(1) is not diagonalizable: it has an eigenvalue of 1 with algebraic multiplicity of 2 but geometric multiplicity of 1 (i.e., the dimension of the eigenspace of 1 is 1). We conclude that V is not the direct sum of two one-dimensional representations. Thus V is not completely reducible. Exercise: 17 Section 8.6 Question: Let ρ : G → GL(V ) be a representation of a group G. Prove that ρ gives a faithful representation of the group G/ Ker ρ. Solution: Let ρ : G → GL(V ) be a representation of a group G. We can define a representation ρ̄ : G/ Ker ρ → GL(V ) by ¯(ρ)(ḡ) = ρ(g), where ḡ = g Ker ρ in G/ Ker ρ. This is well-defined: The elements g1 and g2 represent the same coset of Ker ρ if and only if g2−1 g1 ∈ Ker ρ and furthermore idV = ρ(g2−1 g1 ) = ρ(g2 )−1 ρ(g1 ) =⇒ ρ(g2 ) = ρ(g1 ). We now calculate that Ker ρ̄ = {ḡ ∈ G/ Ker ρ | ρ̄(ḡ) = ρ(g) = idV } = Ker ρ = 1̄, which is the identity in G/ Ker ρ. Thus the homomorphism ρ̄ : G/ Ker ρ → GL(V ) is injective, which we restate to say that the representation is injective. Exercise: 18 Section 8.6 Question: (Schur’s Lemma) Let G be a finite group and let V and W be irreducible representations of G over the field C. a) Prove that a G-representation homomorphism ϕ : V → W is either trivial (maps to 0) or is an isomorphism. b) Prove all isomorphisms ϕ : V → V are of the form λidV . [Hint: Let λ be an eigenvalue of ϕ and consider the linear transformation ϕ − λidV .] Solution: [Note: the original text has a typo in that it missed the key word that V and W are irreducible representations of G.] a) For any G-representation homomorphism ϕ : V → W , by Proposition 8.6.15 Ker ϕ is a subrepresentation of V and that Im ϕ is a subrepresentation of W . Since V and W are irreducible, we deduce that Ker ϕ = V , in which case the homomorphism is trivial or Ker ϕ = {0}, in which case the homomorphism is injective. Now if ϕ is not the trivial homomorphism, then there exists v ∈ V with ϕ(v) 6= 0. Thus Im ϕ 6= {0} and since W is irreducible, we deduce that Im ϕ = W . Thus, ϕ is also injective, so ϕ is an isomorphism. [Note that this portion did not use the factor that the field is C. b) Now an isomorphism ϕ : V → V . As a linear transformation between vector spaces over C, the transformation ϕ must have at least one eigenvalue λ. (This is because C is algebraically closed so the characteristic polynomial of ϕ has a root in C.) Consider the eigenspace Eλ = Ker(ϕ − λ idv ). This is a vector subpace of g. For all v ∈ V and g ∈ G, we have ϕ(gv) − λ idV (gv) = gϕ(v) − g idV (v) = g(ϕ(v) − idV (v)), so ϕ − λ idV is representation homomorphism. Thus, as the kernel of a representation homomorphism, the eigenspace is a nontrivial subrepresentation of an irreducible representation V . Thus Eλ = V , which implies that ϕ(v) − λ idV (v) = 0 for all v ∈ V , which means that ϕ = λ idV .

8.6. A BRIEF INTRODUCTION TO REPRESENTATIONS OF GROUPS

459

Exercise: 19 Section 8.6 Question: Use Schur’s Lemma (Exercise 8.6.18) to prove the following result. Let V be a representation of G over a field C and suppose that V = W1 ⊕ W2 ⊕ · · · ⊕ Ws with Wi irreducible subrepresentations. Prove that every irreducible U of V is isomorphic to Wi for some i. Solution: We first prove a lemma. Let G be a group and let ϕ : V → W be a homomorphism between two representations of G. If U is a subrepresentation of W , then ϕ−1 (U ) is a subrepresentation of V . Let v1 , v2 ∈ ϕ−1 (U ), let c ∈ C and let g ∈ G. Then ϕ(v1 ), ϕ(v1 ) ∈ U . Since U is a subspace of W , we know that ϕ(v1 ) + ϕ(v2 ) = ϕ(v1 + v2 ) ∈ U . Thus, v1 + v2 ∈ ϕ−1 (U ). we also know that ϕ(cv1 ) = cϕ(v1 ) ∈ U . Thus, cv1 ∈ ϕ−1 (U ). This shows that ϕ−1 (U ) is a subspace of V . Finally, ϕ(g · v1 ) = g · ϕ(v1 ) ∈ U because ϕ(v1 ) ∈ U and U is a subrepresentation of W . This proves the lemma. Now consider the situation of our exercise. If U is an irreducible subrepresentation of V , then the injection φ : U → V is an injective representation homomorphism. We can consider the inverse images φ−1 (Wi ) in U . Since U is irreducible and φ−1 (Wi ) is a subrepresentation of U , then φ−1 (Wi ) = {0} or all of U . Assume that φ−1 (Wi ) = {0} for all i. Then φ−1 (V ) = {0} but this contradicts the assumptions on U . Hence, φ−1 (Wi ) = U for some i. This implies that φ(U ) = Im φ. Then φ is a nontrivial representation homomorphism betweenirreducible representations U and Wi . By Schur’s Lemma U and Wi are isomorphic. Exercise: 20 Section8.6 Question: Use Schur’s Lemma (Exercise 8.6.18) to prove that every irreducible representation over C of a finite abelian group is one dimensional. Solution: Let V be an irreducible representation over C of a finite abelian group. Let v ∈ V and g, x ∈ G. Since G is abelian, gx = xg and in particular g(xv) = x(gv). This implies that the mapping ϕx : V → V is representation homomorphism. Since ϕx is an isomorphism with ϕ−1 x = ϕx−1 , it is a nontrivial homomorphism on an irreducible representation over C. By Schur’s Lemma, there exists a constant λx (that depends on x) such that ϕx (v) = λx v. Take v0 ∈ V − {0} and nonzero vector. Since xv0 ∈ Span(v0 ) for all x ∈ G and thus x(cv0 ) ∈ Span(v0 ), we see that Span(v0 ) is a nontrivial representation of V . Since V is irreducible, then V = Span(v0 ). Hence, V is one dimensional

9 | Classification of Groups 9.1 – Composition Series and Solvable Groups Exercise: 1 Section 9.1 Question: Exhibit all the composition series of Z20 and determine the composition factors associated to each series. Solution: Consider Z20 = hz | z 20 = 1i. Recall that every subgroup of a cyclic group is cyclic. All the composition series have the form {1} E H1 E H2 E G and the composition factors are H1 , H2 /H1 , and G/H2 . The following gives the different composition series of Z20 , with their respective factors listed in order: {1} E hz 4 i E hz 2 i E Z20 {1} E hz 10 i E hz 2 i E Z20 {1} E hz 10 i E hz 5 i E Z20

Z5 , Z 2 , Z 2 Z2 , Z 5 , Z 2 Z2 , Z 2 , Z 5 .

Exercise: 2 Section 9.1 Question: Exhibit all the composition series of A4 and determine the composition factors associated to each series. Solution: Consider the alternating group A4 . All the composition series have the form {1} E H1 E H2 E G and the composition factors are H1 , H2 /H1 , and G/H2 . The following gives the different composition series of A4 , with their respective factors listed in order: {1} E h(1 2)(3 4)i E h(1 2)(3 4), (1 3)(2 4)i E A4 {1} E h(1 3)(2 4)i E h(1 2)(3 4), (1 3)(2 4)i E A4 {1} E h(1 4)(2 3)i E h(1 2)(3 4), (1 3)(2 4)i E A4

Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 .

Exercise: 3 Section 9.1 Question: Exhibit all the composition series of D6 and determine the composition factors associated to each series. Solution: Consider the alternating group D6 = hr, s | r6 = s2 = 1, rs = sr−1 i. All the composition series have the form {1} E H1 E H2 E G and the composition factors are H1 , H2 /H1 , and G/H2 . The following gives the different composition series of 461

462

CHAPTER 9. CLASSIFICATION OF GROUPS

D6 , with their respective factors listed in order: {1} E hr2 i E hri E D6 {1} E hr3 i E hri E D6 {1} E hr2 i E hr2 , si E D6 {1} E hsi E hr2 , si E D6 {1} E hsr2 i E hr2 , si E D6 {1} E hsr4 i E hr2 , si E D6 {1} E hr2 i E hr2 , sri E D6 {1} E hsri E hr2 , sri E D6 {1} E hsr3 i E hr2 , sri E D6 {1} E hsr5 i E hr2 , sri E D6 {1} E hr3 i E hr3 , si E D6 {1} E hsi E hr3 , si E D6 {1} E hsr3 i E hr3 , si E D6 {1} E hr3 i E hr3 , sri E D6 {1} E hsri E hr3 , sri E D6 {1} E hsr4 i E hr3 , sri E D6 {1} E hr3 i E hr3 , sr2 i E D6 {1} E hsr2 i E hr3 , sr2 i E D6 {1} E hsr5 i E hr3 , sr2 i E D6

Z3 , Z2 , Z2 Z2 , Z3 , Z2 Z3 , Z2 , Z2 Z2 , Z3 , Z2 Z2 , Z 3 , Z 2 Z2 , Z 3 , Z 2 Z3 , Z 2 , Z 2 Z2 , Z 3 , Z 2 Z2 , Z 3 , Z 2 Z2 , Z 3 , Z 2 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3 Z2 , Z 2 , Z 3

Exercise: 4 Section 9.1 Question: Exhibit all the composition series of F20 (Exercise 4.3.16) and determine the composition factors associated to each series. Solution: Consider the alternating group F20 . We note that in F20 , the powers of the following elements are xy, (xy)2 = x2 y 3 ,

(xy)3 = x3 y 2 ,

(xy)4 = x4 y 10 = 1

xy 2 , (xy 2 )2 = x2 y,

(xy 2 )3 = x3 y 4 ,

(xy 2 )4 = 1

xy 3 , (xy 3 )2 = x2 y 4 ,

(xy 3 )3 = x3 y,

(xy 3 )4 = 1

xy 4 , (xy 4 )2 = x2 y 2 ,

(xy 4 )3 = x3 y 3 ,

(xy 4 )4 = 1.

Along with knowing the order of x and of y, we have found every element in F20 and what subgroups they lie in. All the composition series have the form {1} E H1 E H2 E G and the composition factors are H1 , H2 /H1 , and G/H2 . The following gives the different composition series of F20 , with their respective factors listed in order: {1} E hyi E hy, x2 i E F20 {1} E hyi E hy, x2 yi E F20 {1} E hyi E hy, x2 y 2 i E F20 {1} E hyi E hy, x2 y 3 i E F20 {1} E hyi E hy, x2 y 4 i E F20 {1} E hx2 i E hxi E F20 {1} E hx2 y 3 i E hxyi E F20 {1} E hx2 yi E hxy 2 i E F20 {1} E hx2 y 4 i E hxy 3 i E F20 {1} E hx2 y 2 i E hxy 4 i E F20 {1} E hx2 i E hx2 , yi E F20 {1} E hx2 i E hx2 y 3 , yi E F20 {1} E hx2 i E hx2 y, yi E F20 {1} E hx2 i E hx2 y 4 , yi E F20 {1} E hx2 i E hx2 y 2 , yi E F20

Z5 , Z 2 , Z 2 Z5 , Z 2 , Z 2 Z5 , Z 2 , Z 2 Z5 , Z 2 , Z 2 Z5 , Z 2 , Z 2 Z2 , Z 2 , Z 5 Z2 , Z 2 , Z 5 Z2 , Z 2 , Z 5 Z2 , Z 2 , Z 5 Z2 , Z 2 , Z 5 Z2 , Z 5 , Z 2 Z2 , Z 5 , Z 2 Z2 , Z 5 , Z 2 Z2 , Z 5 , Z 2 Z2 , Z 5 , Z 2 .

9.1. COMPOSITION SERIES AND SOLVABLE GROUPS

463

Exercise: 5 Section 9.1 Question: Prove that if N E G, then G has a composition series with N as a term. Solution: By Jordan-Hölder, N has a composition series {1} = A0 ≤ A1 ≤ . . . Ak = N . As does G/N : {1} = B0 ≤ B1 ≤ . . . Bl = G/N . By the fourth isomorphism theorem we know there is a chain of subgroups N = C0 ≤ C1 ≤ . . . Cl = G such that Ci /N = Bi and Ci E Ci+1 . Then we can form the entire composition series which has N as a term. {1} = A0 ≤ A1 ≤ . . . Ak = N = C0 ≤ C1 ≤ . . . Cl = G.

Exercise: 6 Section 9.1 Question: Suppose that G and H are groups with composition series of length r and s, respectively. Show that G ⊕ H has a composition series of length r + s and that the composition factors of G ⊕ H are of the form M 0 ⊕ N 0 , where M 0 and N 0 are composition factors of G and H, respectively. Solution: [Correction: the composition series of G ⊕ H has length r + s.] Suppose that G and H have composition series of G : {1} = M0 E M1 E M2 E · · · E Mr H : {1} = N0 E N1 E N2 E · · · E Ns . We identify G with the subgroup {(g, 1H ) ∈ G ⊕ H | g ∈ G} of G ⊕ H and similarly H with the subgroup {(1G , h) ∈ G ⊕ H | h ∈ H} of G ⊕ H. Consider the following chain of subgroups in G ⊕ H: {1}

≤ {1} ⊕ N1 ≤ M1 ⊕ Ns

≤ {1} ⊕ N2 ≤ M 2 ⊕ Ns

≤ ··· ≤ ≤ ··· ≤

{1} ⊕ Ns M r ⊕ Ns .

We note that in the first row {1} ⊕ Nj−1 E {1} ⊕ Nj with ({1} ⊕ Nj )/({1} ⊕ Nj−1 ) ∼ = Nj /Nj−1 . In the second row and also going from the first row to the second row, Mi−1 ⊕Ns EMi ⊕Ns with (Ns ⊕Mi )/(Ns ⊕Mi−1 ) ∼ = Mi /Mi−1 . So in fact, we obtain a composition series of G ⊕ H of length r + s where the composition factors are the of the form {1} ⊕ N 0 or M 0 ⊕ {1} where M 0 is a composition factor of G and N 0 a composition factor of H. Exercise: 7 Section 9.1 Question: Prove that if G is finite, then it is solvable if and only if each of its composition factors is a finite cyclic group of prime order. Solution: Lemma: Every finite abelian group only has cyclic groups of prime order as composition factors. Let H be a finite abelian group. We prove be induction on the number n of prime factors (not distinct) of |H|, then H has a composition series where all the composition factors are cyclic groups of prime order. (To clarify, αs 2 if |H| = p1α1 pα 2 · · · ps , then n = α1 + α2 + · · · + αs .) This is obviously true if |H| has only one prime factor, since then H = Zp for some prime p. Now suppose that the result is true for all groups H such that |H| has n prime factors. Let K be an abelian group with n + 1 prime factors. Let p be a prime dividing |K|. By Cauchy’s Theorem, there exists an element x of order p in K. Since K is abelian, hxi E K and we note that |K/hxi| has n prime factors. By Exercise 9.1.5, K has a composition series with hxi ∼ = Zp as a composition factor, and in fact, that composition series can begin as {1} E hxi E · · · . The remaining composition factors are the composition factors of K/hxi. However, by the induction hypothesis, all of these composition factors are finite cyclic groups of prime order. Hence, by induction, every abelian group has composition factors that are cyclic groups of prime order. Now let G be a finite group and suppose that G is solvable. Then there exists a chain of subgroups {1} = M0 E M1 E M2 E · · · E Mr = G such that Mi /Mi−1 is abelian for all 1 ≤ i ≤ r. Consider a composition series of Mi /Mi−1 . By the above lemma, this series have the form {1} = Mi−1 /Mi−1 E N̄1 E N̄2 E · · · E N̄n = Mi /Mi−1 , where N̄j /N̄j−1 is a finite cyclic group. By the Fourth Isomorphism Theorem, there exist groups Nj in Mi , containing Mi−1 such that Nj /Mi−1 = N̄j . Furthermore, by the Third Isomorphism Theorem, N̄j /N̄j−1 = (Nj /Mi−1 )/(Nj−1 /Mi−1 ) ∼ = Nj /Nj−1 .

464

CHAPTER 9. CLASSIFICATION OF GROUPS

Thus, in the M -chain of subgroups of G, between any Mi−1 E Mi , we can insert Mi−1 = N0 E N1 , N2 , . . . , NE n = Mi , in which the composition factors that arise here are finite cyclic groups of prime order. Since this occurs between every successive subgroup pair in the M -chain, we deduce that all the composition factors of an abelian group are finite cyclic of prime order. The converse is trivial. If each of the composition factors of a finite group is finite and cyclic, then any composition series provides a chain of subgroups with the defining property of solvable groups. Exercise: 8 Section 9.1 Question: Prove that subgroups and quotient groups of solvable groups are solvable. Solution: Let H be a subgroup of a solvable group G and let {1} = M0 E M1 E M2 E · · · E Mr = G be a chain of subgroups of G such that Mi−1 E Mi and Mi /Mi−1 is abelian. Taking the intersection of all these subgroups with H, we get {1} = H ∩ M0 E H ∩ M1 E H ∩ M2 E · · · E H ∩ Mr = H as a chain of subgroups within H. For any index i with 1 ≤ i ≤ r, we have the following sublattice Mi Mi−1 (H ∩ Mi ) Mi−1

H ∩ Mi−1

H ∩ Mi−1 By Exercise 4.2.14, H ∩ Mi−1 E H ∩ Mi . From the lattice diamond shown above, the Second Isomorphism Theorem affirms that (H ∩ Mi )/(H ∩ Mi−1 ) ∼ = Mi−1 (H ∩ Mi )/Mi−1 . However, Mi−1 (H ∩ Mi )/Mi−1 is a subgroup of Mi /Mi−1 and thus is abelian. Consequently, H ∩ Mi−1 E H ∩ Mi is abelian, and we deduce that the above chain of subgroups of H is such that successive quotients are abelian. Hence, H is a sovlable group. To prove the second part of the exercise, we will use the result of Exercise 9.1.7. Let N be a normal subgroup of a solvable group G. Exercise 9.1.5 shows that G has a composition series with N as a term. Let {1} = K̄0 E K̄1 E K̄2 E · · · E K̄s = G/N be a composition series of G/N . By the Fourth Isomorphism Theorem, there exist subgroups Kj of G containing N such that K̄j = Kj /N . Furthermore, this correspondence preserves containment and normality. Hence, N = K0 E K1 E K2 E · · · E Ks = G is a portion of a composition series of G that has N as a term. Since G is solvable, then K̄j /K̄j−1 ∼ = Kj /Kj−1 is a finite cyclic group of prime order. Thus, G/N is solvable. Exercise: 9 Section 9.1 Question: Suppose that if G is a group with a normal subgroup N . Prove that if N and G/N are both solvable, then G is solvable. Solution: We first write the chain of subgroups for each. {1} = A0 E A1 E . . . E Ak = N and {1} = B0 E B1 E . . . E Bl = G/N . Then by the Fourth Isomorphism Theorem there exist a chain of subgroups,

9.1. COMPOSITION SERIES AND SOLVABLE GROUPS

465

N = C0 E C1 E . . . E Cl = G where Ci /N = Bi and by the Third Isomorphism Theorem, we have Ci+1 /Ci ∼ = (Ci+1 /N )/(Ci /N ) so that Ci+1 /Ci is abelian. Then we can write the chain of subgroups, {1} = A0 E A1 E . . . E Ak = N = C0 E C1 E . . . E Cl = G, where the quotient of two consecutive factors is an abelian group. This shows that G is solvable. Exercise: 10 Section 9.1 Question: Let G be a solvable group and let ϕ : G → H be a surjective homomorphism. Prove that H is solvable. Solution: If ϕ : G → H is a surjective homomorphism, then by the First Isomorphism Theorem, H ∼ = G/ Ker ϕ. By Exercise 9.1.8, we deduce that H is a solvable group. Exercise: 11 Section 9.1 Question: Prove that every p-group is solvable. Solution: Recall that a p-group is a group G of order pn , where p is prime and n is a positive integer. We prove this by induction on n. If n = 1, then G ∼ = Zp , which is solvable. Suppose that the result is true for all n with 1 ≤ n < m and consider a group of order pm . By Exercise 8.4.14, we know that every p-group has a nontrivial center Z(G). Then we have the chain of subgroups {1} E Z(G) E G Obviously Z(G) is abelian so it is solvable. However, G/Z(G) is a p group of order strictly less than pm . Using the induction hypothesis, we deduce that G/Z(G) is solvable. Now we use Exercise 9.1.8, and deduce that G is also solvable. By induction, we conclude that for all positive integers n (and any prime p) a group of order pn is solvable. Exercise: 12 Section 9.1 Question: Prove that if H E G, then [H, G] ≤ H. Solution: Suppose that H E G. Recall that [H, G] is the subgroup generated by elements of the form [h, g] = h−1 g −1 hg for h ∈ H and g ∈ G. Now since H E G, then g −1 hg = h0 for some element h0 ∈ H. Then [h, g] = h−1 h0 , so [h, g] ∈ H. We deduce that [H, G] ≤ H. Exercise: 13 Section 9.1 Question: Prove that the commutator subgroup of A4 is (A4 )0 = h(1 2)(3 4), (1 3)(2 4)i. Solution: Every nonidentity element in A4 can be written as (a b c) or as (a b)(c d). Let H = h(1 2)(3 4), (1 3)(2 4)i By Exercise 9.1.12, [H, A4 ] ≤ H. In other words, in our case [(a b)(c d), g] ∈ H for all g ∈ A4 . On the other hand, any two 3-cycles in A4 that are not powers of each other can be written as (a b c) and (a b d) since such 3-cycles will share two entries. Furthermore, [(a b c), (a b d)] = (a c b)(a d b)(a b c)(a b d) = (a b)(c d). In other words, the commutator of any pair of 3-cycles has the form (a b)(c d). This shows that (A4 )0 ≤ H. Furthermore, the generators (1 2)(3 4) and (1 3)(2 4) of H arise as commutators of specific elements. This now shows that H ≤ (A4 )0 . Hence, we deduce that (A4 )0 = h(1 2)(3 4), (1 3)(2 4)i. Exercise: 14 Section 9.1 Question: Let F be a field with char F 6= 2 and let n ≥ 2. Prove that the commutator subgroup G0 of G = GLn (F ) is a nontrivial strict subgroup of G. Solution: [This exercise has a correction that char F 6= 2.] We know that for all fields F and for all integers n ≥ 2, the group GLn (F ) is nonabelian. By Proposition 9.1.9, the commutator subgroup G0 is not the trivial subgroup. Let A, B ∈ GLn (F ). The commutator of A and B is A−1 B −1 AB. We note that det(A−1 B −1 AB) = det(A)−1 det(B)−1 det(A) det(B) = 1.

466

CHAPTER 9. CLASSIFICATION OF GROUPS

Hence, the commutator subgroup G0 is a subgroup of SLn (F ). If the characteristic of F is not 2, then SLn (F ) is a strict subgroup of GLn (F ), so G0 ≤ SLn (F ) and G0 is a strict subgroup of GLn (F ). Exercise: 15 Section 9.1 Question: Prove that G(i) is a characteristic subgroup of G for all i ≥ 1. Solution: We prove this result by induction on i. If i = 1, then G( i) = G0 = [G, G]. This group is generated by all the elements of the form x−1 y −1 xy for x, y ∈ G. Let ψ ∈ Aut(G) be any automorphism of G. Then ψ([x, y]) = ψ(x−1 y −1 xy) = ψ(x)−1 ψ(y)−1 ψ(x)ψ(y) = [ψ(x), ψ(y)]. Thus, for every generator [x, y] of G0 , we see that ψ([x, y]) ∈ G0 . Hence, ψ(G0 ) ⊆ G0 . However, since ψ is a bijection, its range is in bijection with its domain, so ψ(G0 ) = G0 . Consequently, G0 is a characteristic subgroup of G. Now suppose that G(i) is a characteristic subgroup for some positive integer i. Then G(i+1) = [G(i) , G(i) ]. We saw in the basis step that any ψ ∈ Aut(G) maps G0 to itself so defines an automorphism ψ 0 ∈ Aut(G0 ). By the induction hypothesis, any ψ ∈ Aut(G) defines an automorphism on G(i) . Let x, y ∈ G(i) . Then, as before, ψ([x, y]) = ψ(x−1 y −1 xy) = ψ(x)−1 ψ(y)−1 ψ(x)ψ(y) = [ψ(x), ψ(y)] so under ψ the generators of G(i+1) map to other generators of G(i+1) . Hence, ψ(G(i+1) ) ≤ G(i+1) . Again, since the image of ψ is in bijection with its domain, we deduce that ψ(G(i+1) ) = G(i+1) . Hence, G(i+1) is a characteristic subgroup. By induction, we deduce that G(i) is a characteristic subgroup for all positive integers i.

9.2 – Finite Simple Groups Exercise: 1 Section 9.2 Question: Prove that there is no simple group of order 300. Solution: Let G be a group of order 300. The prime factorization of 300 is 300 = 22 · 3 · 52 . By Sylow’s Theorem, n5 ≡ 1 (mod 5) and n5 | 12. The divisors of 12 are 1, 2, 4, 3, 6, 12. So far, we can deduce that n5 is 1 or 6. Also by Sylow’s Theorem, n3 ≡ 1 (mod 3) and n3 | 100. This means n3 is 1, 4, 25, or 100. Sylow’s Test for Simplicity does not immediately establish that G is not simple. Assume that G is simple. Let P ∈ Syl5 (G). By Sylow’s Theorem, n5 = 6 so NG (P ) has index of 6 in G. Then |G : NG (P )|! = 6! = 720. We note that |G| = 300 so |G| does not divide |G : NG (P )|! so by the Index Theorem, G is not simple. (The action of G on the set of cosets of NG (P ) must have a nontrivial kernel, which is a nontrivial normal subgroup of G.) Exercise: 2 Section 9.2 Question: Prove that there is no simple group of order 936. Solution: Let G be a group of order 936. The prime factorization of 936 is 936 = 23 · 32 · 13. Sylow’s Theorem implies that n13 ≡ 1 (mod 13) and n13 | 72. The only possibility for n13 is 1 so a Sylow 13-subgroup is normal. Hence, G is not simple. Exercise: 3 Section 9.2 Question: Prove that there is no simple group of order 315. Solution: Let G be a group of order 315. The prime factorization of 315 is 315 = 32 · 5 · 7. By Sylow’s Theorem, n7 ≡ 1 (mod 7) and n7 | 45. Then n7 can be 1 or 15. Also n5 ≡ 1 (mod 5) and n5 | 63, so n5 can be 1 or 21. Also n3 ≡ 1 (mod 3) and n3 | 35 so n3 is 1 or 7. Assume that G is simple. Then n7 = 15, n5 = 21, and n3 = 7. Call P an element of Syl7 (G) and Q an element of Syl5 (G). Then the normalizer NG (P ) has index 15 and hence order 21, whereas the normalizer NG (Q) has index 21 and hence order 15. By Example 8.5.10, NG (Q) is a cyclic subgroup isomorphic to Z15 but NG (P ) could be cyclic or could be isomorphic to the unique nonabelian group of order 21. We start counting up elements by their orders.

9.2. FINITE SIMPLE GROUPS

467

• Order 3 or 9: We do not know the structure of the Sylow 3-subgroups, but, to create a minimum number of elements whose order is a power of 3, they could all intersect in the same subgroup of order 3. This would account for 2 + 7 × 6 = 44 elements whose order is 3 or 32 . • Order 5: All of the Sylow 5-subgroups must intersect trivially (because of two of them share the a nonidentity element, they would be both generated by that nonideneity element and we would have fewer than 21 Sylow 5-subgroups) so there are 21 × 4 = 84 elements of order 5. • Order 7: All of the Sylow 7-subgroups must intersect trivially so there are 15 × 6 = 90 elements of order 7. • Order 15: Let Q0 = NG (Q) as defined above. The Sylow 5-subgroups arise as subgroups of cyclic groups of order 15. Furthermore, there must be 21 distinct such subgroups. Indeed, for Q1 , Q2 ∈ Syl5 (G), if Q1 6= Q2 then NG (Q1 ) 6= NG (Q2 ) because otherwise the cyclic subgroup in each would be the same and so Q1 = Q2 . Thus we have 21 distinct cyclic subgroups of order 15. This implies that G must contain 21 × 8 = 168 distinct elements of order 15 (there are 8 elements of order 15 in Z15 ). In our count of elements, we have already found that G must contain 1 + 44 + 84 + 90 + 168 = 387. But |G| = 315 so we arrive at a contradiction. Hence, G cannot be simple. Exercise: 4 Section 9.2 Question: Prove that there is no simple group of order 2784. Solution: The prime factorization of 2784 is 2784 = 25 · 3 · 29. Next we use Sylow’s theorem to examine n29 . Since it must divide 25 · 3 we have that n29 ∈ {1, 2, 3, 4, 6, 8, 12, 16, 18, 24, 32, 48, 96}. Now since n29 ∼ = 1 mod 29, n29 = 1. By Sylow’s theorem, we have a normal group of order 29. So G cannot be simple when it has order 2784. Exercise: 5 Section 9.2 Question: Prove that if G is a group of order pqr with primes p < q < r, then G has a normal subgroup of order p, q, or r. Solution: Let G be a group of order |G| = pqr with primes p < q < r. We prove the result by contradiction. Assume that G has no normal Sylow subgroups. By Sylow’s Theorem, nr ≡ 1 (mod r) and nr | pq. Since nr > 1 (given our assumption), then nr ≥ r + 1 so nr is strictly larger than p or q. Hence, nr = pq. Since any Sylow r-subgroup in Sylr (G) is a cyclic subgroup of order r, then every nonidentity element such a Sylow r-subgroup generates it. Hence, the Sylow r-subgroups intersect pairwise only in the identity. This account for pq(r − 1) elements of order r. By the same reasoning with Sylow’s Theorem, we deduce that nq ≥ p and that np ≥ q. By the same reasoning as for the Sylow r-subgroups, the Sylow q-subgroups intersect pairwise trivially, as do the Sylow p subgroups. These facts account for p(q − 1) elements of order q and q(p − 1) elements of order p. Along with the identity, we have shown that G mus contain at least 1 + pq(r − 1) + p(q − 1) + q(p − 1) = 1 + pqr − pq + pq − p + pq − q = pqr + (p − 1)(q − 1) elements. This is greater than |G| so we arrive at a contradiction. Hence, there exist no simple group whose order is the product of three distinct primes. Exercise: 6 Section 9.2 √ Question: Suppose that an integer n is divisible by a prime number p such that p ≥ n. Prove that a group G with |G| = n is not simple. p √ √ Solution: Suppose that pi · d = n. If i ≥ 3 then we have that p ≤ (pi )1/3 < pi ≤ n or p < n. This is a contradiction and so we must have i = 1 or 2. We examine the case where i = 2. Then we must have that n = p2 and by results from Example 8.4.10, G is isomorphic to Zp2 or Zp ⊕ Zp neither of which is simple. Next we examine the case where i = 1. We know that np must divide d where d · p = n. We also know that np √ = k ·√ p + 1 for some natural number k. So let np = m where m ≤ d. Then we √ √ have that m ≤ d = n/p ≤ n/ n = n so that m ≤ n. Then if k ≥ 1 we have that m = k · p + 1 ≥ p + 1 > n ≥ m or m > m which is a contradiction. So we must have k = 0 and np = 0 · p + 1 = 1. Then G contains a normal subgroup of size p and is not simple.

468

CHAPTER 9. CLASSIFICATION OF GROUPS

Exercise: 7 Section 9.2 Question: 2×Odd Test. Prove that if n > 1 is an odd integer, then there is no simple group of order 2n. Solution: Let G be a group of order 2n where n is odd. Consider the action of G on itself by left multiplication. This action defines a homomorphism ρ : G → S2n . For all g ∈ G, the permutation ρ(g) has no fixed points. By Cauchy’s Theorem (and Sylow’s Theorem), G contains an element x of order 2. Thus ρ(x) is an element of order 2 without any fixed points. Thus ρ(x) is a product of n disjoint transpositions. In particular ρ(x) is an odd permutation. Consider the composition of ρ and the sign function sign : S2n → ({1, −1}, ×), i.e. ϕ : G → ({1, −1}, ×) with ϕ = sign ◦ρ. Since ρ(1) = 1 is even and ρ(x) is an odd permutation, then ϕ is surjective. By the First Isomorphism Theorem, G/ Ker ϕ ∼ = Z2 and hence Ker ϕ has order n. Hence, G contains normal subgroup N , namely Ker ϕ, of order n. Thus N is nontrivial and proper. Thus G is not simple. Exercise: 8 Section 9.2 Question: Prove that if G has a nontrivial conjugacy class K such that |G| does not divide |K|!, then G is not simple. Solution: Let K be a nontrivial conjugacy class of a group G. The action of G on K by conjugation defines a homomorphism ρ : G → SK . Note that this action is transitive by definition of conjugacy classes and hence the kernel Ker ρ cannot be all of G. By the First Isomorphism Theorem, G/ Ker ρ ∼ = Im ρ and by Lagrange’s Theorem |G|/| Ker ρ| divides |SK | = |K|!. If |G| does not divide |K|! then Ker ρ cannot be the trivial subgroup. Since we know that Ker ρ is a proper subgroup, Ker ρ is a nontrivial proper normal subgroup. So we conclude that G is not simple. Exercise: 9 Section 9.2 Question: Prove that G is a finite group such that H is a normal subgroup of maximal order. Prove that G/H is a simple group. Solution: Let G be a group and let H E G. Suppose that G/H has a nontrivial proper normal subgroup N̄ . By the Fourth Isomorphism Theorem, there exists a normal subgroup N of G containing H such that N̄ = G/N . Taking the contrapositive of this implication, we deduce that if H is a normal subgroup whose order is maximal among proper normal subgroups, then G/H contains no nontrivial proper normal subgroups and thus is simple.

Exercise: 10 Section 9.2 Question: Prove that A5 is the only simple group of order 60. Solution: Assume that G is a simple group of order 60. The prime factorization of 60 is 60 = 22 × 3 × 5. By Sylow’s Theorem, n5 = 6 and n3 can be a priori 4 or 10. Also, n2 can be either 3, 5, or 15. From this information, we make an analysis on the number of elements of a given order. • Order 5: Since n5 = 6, and the Sylow 5-subgroups can only intersect at the identity, there are 24 elements of order 5. • Order 3: A priori, n3 = 4 or 10. However, if n3 , the means that a normalizer NG (P3 ) of a Sylow 3-subgroup has index 4. By Proposition 9.2.5, this implies that G is not simple since 60 does not divide 4! = 24. Hence, with the supposition that G is simple, we conclude that n3 = 10. Since none of the 10 Sylow 3-subgroups can intersect except in the trivial subgroup, we deduce that G contains 20 elements of order 3. • Order 2 or 4: By the same reasoning as the previous bullet, since 60 does not divide 3! = 6, assuming that G is simple, we cannot have n2 = 3. Thus, n2 = 5 or 15. For the moment, we point out that the number of elements of order 2 or 4 must be at least 11, since n2 ≥ 5 and the Sylow 2-subgroups could intersect nontrivially in a subgroup of order 2. The elements of order 1, 3, or 5 account for 1 + 24 + 20 = 45 elements in the group G and we have for certain at least 11 more elements of order 2 or 4. Let us consider the possibility that n2 = 5. Then the normalizer N of a Sylow 2-subgroup has index 5 in G. We consider the action of G on the set of cosets of N by left multiplication. Thence, this defines a homomorphism ϕ : G → S5 . The kernel Ker ϕ must be trivial or all of G for G to be simple. The action of G on the cosets is transitive and in particular not trivial, so Ker ϕ = {1}. By the First Isomorphism Theorem, G is isomorphic to a subgroup of S5 . Think of G as a subgroup of S5 . If G is not equal to A5 , then |G : G ∩ A5 | = |Sn : An | = 2 by

9.2. FINITE SIMPLE GROUPS

469

the Second Isomorphism Theorem and thus G ∩ A5 E G, which is a contradiction since G is simple. We deduce that G = A5 . We now consider the possibility that n2 = 15. If the Sylow 2-subgroups are isomorphic to Z4 , then 15 such distinct subgroups contribute 2 × 15 = 30 elements of order 4. Since 30 + 45 > 60, this leads to a contradiction. So for all P ∈ Syl2 (G), we have P ∼ = Z2 ⊕ Z2 . The 15 Sylow 2-subgroups cannot intersect only in the identity since this would imply that G contains 15 × 3 = 45 elements of order 2 or 4. So |G| > 45 + 45 = 90, which is a contradiction. So there exist two Sylow 2-subgroups P and Q such that |P ∩ Q| = 2. Consider the normalizer NG (P ∩ Q). Since P and Q are abelian, both P and Q are subsets of H = NG (P ∩ Q). Then 4 = |P | divides |H| and |H| divides 60 with index greater than 4. The only possibility is |H| = 12. However, since |G : H| = 5, we can consider the action of G on the set of left cosets of H. The reasoning is now identical to the end of the previous paragraph and, which leads to G ∼ = A5 . However, n2 (A5 ) is not 15 so this assumption leads to a contradiction. We know from this section that A5 is a simple group but this exercise now shows that every simple group of order 60 is isomorphic to A5 . Exercise: 11 Section 9.2 Question: Prove that | PSL2 (F5 )| = 60. Without using the previous exercise, show that PSL2 (F5 ) ∼ = A5 . Solution: We know that | GL2 (F5 )| = (52 − 1)(52 − 5) = 480. Since the determinant homomorphism surjects onto (U (F5 ), ×), then as the kernel to the determinant, | SL2 (F5 )| = 120. The center of SL2 (F5 ) consists of diagonal matrices of the form cI, where det(cI) = c2 = 1. For the four possible elements for c ∈ U (F5 ), only c = 1̄ and c = 4̄ work. Hence, 1 0 4 0 Z(SL2 (F5 )) = , , 0 1 0 4 and thus | PSL2 (F5 )| = 60. A key to proving that PSL2 (F5 ) is isomorphic to A5 involves finding a faithful action of PSL2 (F5 ) on a set of 5 elements. In the following calculations, we are working in PSL2 (F5 ), which means that all matrices must have a determinant of 1 in F5 and we consider the matrices equivalent if they differ from each other by a multiple of 1 or 4. Consider the subgroup 0 4 2 0 H= , . 1 0 0 3 This is a Sylow 2-subgroup (of order 4). Consider the normalizer K of H. It contains the generators of H but 1 2 it also contains the matrix because 1 3

2 0 3 1

1 1

and 1 1

2 3

2 0

0 3

1 1

2 3

4 1 0 1

−1 2 2 = 3 0

0 ∈H 3

−1

0 2

4 0

=

2 0

=

0 1

2 0

0 3

∈ H.

2 has order 3 in PSL2 (F5 ). Thus, the group K contains an element of order 3 on top of the subgroup 3 2 1 H. Hence 12 divides K. But H is not normal in PSL2 (F5 ) since for example does not normalize H. 1 1 Thus, 12 divides |K| and K is a strict subgroup of PSL2 (F5 ). Hence, K is a subgroup of order 12 and 0 4 2 0 1 2 K= , , . 1 0 0 3 1 3 But

1 1

Consider the action of PSL2 (F5 ) on the set of left cosets of K by left multiplication. This action induces a homomorphism ϕ : PSL2 (F5 ) → S5 . It is not hard to check that the set of left cosets of K consists of 1 0 1 1 1 2 1 3 1 4 K, K, K, K, K. 0 1 0 1 0 1 0 1 0 1

470

CHAPTER 9. CLASSIFICATION OF GROUPS

To consider the induced homomorphism into S5 ,label these cosets in order by 1, 2, 3, 4, and 5. 1 1 2 1 Under this action, we note that ϕ( ) = (1 2 3 4 5). Also under this action, ϕ( ) = (1 5 3 2 4). 0 1 1 1 Note that (1 5 3 2 4)2 (1 2 3 4 5)−1 = (1 2 3). By Exercise 3.5.44, the image of ϕ contains A5 . However, A5 has order 60 and the image of ϕ has at most 60 elements. Hence, Im ϕ = A5 and Ker ϕ = {1}. Thus, PSL2 (F5 ) ∼ = A5 . Exercise: 12 Section 9.2 Question: Prove that A6 is simple using the strategy given in Proposition 9.2.2. Solution: In order to apply Proposition 9.2.2 to A6 , we need to determine the conjugacy classes of A6 . Since the conjugacy class of x in A6 is {gxg −1 | g ∈ A6 }, while the conjugacy class in S6 is {gxg −1 | g ∈ Sn }, the partition obtained as the conjugacy classes of A6 form a refinement of the partition for conjugacy class in A6 as subsets of S6 , i.e., using g ∈ S6 for conjugates. The S6 -conjugacy classes in A6 are the cycle types, namely cycle type 1 (a b)(c d) (a b c) (a b c)(d e f ) (a b c d)(e f ) (a b c d e) All A5

count 1= 15 × 6/2 = 6 2 3 = 6 2 3 × 2/2 = 6 4 × 3! = 6 5 × 4! =

1 45 40 40 90 144 360

In order to determine the size of the conjugacy classes of elements in A5 (as opposed to in S5 ), we use the Orbit-Stabilizer Theorem. Note that the stabilizer under conjugation of a given element is the centralizer of that element. Hence, the size of a conjugacy class is |A5 : CA5 (σ)|. Using representative elements from the S5 -conjugacy classes, we determine the size of the respective conjugacy classes as follows. g (1 2)(3 4) (1 2 3) (1 2 3)(4 5 6) (1 2 3 4)(5 6) (1 2 3 4 5)

CA5 (g) h(1 2)(3 4), (1 3)(2 4), (1 2)(5 6)i h(1 2 3), (4 5 6)i h(1 2 3), (4 5 6)i h(1 2 3 4)(5 6)i h(1 2 3 4 5)i

|CA5 (g)| 8 9 9 4 5

|A5 : CA5 (g)| 45 40 40 90 72

Thus, the conjugacy classes in A5 are all of the usual cycle-types except that the 5-cycle class divides into two conjugacy classes, each of size 72. A normal subgroup of a group is a union of conjugacy classes that include the identity element class. So we consider ways of adding up the orders of the conjugacy classes and check if we obtain a divisor of 360. The divisors of 360 are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360 By observing the sizes of the conjugacy classes, and since a normal subgroup has to include {1}, we can quickly tell that we only need to consider divisors above 60. We will not obtain the divisor of 72 either since the sum of 1 and any two of the conjugacy classes that are not 5-cycles gives more than 72. All the divisors of 360 that are greater than 72 are divisible by 10. In order to include the identity and also obtain the sum of conjugacy classes to get a number divisible by 10, we need to include the identity, the (a b)(c d) class and both 5-cycle classes. This already accounts for 1 + 144 + 45 = 190 elements, which is greater than the second smallest divisor of 360. Hence, the only unions of conjugacy classes to produce a divisor of 360 is either the trivial subgroup or all of A5 . This shows that A5 is simple. Exercise: 13 Section 9.2 Question: Let G be a finite simple group. Suppose that H, K ≤ G such that |G : H| = p and |G : K| = q are prime numbers. Prove that p = q.

9.2. FINITE SIMPLE GROUPS

471

Solution: Assume that p 6= q and, without losing generality, assume p < q. Consider |G : H|! = p!. Since p < q, then q - p!, and because q divides |G| then |G| cannot divide |G : H|!. By the Index Theorem, G is not simple. This is a contradiction, so p must equal q. Exercise: 14 Section 9.2 Question: If F is a finite field of order |F | = q, determine a formula for | PSLn (F )|. Solution: An n×n matrix with coefficients in F is invertible if and only if its columns are linearly independent. We interpret what this means for the columns considering them one at a time. The first column cannot be the ~0 vector in F n so there are q n − 1 options. The second column cannot be a multiple (by a scalar in F ) of the first column so there are q n − q options. For the kth column with k > 1, the column cannot be in the k − 1 dimensional subspace spanned by the previous k − 1 column vectors, so there are q n − q k−1 options. Hence, | GLn (F )| = (q n − 1)(q n − q)(q n − q 2 ) · · · (q n − q n−1 ) We know that SLn (F ) = Ker det where the image group of the determinant homomorphism is (U (F ), ×), which has order q − 1. The First Isomorphism Theorem shows that | SLn (F )| =

(q n − 1)(q n − q) · · · (q n − q n−1 ) = q n−1 (q n − 1)(q n − q) · · · (q n − q n−2 ). q−1

The center of SLn (F ) consists of diagonal matrices (whose determinant is 1), i.e., matrices of the form cI with c ∈ U (F ) with cn = 1. In other words, the order of c in the group U (F ) divides n. From Proposition 7.5.2, the multiplicative group (U (F ), ×) is cyclic of order |F | − 1 = q − 1. In a cyclic group of order m, there are d elements whose orders divide d when d is a divisor of m. Hence, in Zq−1 , there are gcd(q − 1, n) elements g such that g n = 1. Thus q n−1 (q n − 1)(q n − q) · · · (q n − q n−2 ) | PSLn (F )| = . gcd(q − 1, n)

Exercise: 15 Section 9.2 Question: In this exercise, we prove that the group of symmetries of the Fano plane is isomorphic to PSL3 (F2 ). (See Exercise 8.3.17.) a) Show that there are 7 points in the projective space P(F32 ). b) Given that “lines” in P(F32 ) arise from subspaces in F32 , prove that the lines in the projective space P(F32 ) correspond to triples (a1 : a2 : a3 ), (b1 : b2 : b3 ), and (c1 : c2 : c3 ) such that ai + bi + ci = 0 in F2 . c) Show that the space P(F32 ) along with its set of lines has the geometry of the Fano plane. d) The group of symmetries is the subgroup of S7 (acting on the points Fano plane) that preserves collineations. Prove that this group of symmetries is PGL3 (F2 ) and observe that this group is equal to PSL3 (F2 ). Solution: Consider the 3 dimensional vector space over F2 , namely the 8 element set F32 . a) Elements in the projective space P(F32 ) consist of triples (a1 , a2 , a3 ) ∈ F32 − {(0, 0, 0)} that are identified if they differ by a constant multiple by an element in U (F2 ). However, the group of units in F2 is the trivial group and hence P(F32 ) = F32 − {(0, 0, 0)}. Thus, |P(F32 )| = 7. Note that we denote elements in P(F32 ) as (a1 : a2 : a3 ). b) Consider two points p1 , p2 in P(V ), where V is a vector space over a field F . There is a uniquely line passing through p1 and p2 . However, the point p1 ∈ P(V ) corresponds to all nonzero multiples of p1 (with p1 viewed as a vector). Hence, the points on the line passing through p1 and p2 correspond in V − {~0} to the points (except ~0) of the span of vectors corresponding to p1 and p2 . In F32 a subspace with the ~0 removed will contain exactly 4 − 1 = 3 points. By considering all possibilities of coefficients, the span of two vectors ~a and ~b in F32 is the set {~0, ~a, ~b, ~a + ~b} and in particular, we note that ~a + ~b + (~a + ~b) = ~0. Thus, three points (a1 : a2 : a3 ), (b1 : b2 : b3 ), and (c1 : c2 : c3 ) in P(F32 ) are collinear if and only if ~a + ~b + ~c = ~0.

472

CHAPTER 9. CLASSIFICATION OF GROUPS

c) Let us label a point (a1 : a2 : a3 ) ∈ P(F32 ) with the integer n that has (a3 a2 a1 )2 as its binary representation, so (0 : 0 : 1) is labeled with 1, (0 : 1 : 0) is labeled with 2 and so forth. The following gives a list of all the lines in P(F32 ) according to the rule given in the previous part: {1, 2, 3}, {1, 4, 5}, {1, 6, 7}, {2, 4, 6}, {2, 5, 7}, {3, 4, 7}, and {3, 5, 6}. This does not give the same labeling as the Fano plane depicted in Figure 8.7 but under the relabeling (from the Fano plane labels) given by the permutation (3 7 5), we recover the same incidence relationship. Hence, the set of lines in P(F32 ) corresponds with the points and lines of the Fano plane. d) The group of symmetries is the subgroup of S7 that preserves the collinear relationships. Now | GL3 (F2 )| = (8 − 1)(8 − 2)(8 − 4) = 168 and since the center of this group consists of matrices of the form cI with c 6= 0, with the base field of F2 , we note that GL3 (F2 ) = PGL3 (F2 ). Furthermore, GL3 (F2 ) means invertible matrices, so nonzero determinant, but again withe the base field of order 2, this group is equal to SL3 (F2 ). Hence, PGL3 (F2 ) = PSL3 (F2 ). In Exercise 8.3.17, we calculated that the order of the group of symmetries of the Fano plane is 168. A group of symmetries on the Fano plane is isomorphic to a group of symmetries of P(F32 ) that preserves the property ~a + ~b + ~c = ~0 where ~a = (a1 : a2 : a3 ) and similarly for ~b and ~c. However, as we have seen, PSL3 (F2 ) = GL3 (F2 ) and for all M ∈ GL3 (F2 ), we have (M~a) + (M~b) + (M~c) = M (~a + ~b + ~c) = M ~0 = ~0. Hence, every element in PSL3 (F2 ) acting on P(F32 ) preserves lines. Hence, PSL3 (F2 ) is isomorphic to the group of symmetries of the Fano plane.

9.3 – Semidirect Product Exercise: 1 Section 9.3 Question: Prove that Sn = An oϕ Z2 for some appropriate ϕ. Solution: Write Z2 = hx | x2 = 1i and consider the homomorphism ϕ : Z2 → Aut(An ) defined by ϕ(xa )(σ) = (1 2)a σ(1 2)−a for all σ ∈ An . Since An ESn , conjugation on An by an element in Sn is an automorphism of An . Then elements of the semidirect product An oϕ Z2 are pairs (σ, xa ) ∈ An × Z2 with the operation (σ, xa )(τ, xb ) = (σϕ(xa )(τ ), xa+b ). Consider the following function f : Sn → An oϕ Z2 defined by ( (σ, 1) if σ ∈ An f (σ) = (σ(1 2), x) if σ ∈ Sn − An . It is easy to see that this is a bijective function. Setting τ 0 = (1 2)τ (1 2), we notice that  (στ, 1)    (στ (1 2), x) f (στ ) =  (στ 0 (1 2), x)    (στ 0 , 1)

if σ ∈ An and τ ∈ An if σ ∈ An and τ ∈ / An if σ ∈ / An and τ ∈ An if σ ∈ / An and τ ∈ / An

while   (στ, 1)  (στ (1 2), x · 1) f (σ)f (τ ) = (σ(1 2)τ, 1 · x)    (σ(1 2)τ (1 2), x · x)

 if σ ∈ An and τ ∈ An (στ, 1)    (στ (1 2), x) if σ ∈ An and τ ∈ / An =  if σ ∈ / An and τ ∈ An (στ 0 (1 2), x)    if σ ∈ / An and τ ∈ / An (στ 0 , 1)

Hence, f is a homomorphism and thus a bijection.

if σ ∈ An and τ ∈ An if σ ∈ An and τ ∈ / An if σ ∈ / An and τ ∈ An if σ ∈ / An and τ ∈ / An .

9.3. SEMIDIRECT PRODUCT

473

Exercise: 2 Section 9.3 Question: Find a nontrivial homomorphism ϕ : Z3 → Aut(Z2 ⊕ Z2 ). Prove that the resulting semidirect product is (Z2 ⊕ Z2 ) o Z3 ∼ = A4 . 0 1 ∼ Solution: The automorphism group of Z2 ⊕Z2 is Aut(Z2 ⊕Z2 ) = GL2 (F2 ). Consider the element A = . 1 1 It is easy to check that this matrix has order 3 in GL2 (F2 ). Hence, there is a homomorphism ϕ : Z3 → GL2 (F2 ) defined by ϕ(z k ) = Ak , where z is the generator of Z3 . Since Z2 ⊕ Z2 ∼ = hx, y | x2 = y 2 = 1, xy = yxi, the semidirect product induced by ϕ is (Z2 ⊕ Z2 ) oϕ Z3 = hx, y | x2 = y 2 = z 3 = 1, xy = yx, zxz −1 y, zyz −1 = xyi. We can check that this is isomorphic to A4 by citing the table in Appendix A.2. We note that there are only 3 nonabelian groups of order 12, namely D6 , A4 , and Z3 o Z4 . In the semidirect product (Z2 ⊕ Z2 ) oϕ Z3 , the Sylow 2-subgroup is normal and not cyclic while the Sylow 3-subgroup is not normal. Among, D6 , A4 , and Z3 o Z4 , only A4 satisfies those two properties about its Sylow subgroups. Hence, (Z2 ⊕ Z2 ) oϕ Z3 ∼ = A4 . Another approach is to point out that an explicit isomorphism ψ from (Z2 ⊕ Z2 ) oϕ Z3 to A4 has ψ(x) = (1 2)(3 4) ψ(y) = (1 3)(2 4) ψ(z) = (2 3 4) and that these images in A4 generate all of A4 and satisfy the same relations as x, y, and z. Hence, ψ defines an isomorphism by the Extension Theorem on Generators. Exercise: 3 Section 9.3 Question: Let G be an arbitrary group and let n be a positive integer. Let ϕ1 and ϕ2 be homomorphisms Zn → Aut(G) such that ϕ1 (Zn ) and ϕ2 (Zn ) are conjugate subgroups in Aut(G). Suppose that Zn is generated by z. a) Prove that there exists an automorphism ψ ∈ Aut(G) and a ∈ U (n) such that ϕ2 (z)a = ψ ◦ ϕ1 (z) ◦ ψ −1 . b) Prove that the function f : G oϕ1 Zn → G oϕ2 Zn defined by f (g, x) = (ψ(g), xa ) is an isomorphism. Solution: Let G be an arbitrary group and let n be a positive integer. a) The subgroup ϕ1 (Zn ) is generated by ϕ1 (z) and similarly for ϕ2 . If ϕ1 (Zn ) and ϕ2 (Zn ) define conjugate subgroups in Aut(G), then there exists a conjugate of ϕ1 (z) that generates ϕ2 (Zn ). Thus ψ ◦ ϕ1 (z) ◦ ψ −1 = ϕ2 (z)a for some a that is relatively prime to |ϕ1 (z)|. Without knowing more about ϕ1 , we do nonetheless know that the order of ϕ1 (z)| divides n. Hence, imposing the requirement that a is relatively prime to n is sufficient, so we can consider a as an element on U (n). b) Consider the function f : G oϕ1 Zn → G oϕ2 Zn defined by f (g, x) = (ψ(g), xa ). Since ψ ∈ Aut(G) and a is relatively prime to n, the function f is a bijection. Let (g, x) and (h, y) be in : G oϕ1 Zn . Then f ((g, x)(h, y)) = f (gϕ1 (x)(h), xy) = (ψ(gϕ1 (x)(h)), (xy)a ) = (ψ(g)(ψ ◦ ϕ1 )(x)(h), xa y a ) = (ψ(g)ϕ2 (x)a (ψ(h)), xa y a ) = (ψ(g)ϕ2 (xa )(ψ(h)), xa y a ) = f (g, x)f (h, y). Hence, the function f is a bijection that is also a homomorphism and so provides an isomorphism between G oϕ1 Zn and G oϕ2 Zn . Exercise: 4 Section 9.3 Question: Let P and Q be groups. Let ϕ1 and ϕ2 be homomorphisms Q → Aut(P ) such that there exists an automorphism ψ ∈ Aut(Q) such that ϕ1 ◦ ψ = ϕ2 . Show that the function Ψ : P oϕ1 Q → P oϕ2 Q defined by Ψ(a, b) = (a, ψ −1 (b)) is an isomorphism. Solution: Consider the function Ψ : P oϕ1 Q → P oϕ2 Q defined by Ψ(a, b) = (a, ψ −1 (b)). This function is obviously a bijection with inverse function Ψ−1 (a, b) = (a, ψ(b)). Now let (p1 , q1 ) and (p2 , q2 ) be in P oϕ1 Q. We have Ψ((p1 , q1 )(p2 , q2 )) = Ψ(p1 ϕ1 (q1 )(p2 ), q1 q2 ) = (p1 ϕ1 (q1 )(p2 ), ψ −1 (q1 q2 )) = (p1 ϕ2 (ψ −1 (q1 ))(p2 ), ψ −1 (q1 )ψ −1 (q2 )) = (p1 , ψ −1 (q1 ))(p2 , ψ −1 (q2 )) = Ψ(p1 , q1 )Ψ(p2 , q2 ).

474

CHAPTER 9. CLASSIFICATION OF GROUPS

Hence, Ψ is also a homomorphism and thus an isomorphism. Consequently, P oϕ1 Q ∼ = P oϕ2 Q. Exercise: 5 Section 9.3 Question: Suppose that p < q are primes with p | (q − 1). Prove that all nontrivial homomorphisms ϕ : Zp → Aut(Zq ) lead to isomorphic semidirect product Zq oϕ Zp . [That is why the unique nonabelian group of order pq is written Zq o Zp .] Solution: Let p < q be primes with p | (q − 1). Recall from Example 8.5.10, that this is the situation when there exist nonabelian groups of order pq. Let ϕ1 and ϕ2 be nontrivial homomorphisms from Zp into Aut(Zq ). Since Zp is a simple group, since ker ϕi 6= Zp then ker ϕi = {1} (since ker ϕi is a normal subgroup of Zp ). Thus ϕ1 (Zp ) and ϕ2 (Zp ) are subgroups of order p. By Sylow’s Theorem, ϕ1 (Zp ) and ϕ2 (Zp ) are in conjugate Sylow p-subgroups. Consequently, ϕ1 (Zp ) and ϕ2 (Zp ) are conjugate groups in Aut(Zq ). By Exercise 9.3.3, there is an isomorphism between Zq oϕ1 Zp and Zq oϕ2 Zp . Exercise: 6 Section 9.3 Question: Prove Proposition 9.3.6. Solution: Suppose that G = H oϕ K, where ϕ is a homomorphism ϕ : K → Aut(H). Consider the subset H̃ = {(h, 1) | h ∈ H} and the function ψH : H̃ → H defined by ψH (h, 1) = h. The function ψH is a bijection. Note that ϕ(1) must be the identity function in Aut(H). Furthermore, for h1 , h2 ∈ H, we have ψH ((h1 , 1)(h2 , 1)) = ψH (h1 ϕ(1)(h2 ), 1 · 1) = ψH (h1 idH (h2 ), 1) = ψH (h1 h2 , 1) = h1 h2 = ψH (h1 , 1)ψH (h2 , 1). Thus, ψH is a homomorphism as well. Thus H̃ ∼ = H. Consider the subset K̃ = {(1, k) | k ∈ K} and the function ψK : K̃ → K defined by ψK (1, k) = k. The function ψK is a bijection. Furthermore, for k1 , k2 ∈ H, we have ψK ((1, k1 )(1, k2 )) = ψK (1 · ϕ(k1 )(1), k1 k2 ) = ψK (1, k1 k2 ) = k1 k2 = ψK (1, k1 )ψK (1, k2 ). Thus, ψK is a homomorphism as well. Thus K̃ ∼ = K. Now, let (h1 , k1 ) ∈ H̃ E H oϕ K and let (h, 1) ∈ H̃. We have −1 (h1 , k1 )(h, 1)(h1 , k1 )−1 = (h1 ϕ(k1 )(h), k1 )(ϕ(k1−1 )(h−1 1 ), k1 ) −1 = (h1 ϕ(k1 )(h)(ϕ(k1 ) ◦ ϕ(k1−1 ))(h−1 1 ), k1 k1 )

= (h1 ϕ(k1 )(h)h−1 1 , 1), because ϕ(k1−1 ) is the inverse automorphism from ϕ(k1 ) so that ϕ(k1 ) ◦ ϕ(k1−1 ) = idH . In particular, we see that (h1 , k1 )(h, 1)(h1 , k1 )−1 ∈ H̃, which shows that H̃ E G. Furthermore, the function f : G → K defined by f (h, k) is a homomorphism whose kernel is H̃. By the First Isomorphism Theorem, G/H̃ ∼ = K. Exercise: 7 Section 9.3 Question: Fix a positive integer n and let F be a field. Let T be the subgroup of GLn (F ) of upper triangular matrices. Let D ≤ T be the subgroup of diagonal matrices and let U = {g ∈ T | gii = 1 for all i}. Prove that T is a semidirect product U o D. Explicitly describe the relevant homomorphism ϕ : D → Aut(U ) for n = 2 and n = 3. Solution: In Exercise 4.2.8, we saw that U E T with U arising as the kernel of a homomorphism. Furthermore, U ∩ T = {I} and also U D = T because 1 0  0  .  .. 0 

a12 a22

1 0 .. . 0

a13 a33 a23 a33

1 .. . 0

··· ··· ··· .. . ···

a1n   a11 ann a2n   ann   0 a3n   0 ann  

 .. ..  .  . 0 1

0 a22 0 .. .

0 0 a33 .. .

··· ··· ··· .. .

0

0

···

  0 a11  0    0  0  = 0 ..   .. .   . ann 0

a12 a22 0 .. .

a13 a23 a33 .. .

··· ··· ··· .. .

0

0

···

 a1n a2n   a3n  . ..  .  ann

By Proposition 9.3.9, T = U oϕ D where for all k ∈ D and all h ∈ U , the homomorphism ϕ : D → Aut(U ) is given by ϕ(k)(h) = khk −1 .

9.3. SEMIDIRECT PRODUCT

475

With n = 2, the ϕ homomorphism is

s 0

s 0

1 0

= =

0 1 b s t 0 1 0 1 0 sb s t 0 1t sb t . 1

−1 0 t

With n = 3, the ϕ homomorphism is 

s 0 0  s = 0 0  1 = 0 0

  0 1 b c s 0  0 1 f  0 u 0 0 1 0   1 0 0 sb sc s t tf   0 1t 0  0 u 0 0 u1  sb sc

0 t 0

t

1 0

0 t 0

−1 0 0 u

u tf  . u

1

Exercise: 8 Section 9.3 Question: Prove that the symmetries of the cube (including reflections) is a group of the form S4 o Z2 . Solution: Let G be the groups of symmetries of a cube. These include reflections through planes. The rigid motions, a subgroup R of G, only are compositions of rotations. We can count the number of symmetries of a cube by first deciding the number of vertices to which 1 could go (8 possibilities) and multiplying this by the possibilities for the 3 vertices that are adjacent to 1 (a total of 3! = 6) because the vertices adjacent to 1 must be mapped to the vertices adjacent to the image of 1. Once the images of these four vertices are decided, the rest of the image of the cube follows. Hence, |G| = 48 and G contains the group of rigid motions of order 24, isomorphic to S4 as a subgroup. The group of symmetries of the cube is a subgroup of O(3), the group of orthogonal transformations on R3 . The determinant function det : G → {−1, 1} is a surjective homomorphism, where the latter set is equipped with multiplication. We point out that the group of rigid motions of the cube correspond to Ker det and hence is a normal subgroup of G, and by the First Isomorphism Theorem, G/ Ker det ∼ = Z2 . Let ` be a reflection through a plane a consider the 2 elements subgroup H = h`i. Since H ∩ R = {1}, we see that |RH| = |R| |H| = 48. Hence, RH = G. Hence, by Proposition 9.3.9, G ∼ =RoH ∼ = S4 o Z2 . Exercise: 9 Section 9.3 Question: Prove that there are 4 distinct homomorphisms from Z2 into Aut(Z8 ). Show that the resulting semidirect products are Z8 ⊕ Z2 , D8 , the quasidihedral group QD16 (Exercise 3.8.9) and the modular group (Exercise 4.3.17). Solution: If Z8 is generated by the cyclic element of order 8, the automorphisms of Z8 consist of functions defined by ψa (z) = z a , where a ∈ U (8). Now, U (8) = {1, 3, 5, 7}, and it is easy to check that U (8) is isomorphic to Z2 ⊕ Z2 . Let Z2 be generated by y. We note that ψa ◦ ψa = ψ1 = id for all a ∈ U (8). Consequently, there are four homomorphisms, ϕ : Z2 → Aut(Z8 ) defined by ϕ(y) = ψ1 , ϕ(y) = ψ3 , ϕ(y) = ψ5 , and ϕ(y) = ψ7 . These four homomorphisms lead respectively to the following four groups hy, z | y 2 = z 8 = 1, yzy −1 = zi ∼ = Z8 ⊕ Z2 2 8 −1 3 ∼ hy, z | y = z = 1, yzy = z i = QD16 hy, z | y 2 = z 8 = 1, yzy −1 = z 3 i the modular group hy, z | y 2 = z 8 = 1, yzy −1 = z 7 i ∼ = D8 We compare this situation to the result of Exercise 9.3.3. We point out that since Aut(Z8 ) is an abelian group, then every subgroup is normal. In other words, no subgroup is conjugate to a distinct subgroup. Consequently,

476

CHAPTER 9. CLASSIFICATION OF GROUPS

the four possible described subgroups in Aut(Z8 ) are not conjugate to each other, so that the semidirect products are not necessarily isomorphic to each other under the result of Exercise 9.3.3. In fact, it is not hard to check that the resulting subgroups are indeed not isomorphic to each other. Exercise: 10 Section 9.3 Question: In this exercise, we prove that is p is an odd prime then U (pk ) is a cyclic group of order pk−1 (p − 1). k−2

a) Prove that if k ≥ 2 and a ∈ Z with p - a, then (1 + ap)p ≡ 1 + apk−1 (mod pk ). b) Deduce that for any a with p - a, the element 1 + ap has order pk−1 in U (pk ). c) By Proposition 7.5.2, U (Fp ) = U (p) is a cyclic group. Show that there exists g ∈ Z that is a generator of U (p) and such that g p−1 6≡ 1 (mod p2 ). d) Prove that a g found in the previous part generates U (pk ) to deduce that U (pk ) is cyclic. Solution: Let p be an odd prime and consider the group U (pk ). We remark right away that U (pk ) has order φ(pk ) = pk−1 (p − 1). a) To solve this first part, we consider n p (pn )! . ordp = ordp (pn − j)! j! j pn j

n for 1 ≤ j ≤ p − 1. More generally, ordp pj remains constant for j between n multiples of p. Hence, to decide the values of ordp pj , we only need to consider how it changes through multiples of p namely, when j = pm. Suppose that j = pα m with p - m. Then n p pn (pn − 1) · · · (pn − j + 2) = 1 · 2 · 3···j − 1 j−1 n n n n n p (p − pα m + 1) p p (p − 1) · · · (pn − j + 1) = = 1 · 2 · 3···j j−1 pα m j n p pn (pn − pα m + 1)(pn − pα m) pn (pn − 1) · · · (pn − j) = = j+1 1 · 2 · 3···j + 1 j−1 pα m(pα m + 1) n n p (p − pα m + 1)(pn−α − m) = . j−1 m(pα m + 1)

We see that pn divides

We deduce that ordp

n p = n − ordp j. j

Now, let k ≥ 2 and let a ∈ Z with p - a. Then pk−2

(1 + ap)

=

k−2 pX k−2

p

j=0

j

j j

a p = 1 + ap

k−1

+

k−2 pX k−2

p

j=2

j

aj pj

However, from our above calculation, k−2 p ordp aj pj = k − 2 − ordp (j) + j. j However, since p is an odd prime (namely, p ≥ 3) and with j ≥ 2, we have j − ordp (j) ≥ 2. We deduce that k−2 (1 + ap)p ≡ 1 + apk−1 (mod pk ). k−1 ∼ b) By the same reasoning as in the previous part, it is easy to see that (1 + ap)p = 1 (mod pk ). Thus, k−2 the order of 1 + ap ∈ U (pk ) divides pk−1 . However, we saw in part (a) that (1 + ap)p ≡ 1 + apk−1 6≡ 1 k k−1 (mod p ). Thus, the order of 1 + ap is exactly p . Show that there exists g ∈ Z that is a generator of U (p) and such that g p−1 6≡ 1 (mod p2 ). Let c be an integer such that c̄ is a generator of U (p). Suppose that c is such that cp−1 ≡ 1 (mod p2 ). Then the integer

9.3. SEMIDIRECT PRODUCT

477

c(1 + p) is still (in congruence) a generator of U (p) and p−1 2 (c(1 + p))p−1 ≡ cp−1 (1 + (p − 1)p + p + · · · + pp−1 ) 2

(mod p2 )

≡ 1 + (p − 1)p ≡1−p and this is not congruent to 1 modulo p2 . Thus, we can always find an integer g such that ḡ is a generator of U (p) and such that g p−1 6≡ 1 (mod p2 ) c) Since g p−1 6≡ 1 (mod p2 ) but g p−1 ≡ 1 (mod p), then g p−1 can be written as g p−1 = 1 + ap for some a that is not divisible by p. Then, by part (b), g p−1 has order pk−1 in U (p). Since gcd(p − 1, pk−1 ) = 1, then g itself has order pk−1 (p − 1). Hence, g is a generator of U (pk ). Exercise: 11 Section 9.3 Question: Determine the isomorphism type of Aut(Z40 ) and express the result in invariant factors form. Solution: We know that Aut(Z40 ) ∼ = U (40) = U (Z/40Z). By the Chinese Remainder Theorem and Corollary 5.6.19, we have Aut(Z40 ) ∼ = U (8) ⊕ U (5) ∼ = Z4 ⊕ Z2 ⊕ Z2 , by the propositions in this section. Exercise: 12 Section 9.3 Question: Determine the isomorphism type of Aut(Z210 ) and express the result in invariant factors form. ∼ U (210) = U (Z/210Z). By the Chinese Remainder Theorem and Solution: We know that Aut(Z210 ) = Corollary 5.6.19, we have Aut(Z20 ) ∼ = U (2) ⊕ U (3) ⊕ U (5) ⊕ U (7) ∼ = Z2 ⊕ Z4 ⊕ Z6 , by the propositions in this section. Exercise: 13 Section 9.3 Question: This exercise guides a proof that Aut(Sn ) = Sn for all n 6= 6. a) Prove that for all ψ ∈ Aut(Sn ) and all conjugacy classes K of Sn , the subset ψ(K) is another conjugacy class. b) Let K be the conjugacy class of transpositions and let K0 be another conjugacy class of elements of order 2 (e.g., cycle type like (a b)(c d)). Prove that |K| 6= |K0 |, unless possibly if n = 6. c) Prove that for each ψ ∈ Aut(Sn ) and for all k with 2 ≤ k ≤ n, we have ψ((1 k)) = (a bk ) for some distinct integers a, b2 , b3 , . . . , bn in {1, 2, . . . , n}. d) Show that the transpositions (1 2), (1 3), . . . , (1 n) generate Sn . e) Deduce that Aut(Sn ) = Inn(Sn ) ∼ = Sn . Solution: Let n be a positive integer different from 6. a) Let ψ ∈ Aut(Sn ) and let K be a conjugacy class of Sn . Each conjugacy class is the orbit of some permutation σ under the action of Sn itself by conjugation. So K = {gxg −1 | g ∈ G}. Then ψ(K) = {ψ(gxg −1 ) | g ∈ G} = {ψ(g)ψ(x)ψ(g)−1 | g ∈ G} = {hψ(x)h−1 | h ∈ G}, since ψ is a bijection on G. Thus ψ(K) is the conjugacy class of ψ(x), another conjugacy class. b) Let K be the conjugacy class of transpositions. Then |K| = n2 . Let K0 be the conjugacy class consisting of a product of k disjoint transpositions (with k ≤ n2 ), then |K0 | =

n(n − 1)(n − 2) · · · (n − 2k + 1) . 2k k!

We have an equality of sizes of these conjugacy classes when n n(n − 1)(n − 2) · · · (n − 2k + 1) = ⇐⇒ 2k−1 k! = (n − 2)(n − 3) · · · (n − 2k + 1). 2 2k k!

478

CHAPTER 9. CLASSIFICATION OF GROUPS We can write this equality as n−2 n−3 n−k n−k−1 n−k−2 n − 2k − 1 ··· ··· = 1. k k−1 2 2 2 2 If k 6= n2 so that n − 2k + 1, then the equality cannot hold because all of the fractions are greater than 1. So now we suppose that n = 2k. Then the equality we are trying to solve is 2k−1 k! = (2k −2)(2k −3) · · · 2·1 ⇐⇒ k2k−1 (k −1)! = (2k −2)(2k −3) · · · 2·1 ⇐⇒ k = (2k −3)(2k −5) · · · 3·1.

We find that this leads to an equality for k = 3 but that the right-hand side grows much faster than the left-hand side and hence, the equality never holds for k > 3. This proves that if n 6= 6, then |K0 | = 6 |K|. c) Let ψ ∈ Aut(Sn ) and let k be an integer 2 ≤ k ≤ n. By part (b), working with the assuming that n 6= 6, the ψ maps the conjugacy class of transpositions back into itself bijectively. Hence, ψ maps a transposition to another transposition. Furthermore, if (a b) and (c d) are disjoint, then ψ((a b)) and ψ((c d)) must also be disjoint because otherwise, ψ would map the element (a b)(c d) to ψ((a b))ψ((c d)), which is a 3-cycle. Consequently, ψ must map disjoint transpositions to disjoint transpositions and also transpositions that share an entry to transpositions that also share an entry. Hence, all the image transpositions ψ((1 k)) must share a common entry, and common to all image transpositions because we can compare all of them ψ((1 k)) pairwise. We deduce that ψ((1 k)) = (a bk ) for some distinct integers a, b2 , b3 , . . . , bn in {1, 2, . . . , n}. d) Every k-cycle (a1 a2 · · · ak ) can be written as (a1 a2 · · · ak ) = (1 a1 )(1 ak )(a1 ak−1 ) · · · (1 a2 )(1 a1 ). Thus, since every permutation is a product of disjoint cycles, it can be written as a product of transpositions of the form (1 k) with 2 ≤ k ≤ n. e) By part (d), an automorphism ψ ∈ Aut(Sn ) is completely determined by where it maps the n − 1 transpositions (1 2), (1 3), . . . , (1 n). Part (c) showed that Aut(Sn ) is isomorphic to a subgroup of Sn since ψ is completely determined by a permutation on {1, 2, . . . , n}. However, we know that the permutation τ such that τ (1) = a, and τ (k) = bk for all 2 ≤ k ≤ n induces an inner automorphism on Sn via σ 7→ τ στ −1 and also has the effect of mapping (1 k) to (a bk ). Hence, Aut(Sn ) ≤ Inn(Sn ), which means that Aut(Sn ) = Inn(Sn ) = Sn . Exercise: 14 Section 9.3 Question: Let G be a group. Consider the homomorphism ϕ : G → Aut(G) defined by ϕ(g)(x) = gxg −1 . Prove that the resulting semidirect product G oϕ G is equal to G ⊕ G if and only if G is abelian. Find a presentation for D3 oϕ D3 . Solution: By Proposition 9.3.8, the semidirect product G oϕ G with the given ϕ is isomorphic to G ⊕ G if and only if ϕ is the identity homomorphism. However, ϕ(g)(x) = gxg −1 = x for all g, x ∈ G if and only if G is abelian. With this definition of ϕ, we have D3 oϕ D3 = hr, ρ, s, σ | r3 = ρ3 = s2 = σ 2 = 1, srs = r2 , σρσ = ρ2 , ρrρ2 = r, σsσ = s, ρsρ2 = sr, σrσ = r2 i.

Exercise: 15 Section 9.3 Question: Give a presentation for a nonabelian semidirect product (Z7 ⊕ Z7 ) oϕ Z3 . Solution: We know from Proposition 9.3.16 that Aut(Z7 ⊕ Z7 ) ∼ = GL2 (F7 ). There are a few options for a nonabelian semidriect product as desired. Let Z3 be generated by the element y. Then with ϕ defined by 2 0 ϕ(y) = 0 1 it is easy to see that this matrix has order 3 in GL2 (F7 ). Then the resulting semidirect product has the presentation hz1 , z2 , y | z17 = z27 = y 3 = 1, z1 z2 = z2 z1 , yz1 y −1 = z12 , yz2 y −1 = z2 i.

9.3. SEMIDIRECT PRODUCT

479

As another example, with ϕ defined by ϕ(y) =

6 1

1 4

it is easy to see that this matrix has order 3 in GL2 (F7 ). Then the resulting semidirect product has the presentation hz1 , z2 , y | z17 = z27 = y 3 = 1, z1 z2 = z2 z1 , yz1 y −1 = z16 z2 , yz2 y −1 = z1 z24 i. 2 0 6 1 However, in this particular case, since and are conjugate matrices in GL2 (F7 ), by Exercise 9.3.3, 0 1 1 4 these two groups are isomorphic. Exercise: 16 Section 9.3 Question: Let ϕ : Z4 → S5 be the homomorphism that sends the generator x of Z4 to ϕ(x) = (1 2 3 4). Exercise 9.3.13 showed that Aut(S5 ) = Inn(S5 ) = S5 . Let G = S5 oϕ Z4 . Perform the following calculations in G. a) ((1 4 3)(2 5), x2 ) · ((2 4 5 3), x). b) ((1 4 3)(2 5), x2 )−1 . c) ((1 3 5 2 4), x3 ) · ((1 3)(2 4), 1) · ((1 3 5 2 4), x3 )−1 . Solution: Consider the group G = S5 oϕ Z4 with ϕ : Z4 → Aut(S5 ) ∼ = S5 defined by ϕ(x) = (1 2 3 4), where x is the generator of Z4 . Since Aut(Sn ) = Inn(Sn ), this means more precisely that ϕ(x)(σ) = (1 2 3 4)σ(1 2 3 4)−1 for all σ ∈ S5 . We perform the following calculations a) Since ϕ(x2 ) = ϕ(x)2 = ϕ(x) ◦ ϕ(x), then ϕ(x2 )(σ) = (1 2 3 4)2 σ(1 2 3 4)−2 . Thus ((1 4 3)(2 5), x2 ) · ((2 4 5 3), x) = ((1 4 3)(2 5)(1 3)(2 4)(2 4 5 3)(1 3)(2 4), x3 ) = ((1 3)(4 5), x3 ). b) We use the proof of Proposition 9.3.4 which provides a formula for the inverse element so ((1 4 3)(2 5), x2 )−1 = (ϕ(x−2 )((1 3 4)(2 5)), x−2 ) = (ϕ(x2 )((1 3 4)(2 5)), x2 ) = ((1 3)(2 4)(1 3 4)(2 5)(1 3)(2 4), x2 ) = ((1 2 3)(4 5), x2 ). c) Since ϕ(x3 ) = ϕ(x)3 = ϕ(x) ◦ ϕ(x) ◦ ϕ(x), then ϕ(x3 )(σ) = (1 2 3 4)3 σ(1 2 3 4)−3 = (1 4 3 2)σ(1 2 3 4). Thus ((1 3 5 2 4), x3 ) · ((1 3)(2 4), 1) · ((1 3 5 2 4), x3 )−1 = ((1 3 5 2 4)(1 4 3 2)(1 3)(2 4)(1 2 3 4), x3 ) · (ϕ(x)((1 3 5 2 4)), x) = ((1 5 2), x3 ) · ((1 2 3 4)(1 3 5 2 4)(1 4 3 2), x) = ((1 5 2), x3 ) · ((1 2 4 5 3), x) = ((1 5 2)(1 4 3 2)(1 2 4 5 3)(1 2 3 4), 1) = ((1 3 2 4 5), 1).

Exercise: 17 Section 9.3 Question: Let G be any group. We define the holomorph of G as the group Hol(G) = G o Aut(G), where the semidirect product is the natural one where ϕ : Aut(G) → Aut(G) is the identity (not trivial) homomorphism. a) Prove that the holomorph of Zp is a nonabelian group and give a presentation of it. b) Prove that Hol(Z2 × Z2 ) ∼ = S4 . Solution: a) Recall that Aut(Zp ) = U (p) and that this is isomorphic to a cyclic group, Zp−1 . Let a be a generator of U (p). This generator corresponds to the homomorphism ψ(g) = g a on Zp . Then a presentation of Hol(Zp ) is Hol(Zp ) = hx, y | xp = y p−1 = 1, yxy −1 = xa i.

480

CHAPTER 9. CLASSIFICATION OF GROUPS

b) Now consider Hol(Z2 × Z2 ). We know that Aut(Z2 ⊕ Z2 ) ∼ = GL2 (F2 ). This automorphism group has order 6, so the holomorph Hol(Z2 × Z2 ) is a nonabelian group of order 24. In setting up an isomorphism between (Z2 ⊕ Z2 ) o GL2 (F2 ), we wish to make the normal subgroup Z2 ⊕ Z2 correspond to h(1 2)(3 4), (1 3)(2 4)i. We first write Z2 ⊕ Z2 as hx, y | x2 = y 2 = 1, xy = yxi and establish the following mapping f : (Z2 ⊕ Z2 ) o GL2 (F2 ) by f : (1, 1) −→ 1 (x, 1) −→ (1 2)(3 4) (y, 1) −→ (1 3)(2 4) (xy, 1) −→ (1 4)(2 3) Now consider the element of the form (1, g) with g ∈ GL2 (F2 ) and how they act on (x, 1), (y, 1) and (xy, 1). 0 1 The element g = satisfies 1 0 (1, g)(1, 1)(1, g)−1 = (1, g) (1, g)(x, 1)(1, g)−1 = (ϕ(g)(x), 1) = (y, 1) (1, g)(y, 1)(1, g)−1 = (ϕ(g)(y), 1) = (x, 1) (1, g)(xy, 1)(1, g)−1 = (ϕ(g)(xy), 1) = (yx, 1) = (xy, 1) The element in S4 that acts on h(1 2)(3 4), (1 3)(2 4)i in the described way is (2 3). Hence, f (g 1) = (2 3). 0 1 Now, consider g = satisfies 1 1 (1, g)(1, 1)(1, g)−1 = (1, g) (1, g)(x, 1)(1, g)−1 = (ϕ(g)(x), 1) = (y, 1) (1, g)(y, 1)(1, g)−1 = (ϕ(g)(y), 1) = (xy, 1) (1, g)(xy, 1)(1, g)−1 = (ϕ(g)(xy), 1) = (x, 1) An element in S4 that acts on h(1 2)(3 4), (1 3)(2 4)i in the described way is (1 2 4). 0 1 0 1 Since (Z2 ⊕Z2 )oGL2 (F2 ) is generated by the elements (x, 1), (y, 1), 1, and 1, and by 1 0 1 1 0 1 0 1 construction, the stated values f (x, 1), f (y, 1), f 1, and f 1, satisfy the operational 1 0 1 1 properties in S4 as their preimages, then by the Extension Theorem on Generators, we have defined a homomorphism f : Hol(Z2 ⊕ Z2 ) → S4 . We note that Im f contains (1 2 4)(2 3)(1 2 4)−1 = (3 4), (1 2)(3 4)(2 3)(1 2)(3 4) = (1 4), (1 4)(3 4)(1 4) = (1 3), and (1 3)(2 3)(1 3) = (1 2). Since Im f contains (1 2), (1 3), and (1 4) and since we know from previous exercises that S4 can be generated by these three elements, then f is a surjective homomorphism. Since the groups are the same size, we deduce that Hol(Z2 ⊕ Z) ∼ = S4 . Exercise: 18 Section 9.3 Question: Let ρ be the standard permutation representation of S3 acting on {1, 2, 3} and let G = Z5 oρ S3 . Use the presentation of Z5 = hx | x5 = 1i. a) Calculate the product in G of (x, x2 , x, (1 2)) · (x3 , 1, x2 , (1 2 3)). b) Calculate the inverse (x, x2 , x4 , (1 3))−1 . c) Calculate the general conjugate (xa , xb , xc , σ) · (xp , xq , xr , 1) · (xa , xb , xc , σ)−1 . Solution: Let ρ be the standard permutation representation of S3 acting on {1, 2, 3} and let G = Z5 oρ S3 . In particular, this means that in the Wreath product Z5 oρ S3 , we have ((a, b, c), σ) · ((d, e, f ), τ ) = ((a b c)(σ · (d, e, f )), στ ), where σ · (d, e, f ) is the permuted triple where d is put in the σ(1)th entry, e is put in the σ(2)th entry, and f is put in the σ(3)th entry.

9.3. SEMIDIRECT PRODUCT

481

a) We have in G, (x, x2 , x, (1 2)) · (x3 , 1, x2 , (1 2 3)) = ((x, x2 , x)(1, x3 , x2 ), (1 2)(1 2 3)) = (x, 1, x3 , (2 3)). b) In G, we have (x, x2 , x4 , (1 3))−1 = ((1 3) · (x4 , x3 , x), (1 3)) = (x, x3 , x4 , (1 3)). c) For the general conjugate, we have (xa , xb , xc , σ) · (xp , xq , xr , 1) · (xa , xb , xc , σ)−1 = ((xa , xb , xc )(σ · (xp , xq , xr )), σ) · ((σ −1 · (x−a , x−b , x−c )), σ −1 ) = ((xa , xb , xc )(σ · (xp , xq , xr ))(x−a , x−b , x−c ), 1) = ((σ · (xp , xq , xr )), 1), where the last inequality follows because Z5 ⊕ Z5 ⊕ Z5 is abelian. Hence, this conjugate simply permutes the elements in the first part of the pair according to the permutation σ. Exercise: 19 Section 9.3 Question: Give a presentation for the group Z5 o Z3 . Solution: The notation Z5 o Z3 assume that we view Z3 as acting on itself by left multiplication. This is the same as Z3 acting transitively on a set of 3 elements. So Z5 o Z3 as the form (Z5 ⊕ Z5 ⊕ Z5 ) o Z3 where the generator in Z3 acts on the triple (a, b, c) ∈ Z53 by permuting through the entries with a 3-cycle. We can give the following presentation Z5 o Z3 = hx y z t | x5 = y 5 = z 5 = t3 = 1, xy = yx, yz = zy, xz = zx, txt−1 = y, tyt−1 = z, tzt−1 = xi.

Exercise: 20 Section 9.3 Question: Let n be a positive integer and let d be a nontrivial divisor. Show that the largest subgroup of Sn acting naturally on X = {1, 2 . . . , n} that has n/d blocks of size d is a wreath product Sd o Sn/d . Calculate the order of this group. Solution: Let G be a group that acts on {1, 2, . . . , n} that has n/d blocks of size d. The largest group (viewed as a subgroup of Sn ) with this property, must have the full symmetric group on each block and involve the full symmetric group to permute the blocks. The group that gives the full symmetric group within each block is n/d Sd = Sd ⊕ Sd ⊕ · · · ⊕ Sd . To change between blocks, the group is (Sd ⊕ Sd ⊕ · · · ⊕ Sd ) o ϕSn/d n/d

where ϕ : Sn/d → Aut(Sd

) is the homomorphism defined by ϕ(σ)(τ1 , τ2 , . . . , τn/d ) = (τσ−1 (1) , τσ−1 (2) , . . . , τσ−1 (n/d) ).

This has the structure of a wreath product. The order of this group is (d!)n/d (n/d)!. (The reader should observe that this is less than n!.) Exercise: 21 Section 9.3 Question: Prove that Zp o Zp is a nonabelian group of order pp+1 that is isomorphic to the Sylow p-subgroup of Sp2 . (See Exercise 8.5.3.) Solution: The group Zp oZp is a semidirect product of the form Zpp oϕ Zp for a certain ϕ. Hence, |Zp oZp | = pp+1 and is generated by p + 1 elements of order p, with p of the generators commuting among each other. Let us use z as the generator for Zp and let us write elements in Zp o Zp as ((xa0 , xa1 , . . . , xap−1 ), xs ) and the homomorphism ϕ satisfies ϕ(xs )(xa0 , xa1 , . . . , xap−1 ) = (xa0−s mod p , xa1−s mod p , . . . , xap−1−s mod p ). In other words, it moves the ith entry forward s entries in the p-tuple of Zpp .

482

CHAPTER 9. CLASSIFICATION OF GROUPS

We refer now to the solution of Exercise 8.5.3. Call P the Sylow p-subgroup of Sp2 as described in that exercise. We know that Zpp oϕ Zp is generated by ((x, 1, 1, . . . , 1), 1), ((1, x, 1, . . . , 1), 1), . . . ((1, 1, 1, . . . , x), 1), ((1, 1, 1, . . . , 1), x). We define a homomorphism ψ : Zpp oϕ Zp → P by how it maps the generators, namely ψ((x, 1, 1, . . . , 1), 1) = σ0 ψ((1, x, 1, . . . , 1), 1) = σ1 .. . ψ((1, 1, 1, . . . , x), 1) = σp−1 ψ((1, 1, 1, . . . , 1), x) = τ. It is clear that all the image elements have order p and that σi σj = σj σi for all 0 ≤ i, j ≤ p − 1. Furthermore, τ was defined so that τ σi τ −1 = σi+1 mod p . This is the same relation as how ((1, 1, . . . , 1), x) acts on the other given generators of Zpp o Zp . Hence, we have defined a homomorphism. In fact, this homomorphism is surjective between groups of the same size so it is an isomorphism.

9.4 – Classification Theorems Exercise: 1 Section 9.4 Question: Prove that there is only one nonabelian group of order 1183 = 7 · 132 . Give an explicit presentation of it. Solution: By Sylow’s Theorem, n1 3 ≡ 1 (mod 13) and n13 divides 7. Thus n13 = 1. Call P the unique Sylow 13-subgroup, which is normal. Also, n7 ≡ 1 (mod 7) and n7 divides 132 , so n7 is either 1 or 132 . If Q is a Sylow 7-subgroup, then P ∩ Q = {1} by Lagrange’s Theorem and thus |P Q| = 1183, so P Q is the whole group. By Proposition 9.3.9, every group of order 1183 is a semidirect product P oϕ Q. Now, there exist two nonisomorphic groups of order 169 = 132 , namely Z169 and Z13 ⊕ Z13 . From Proposition 9.3.14, U (Z169 ) is a cyclic group of order 13 · 12 = 156, which is not divisible by 7. Hence, the only homomorphism from Z7 into Aut(Z169 ) is trivial. Hence, the only semidirect product between Z169 and Z7 2 is the direct sum, thereby giving an abelian group. From Proposition 9.3.16, Aut(Z13 ) ∼ = GL2 (F13 ). Now, 2 2 5 2 | GL2 (F1 3)| = (13 − 1)(13 − 13) = 26208 = 2 · 3 · 7 · 13. By Sylow’s Theorem, all the Sylow 7-subgroups of GL2 (F13 ), which are cyclic subgroups of order 7, are all conjugate to each other. By Exercise 9.3.3, nontriv2 ial homomorphisms ϕ : Z7 → Aut(Z13 ) induce isomorphic semidirect products. Hence, there exists only one 2 nonisomorphic nonabelian semidirect product (Z13 ) o Z7 . 2 In order the give an explicit presentation of (Z13 ) o Z7 , we need to find a matrix of order 7 in GL2 (F13 ). In searching for such a matrix, we can point out that we should only search for such a matrix in SL2 (F13 ), because the homomorphism det : GL2 (F13 ) → U (F13 ) is surjective onto a group of order 12. Hence, an element of order 7 must map to an element of order 1, i.e., to the identity. In our calculations, we find that 2 1 1 1 is a matrix of order 14, so that its square

2 1

2 1 5 = 1 3

3 2

has order 7. This leads to the following presentations of the nonabelian group of order 1183 G = hx, y, z | x13 = y 13 = z 7 = 1, xy = yx, zxz −1 = x5 y 3 , zyz −1 x3 y 2 i.

Exercise: 2 Secion 9.4 Question: Prove that all groups of order 4225 are abelian.

9.4. CLASSIFICATION THEOREMS

483

Solution: Let G be a group of order 4225 = 52 · 132 . By Sylow’s theorem n13 ≡ 1 (mod 13) and n13 | 25, so n13 = 1. Also, n5 ≡ 1 (mod 5) and n5 | 169 so n5 = 1. We deduce that G contains both a normal Sylow 13-subgroup P and a normal Sylow 5-subgroup Q. We know from earlier exercises that groups of order p2 , where p is a prime, are abelian. Furthermore, by the Direct Sum Decomposition Theorem, since P ∩ Q = {1} and P Q = G, then G ∼ = P ⊕ Q. Consequently, G is abelian and there are precisely four options for the group G, given by FTFGAG. Exercise: 3 Section 9.4 Question: Show that every group of order 30 has a normal subgroup of order 15. Then prove that there are exactly 4 nonisomorphic groups of order 30. Solution: Let G be a group of order 30. We point out that the prime decomposition of 30 is 30 = 2 · 3 · 5. By Sylow’s Theorem, we deduce that n5 = 1 or 6, that n3 = 1 or 10 and that n2 is 1, 3, 5 or 15. We note that if n5 = 6 and n3 = 10, then this accounts for 4 × 6 = 24 elements of order 5 and 2 × 10 = 20 elements of order 3. Since 24 + 20 = 44 > 30, then we deduce that G must have either a normal subgroup of order 3 or a normal subgroup of order 5. Let P ∈ Syl3 (G) and Q ∈ Syl5 (G). Since P or Q is a normal subgroup of G, then P Q ≤ G. However, H = P Q is a subgroup of G of order 15. From Example 8.5.10, we know that H ∼ = Z15 . Since 15 = 30/2, then H E G. Now let K ∈ Syl2 (G). Then H ∩ K = {1} and |HK| = 30, so HK = G. By Proposition 9.3.9, G ∼ = H oϕ K for some ϕ : K → Aut(H). By Exercise 9.3.3, in order to find the nonisomorphic groups of order 30, we must find all the homomorphisms ϕ : Z2 → Aut(Z15 ) that give nonconjugat subgroups in Aut(Z15 ). However, Aut(Z15 ) = U (15) = U (3) ⊕ U (5) ∼ = Z2 ⊕ Z4 . More precisely, we can write U (15) as {(−1)a 2b | a = 0, 1, b = 0, 1, 2, 3}. We consider the four homomorphisms ϕ1 , ϕ2 , ϕ3 , ϕ4 defined so that ϕ1 (Z2 ) = {1}, ϕ2 (Z2 ) = {1, −1}, ϕ3 (Z2 ) = {1, 4}, and ϕ4 (Z2 ) = {1, 11}. Note that since U (15) is isomorphic, every conjugate of every subgroup is itself and hence, in light of Exercise 9.3.3, it is likely that these all give nonisomorphic semidirect products. We have Z15 o1 Z2 = hx, y | x15 = y 2 = 1, yxy = xi ∼ = Z30 = Z15 ⊕ Z2 ∼ 15 2 −1 ∼ Z15 o2 Z2 = hx, y | x = y = 1, yxy = x i = D15 Z15 o3 Z2 = hx, y | x15 = y 2 = 1, yxy = x4 i = ha, b, y | a3 = b5 = y 2 yay = a, yby = b4 i ∼ = D 5 ⊕ Z3 Z15 o4 Z2 = hx, y | x15 = y 2 = 1, yxy = x11 i = ha, b, y | a3 = b5 = y 2 yay = a2 , yby = bi ∼ = D 3 ⊕ Z5 .

Exercise: 4 Section 9.4 Question: Prove that all groups of order 14161 = 72 · 172 are abelian. Solution: Let G be a group of order 14161 = 72 · 172 . By Sylow’s theorem n17 ≡ 1 (mod 17) and n17 | 49, so n17 = 1. Also, n7 ≡ 1 (mod 7) and n7 | 172 so n7 = 1. We deduce that G contains both a normal Sylow 17-subgroup P and a normal Sylow 7-subgroup Q. We know from earlier exercises that groups of order p2 , where p is a prime, are abelian. Furthermore, by the Direct Sum Decomposition Theorem, since P ∩ Q = {1} and P Q = G, then G ∼ = P ⊕ Q. Consequently, G is abelian and there are precisely four options for the group G, given by FTFGAG. Exercise: 5 Section 9.4 Question: Consider groups G of order 351 = 33 · 13. Prove that G has a normal Sylow 13-subgroup or a normal Sylow 3-subgroup. Show that there exists a unique nonabelian group of order 351 with a normal subgroup isomorphic to Z33 and give its presentation. Solution: Let G be a group of order 351 = 33 · 13. By Sylow’s Theorem, n13 ≡ 1 (mod 13) and n13 | 27. Hence, n13 = 1 or 27. Furthermore, n3 ≡ 1 (mod 3) and n3 | 13. Hence, n3 = 1 or 13. Suppose that n13 = 27, then G must contain 27 · 12 = 324 elements of order 13. G must also contain 13 × 2 = 26 elements of order 3 but also necessarily more elements than 26 elements whose order is a power of 3. Note that 324 + 26 + 1 = 351, but this has not accounted for the other elements whose order is a power of 3. Hence, G must contain a normal Sylow 13-subgroup or a normal Sylow 3subgroup.

484

CHAPTER 9. CLASSIFICATION OF GROUPS

There may be a variety of groups of order 351. In particular, there is a nonabelian group of order 27, which could arise as the Sylow 3-subgroup of G. However, we now only consider groups of order 351 whose Sylow 3-subgroup is Z33 . Note that Aut(Z33 ) ∼ = GL3 (F3 ), which has order (33 − 1)(33 − 3)(33 − 32 ) = 11232 = 25 · 33 · 13. By Sylow’s Theorem, all the Sylow 13-subgroups, which is in this case are all isomorphic to Z13 , are conjugate to each other in Aut(Z33 ). By Exercise 9.3.3 all resulting nonabelian semidirect products Z33 oZ13 will be isomorphic. In order to find a presentation for Z33 o Z13 , we need to find a matrix of order 13 in GL3 (F3 ). One that works is   0 1 0 1 0 2 . 0 2 2 This leads to the following presentation Z33 o Z13 ∼ = ha, b, c, x | a3 = b3 = c3 = x13 , ab = ba, ac = ca, bc = cb, xax−1 = y, xbx−1 = ac2 , xcx−1 = b2 c2 i.

Exercise: 6 Section 9.4 Question: Classify the groups of order 105. [There are 2 nonisomorphic groups.] Solution: Let G be a group of order 105 = 3 · 5 · 7. By Sylow’s Theorem, n7 can be 1 or 15 and n5 can be 1 or 21. However, we point that it is impossible for n7 = 15 and n5 = 21 because this would imply that G contains 15 · 6 = 90 elements of order 7 and 21 · 4 = 84 elements of order 5 and 90 + 84 > 105. Let P ∈ Syl7 (G) and G ∈ Syl5 (G). Since P or Q is a normal subgroup of G, then P Q = H is a subgroup. Furthermore, |H| = 35 and |G : H| = 3, which is the smallest prime dividing |G|. By Proposition 9.2.4, H EG. Furthermore, if K ∈ Syl3 (G), then H ∩ K = {1} and also HK = G. Thus, by Proposition 9.3.9, G ∼ = H oϕ K for some ϕ : K → Aut(H). By Example 8.5.10, H ∼ = U (5) ⊕ = Z3 . Furthermore, Aut(Z35 ) = U (35) ∼ = Z35 and we also know that K ∼ U (7) = Z4 ⊕Z6 . In particular, Aut(Z35 ) contains exactly two elements of order 3, which means only one subgroup of order 3, namely h11i. Hence, there are two homomorphisms ϕ : Z3 → Aut(Z35 ) that have nonconjugate image subgroups, namely the trivial homomorphism and the homomorphism ϕ(x) = (y 7→ y 11 ), where x ∈ Z3 and y ∈ Z35 . This leads to the two groups Z35 o1 Z3 = hx, y |, x3 = y 35 = 1, xyx−1 = yi ∼ = Z35 ⊕ Z3 ∼ = Z105 Z35 o1 Z3 = hx, y |, x3 = y 35 = 1, xyx−1 = y 11 i.

Exercise: 7 Section 9.4 Question: Classify the groups of order 20. [There are 5 nonisomorphic groups.] Solution: Let G be a group of order 20. By Sylow’s Theorem, n5 = 1 and n2 = 1 or 5. If n5 = n2 = 1, then the Sylow subgroups are normal in G and by the Direct Sum Decomposition Theorem, G ∼ = P ⊕ Q, where P ∈ Syl5 (G) and Q ∈ Syl2 (G). Since there are two nonisomorphic groups of order 4, this possibility leads to the two abelian groups Z5 ⊕ Z4 ∼ = Z20 and Z5 ⊕ Z2 ⊕ Z2 ∼ = Z10 ⊕ Z2 . The other possibility is that n5 = 1, which means that P ∈ Syl5 (G) is a normal subgroup, and that n2 = 5. Then by Proposition 9.3.9, G ∼ = P oϕ Q, where Q ∈ Syl2 (G). We know that P ∼ = Z5 = hx | x5 = 1i, and that ∼ Aut(P ) = Aut(Z5 ) = U (5), which can be generated by 2. Furthermore, since we have already determined the abelian groups, we only consider nontrivial homomorphisms ϕ. We know that there are only two nonisomorphic groups of order 4, namely Z4 and Z2 ⊕ Z2 . • Suppose that Q = Z4 = hy | y 4 = 1i. Then there are two homomorphisms we can consider that are determined by ϕ1 (y)(x) = x2 and ϕ2 (y)(x) = x−1 . These lead to the semidirect products Z5 o1 Z4 = hx, y | x5 = y 4 = 1, yxy −1 = x2 i Z5 o2 Z4 = hx, y | x5 = y 4 = 1, yxy −1 = x−1 i

9.4. CLASSIFICATION THEOREMS

485

• Suppose that Q = Z2 ⊕ Z2 = hy, z | y 2 = z 2 = 1, yz = zyi. Then there are three nontrivial homomorphisms to consider that are determined respectively by (ϕ1 (y)(x), ϕ1 (z)(x) = (x−1 , x), (ϕ2 (y)(x), ϕ2 (z)(x) = (x−1 , x−1 ), and (ϕ3 (y)(x), ϕ3 (z)(x) = (x, x−1 ). Note that ϕ1 (yz)(x) = (ϕ(y) ◦ ϕ(z))(x) = x−1 ϕ2 (yz)(x) = (ϕ(y) ◦ ϕ(z))(x) = (x−1 )−1 = x ϕ3 (yz)(x) = (ϕ(y) ◦ ϕ(z))(x) = x−1 . This shows that of the three homomorphism, they map exactly two elements of Z2 ⊕ Z2 to x → x−1 . Furthermore, Z2 ⊕ Z2 can be generated by any two of its nonidentity elements. Hence, all of three of the homomorphisms produce an isomorphic group Z5 o (Z2 ⊕ Z2 ) = hx5 = y 2 = z 2 = 1, yz = zy, yxy −1 = x−1 , zxz −1 = xi. This group is isomorphic to one we recognize in another way. Note that in Z5 o (Z2 ⊕ Z2 ), since x and z commute, the element r = xz has order 10 and yry = yxyyzy = x−1 z = (xz)−1 . Thus, an alternate presentation of Z5 o (Z2 ⊕ Z2 ) is hr, y | r10 = y 2 = 1, yry = r−1 i. Hence, Z5 o (Z2 ⊕ Z2 ) ∼ = D10

Exercise: 8 Section 9.4 Question: Classify the groups of order 154. Solution: Let G be a group of order 154 = 2 · 7 · 11. By Sylow’s Theorem, n11 = 1 and n7 = 1 or 22. Let P ∈ Syl11 (G). Then P ∼ = Z11 and P E G. If Q ∈ Syl7 (G), then H = P Q ≤ G since P E G. But H is a group of order 77 and by Example 8.5.10, we deduce that H ∼ = Z77 . Suppose that H is generated by an element x. Since |G : H| = 2, then H E G. Also, if K = hyi ∈ Syl2 (G), then HK = G and by Proposition 9.3.9, G ∼ = H oϕ K, for some ϕ : K → Aut(H). Now Aut(Z77 ) = U (77) ∼ = U (7) ⊕ U (11) ∼ = Z6 ⊕ Z11 . So there are four elements in Aut(Z77 ) whose orders divide 2, namely 1, 34, 43, and 76. Then the four possibilities become Z77 o1 Z2 = hx, z | x77 = z 2 = 1, zxz = xi ∼ = Z77 ⊕ Z2 = Z154 Z77 o34 Z2 = hx, z | x77 = z 2 = 1, zxz = x34 i = ha, b, z | a7 = b =11 = z 2 = 1, ab = ba, zaz = a−1 , zbz = bi ∼ = D7 ⊕ Z11 Z77 o43 Z2 = hx, z | x77 = z 2 = 1, zxz = x43 i = ha, b, z | a7 = b =11 = z 2 = 1, ab = ba, zaz = a, zbz = b−1 i ∼ = D11 ⊕ Z7 77 2 −1 ∼ Z77 o76 Z2 = hx, z | x = z = 1, zxz = x i = D77 .

Exercise: 9 Section 9.4 Question: Classify the groups of order 333. Solution: Let G be a group of order 333 = 32 × 37. By Sylow’s Theorem, n37 = 1 so G has a normal Sylow 37-subgroup P = hxi. Let Q be a Sylow 3-subgroup. It is obvious that P ∩ Q = {1} and that |P Q| = 333 so P Q = G. Hence, by Proposition 9.3.9 G ∼ = P oϕ Q for appropriate Q and appropriate ϕ : Q → Aut(P ). However, we know that Aut(P ) = Aut(Z37 ) ∼ = U (37) ∼ = Z36 . In fact, it is easy to check that 2 is a primitive root modulo 37, i.e., a generator of U (37). We know consider various possibilities for Q and for ϕ. • Let Q ∼ = Z9 with Q = hyi. Then there are three different possible homomorphisms ϕ : Q → Aut(P ), depending on the order of ϕ(y). These are ϕ1 (y) = (x 7→ x16 )

because 24 = 16 has order 9 in U (36)

ϕ2 (y) = (x 7→ x26 )

because 212 = 26 has order 3 in U (36)

ϕ3 (y) = (x 7→ x)

because 236 = 1 has order 1 in U (36)

486

CHAPTER 9. CLASSIFICATION OF GROUPS and they give respectively the following groups Z37 o1 Z9 = hx, y | x37 = y 9 = 1, yxy −1 = x16 i, Z37 o2 Z9 = hx, y | x37 = y 9 = 1, yxy −1 = x26 i, ∼ Z37 ⊕ Z9 ∼ Z37 o3 Z9 = hx, y | x37 = y 9 = 1, yxy −1 = xi = = Z333 . • Let Q ∼ = Z3 ⊕ Z3 with Q = hy, zi. The homomorphisms ψ : Z3 ⊕ Z3 → U (37) are uniquely determine by the images ψ(y) and ψ(z). There are nine total options for these images, depending on whether ψ maps y and z to 1, 10, or 26. This leads to the following 9 options G1 = Z37 o1 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yx = xy, zx = xzi ∼ = Z111 ⊕ Z3 G2 = Z37 o2 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x10 , zxz −1 = xi G3 = Z37 o3 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x26 , zxz −1 = xi G4 = Z37 o4 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x, zxz −1 = x10 i G5 = Z37 o5 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x10 , zxz −1 = x10 i G6 = Z37 o6 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x26 , zxz −1 = x10 i G7 = Z37 o7 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x, zxz −1 = x26 i G8 = Z37 o8 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x10 , zxz −1 = x26 i G9 = Z37 o9 (Z3 ⊕ Z3 ) = hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x26 , zxz −1 = x26 i. However, G2 ∼ = G3 by mapping the generator y ∈ G2 to the element y 2 in G3 and the other generators to their corresponding letters. By a same reasoning, G5 ∼ = G6 and G8 ∼ = G9 . Also by a similar reasoning but ∼ by considering the z element, we deduce that G4 = G7 , G5 ∼ = G8 and G6 ∼ = G9 . Now, also G2 ∼ = G4 by a function that interchanges the generators labeled as y and z. We have found that G2 ∼ = G4 ∼ = G7 = G3 ∼

and G5 ∼ = G6 ∼ = G8 ∼ = G9 .

So in fact, we obtain 3 nonisomorphic groups when Q ∼ = Z3 ⊕ Z3 , namely Z111 ⊕ Z3 hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x10 , zxz −1 = xi hx, y, z | x37 = y 3 = z 3 = 1, yz = zy, yxy −1 = x10 , zxz −1 = x10 i

Exercise: 10 Section 9.4 Question: Let p be an odd prime. Classify the groups of order 4p. Solution: Example 9.4.2 dealt with groups of order 12. So we can assume that p is an odd prime that is greater than 3. Let G be a group of order 4p. By Sylow’s Theorem, np ≡ 1 (mod p) and np | 4, so we deduce that np = 1. Thus, there is a unique Sylow p-subgroup P and it is normal. Let Q be a Sylow 2-subgroup. Since P ∩ Q = {1} and thus |P Q| = 4p = |G|, then P Q = G. So by Proposition 9.3.9, G ∼ = P oϕ Q for some homomorphism ϕ : Q → Aut(P ). Note that Aut(P ) = Aut(Zp ) ∼ = U (p), which is cyclic of order p − 1. Suppose that a is a primitive root modulo p, i.e., generator of U (p). Case 1: Q ∼ = Z4 . Write P = hxi and Q = hyi. Then we need to consider homomorphisms ϕ from Z4 → U (p). These are determined by ϕ(y). However, the order of ϕ(y) must divide 4. If p ∼ = 1 (mod 4), then there does exist an element (in fact two elements) in U (p) of order 4, namely b = a(p−1)/4 and also b0 = a3(p−1)/4 . So the homomorphism ϕ(y) = (g 7→ g b ) gives an automorphism of order 4 in Aut(Zp ). This leads to a semidirect product G1 = hx, y | xp = y 4 = 1, yxy −1 = xb i. 0

We could also consider the homomorphism ϕ(y) = (g 7→ g b ) but this defines an isomorphic group to G1 . Note that this group only exists if p ≡ 1 (mod 4). On the other hand, regardless of the congruence class

9.4. CLASSIFICATION THEOREMS

487

of p modulo 4, there is always an element of order 2 and of order 1 in Aut(Zp ), and mapping y to these respectively leads to the following two semidirect products G2 = hx, y | xp = y 4 = 1, yxy −1 = x−1 i G3 = hx, y | xp = y 4 = 1, yxy −1 = xi ∼ = Zp ⊕ Z4 ∼ = Z4p . Case 2: Q ∼ = Z2 ⊕ Z2 . Write P = hxi and Q = hy, zi, with y and z both of order 2. Then we need to consider homomorphisms ϕ from Z2 ⊕ Z2 → U (p). These are determined by ϕ(y) and ϕ(z). These can take the values of 1 or −1. Obvsiouly, one possibility is the trivial homomorphism with ϕ(y) = ϕ(z) = 1. The resulting semidirect product is G4 = Zp oϕ (Z22 ) = hx, y, z | xp = y 2 = z 2 = 1, zy = yz, yxy = x, zxz = xi ∼ = Zp ⊕ Z2 ⊕ Z2 ∼ = Z2p ⊕ Z2 . For a nontrivial ϕ, we first note that ϕ(yz) = ϕ(y)ϕ(z), so ϕ(y), ϕ(z), and ϕ(yz) cannot all three be −1. Since Q can be generated by any subset of two elements of {y, z, yz}, we can assume without loss of generality that ϕ(y) = −1 and that ϕ(z) = 1. Then G5 = Zp oϕ (Z22 ) = hx, y, z | xp = y 2 = z 2 = 1, zy = yz, yxy = x−1 , zxz = xi ∼ = D2p . We get this last isomorphism by observing that xz is an element of order 2p, that all the powers of x and of z arise as powers of xz, and that yxzy = yxyyzy = x−1 z = (zx)−1 = (xz)−1 . So setting r = xz, we have G5 = hr, y | r2p = y 2 = 1, yry = r−1 i.

Exercise: 11 Section 9.4 Question: Let (p, p + 2) be a prime pair, i.e., both p and p + 2 are primes. Consider groups G of order 2p(p + 2). a) Prove that G contains a normal subgroup N of order p(p + 2). b) Prove that G must be a semidirect product of N with another subgroup. c) Prove that G must be isomorphic to one of the following four groups: Z2p(p+2) , Dp ⊕ Zp+2 , Dp+2 ⊕ Zp , or Dp(p+2) . Solution: Let (p, p + 2) be a prime pair, i.e., both p and p + 2 are primes. Let G be a group of order 2p(p + 2). a) By Sylow’s Theorem, np+2 ≡ 1 (mod p + 2) and np+2 | 2p. It is possible that np+2 = 1 but np+2 cannot be 2 or p. Furthermore, if 1 + k(p + 2) = 2p for some integer, then k must be odd; but 1 + k(p + 2) > 2p if k ≥ 3 and with k = 1, 1 + p + 2 = 2p only when p = 3. The prime p = 3 corresponds to classifying groups of order 30, but this was done in Exercise 9.4.3, where we did find that G has a subgroup of order 15. If we now suppose that p > 3, then we deduce that np+2 = 1. Let P ∈ Sylp+2 (G) and let Q ∈ Sylp (G). Then since P E G, we know that N = P Q ≤ G. Also, |G : N | = 2p(p + 2)/p(p + 2), so since N has index of 2, it is a normal subgroup in G. b) Let K ∈ Syl2 (G). Simply by Lagrange’s Theorem, we deduce that N ∩ K = {1} and hence that |N K| = 2p(p + 2), so HK = G. By Proposition 9.3.9, G is isomorphic to N o ϕK for some ϕ : K → Aut(N ). c) Now by Example 8.5.10, we determine that since p - ((p + 2) − 1), that N ∼ = Zp(p+2) . So Aut(N ) = Aut(Zp(p+2) ) = U (p(p + 2)) ∼ = U (p) ⊕ U (p + 2) ∼ = Zp−1 ⊕ Zp+1 . Suppose that N is generated by the element x and that K is generated by y. The possible homomorphisms ϕ : Z2 → Aut(N ) correspond to mapping the generator y to the various four elements in U (p(p + 2)) whose orders divide 2. To get to the desired result, we write N as N = ha, b | ap = bp+2 = 1, ab = bai. The four homomorphisms are determined by ϕ1 (y) = (a 7→ a; b 7→ b) ϕ2 (y) = (a 7→ a−1 ; b 7→ b) ϕ3 (y) = (a 7→ a; b 7→ b−1 ) ϕ4 (y) = (a 7→ a−1 ; b 7→ b−1 ).

488

CHAPTER 9. CLASSIFICATION OF GROUPS These give the following four semidirect products G1 = ha, b, y | ap = bp+2 = y 2 = 1, ab = ba, yay = a, yby = bi ∼ = Zp ⊕ Zp+2 ⊕ Z2 ∼ = Z2p(p+2) p p+2 2 −1 ∼ G2 = ha, b, y | a = b = y = 1, ab = ba, yay = a , yby = bi = Dp ⊕ Zp+2 G3 = ha, b, y | ap = bp+2 = y 2 = 1, ab = ba, yay = a, yby = b−1 i ∼ = Zp ⊕ Dp+2 G4 = ha, b, y | ap = bp+2 = y 2 = 1, ab = ba, yay = a−1 , yby = b−1 i ∼ hx, y | xp(p+2) = y 2 = 1, yxy = x−1 i = Dp(p+2) . =

Exercise: 12 Section 9.4 Question: Let p be an odd prime. Prove that every element in GL2 (Fp ) of order 2 is conjugate to a diagonal matrix with 1 or −1 on the diagonal. Use this result to classify the groups of order 2p2 . Solution: We consider matrices in the group GL2 (Fp ) of order 2. Let A be such a matrix. By basic linear algebra properties, A has at least one eigenvalue λ in a field extension of degree at most 2, so in Fp2 . Let v ∈ F2p be a nonzero eigenvector so that Av = λv. Thus A2 v = Iv = v = λ2 v. We deduce that λ2 = 1. However, this has only two possible roots, ±1 in Fp (and not in a field extension of Fp ), so this gives the possible eigenvalues. If A = ±I, then obviously it is conjugate to a diagonal matrix with 1 or −1 on the diagonal. Now, suppose that A 6= ±I. Then there exists vectors v, w ∈ F2p such that w = Av and w 6= v and w 6= −v. Since A has order 2, then Aw = A2 v = v. But then A(v + w) = Av + Aw = w + v = v + w A(v − w) = Av − Aw = w − v = −(v − w). 1 0 . 0 −1 We now classify groups of order 2p2 , where p is an odd prime. Obviously, if P ∈ Sylp (G), then |G : P | = 2, so P E G. Also, if Q = hzi ∈ Syl2 (G), then P ∩ Q = {1} and P Q = G, so by Proposition 9.3.9, G ∼ = P oϕ Q, for some ϕ : Q → Aut(P ). Now, there are two possibilities for groups of order p2 and these lead to two cases.

Thus, {v, w} is a basis of F2p with respect to which A acts as the diagonal matrix

Case 1: P ∼ = Zp2 . Then Aut(P ) = Aut(Zp2 ) = U (p2 ) ∼ = Zp(p−1) . Since U (p2 ) is cyclic, there is only one element of order 2, namely −1. So the possible homomorphisms ϕ : Q → Aut(P ) ∼ = U (p2 ), correspond to ϕ(z) = 1 or ϕ(z) = −1. These lead to the following two groups: 2 G1 = hx, z | xp = z 2 = 1, zxz = xi ∼ = Zp2 ⊕ Z2 ∼ = Z2p2 2

G2 = hx, z | xp = z 2 = 1, zxz = x−1 i ∼ = Dp2 . to Case 2: P ∼ = GL2 (Fp ). The homomorphisms ϕ : Q → Aut(P ) correspond = Zp ⊕ Zp . Then Aut(P ) ∼ 1 0 2 ϕ(z) = A, where A = I. By the above discussion, we saw A is conjugate to either I, −I, or . 0 −1 By Exercise 9.3.3, we deduce that there can result at most 3 nonisomorphic semidirect products, one corresponding to each of these three matrices. They give the following groups: G3 = hx, y, z | xp = y p = z 2 = 1, xy = yx, zxz = x, zyz = yi ∼ = Z2p ⊕ Zp G4 = hx, y, z | xp = y p = z 2 = 1, xy = yx, zxz = x−1 , zyz = y −1 i ∼ Dp ⊕ Zp . G5 = hx, y, z | xp = y p = z 2 = 1, xy = yx, zxz = x−1 , zyz = yi = This gives us five nonisomorphic groups of order 2p2 . Exercise: 13 Section 9.4 Question: Recall the Heisenberg group H(Fp ) introduced in Exercise 3.2.25. We observe that it is a group of order p3 . Referring to Example 9.4.7, determine the isomorphism type of H(Fp ). Solution: Recall that the Heisenberg group for a field F is the subgroup of GL3 (F ) consisting of upper triangular matrices with 1s on the diagonal. It is not hard to check that this group is nonabelian, so it must be one of the two nonabelian groups described in Example 9.4.7.

9.4. CLASSIFICATION THEOREMS

489

It is not hard to show that the set N of matrices of the form   1 a b 0 1 a 0 0 1 form a subgroup of the Heisenberg group H(Fp ). Note that  1 0 0

a 1 0

−1  b 1 c  = 0 1 0

From this we can also easily calculate that    1 k ` 1 a b 1 0 1 m 0 1 a 0 0 0 1 0 0 1 0

k 1 0

−a 1 0

 ac − b −c  . 1

−1  ` 1 m = 0 1 0

a 1 0

 (k − `)a + b  a 1

and conclude that N E H(Fp ). We note that N is generated by the two matrices     1 0 1 1 1 0 A = 0 1 1 and B = 0 1 0 0 0 1 0 0 1 since

  1 s 2s + t s . As B t = B t As = 0 1 0 0 1 ∼ Zp ⊕ Zp . Now consider the subgroup We also note that Ap = B p = I. Thus N =  1 K = h0 0

0 1 0

 0 1 . 1

It is easy to see that |K| = p and that N ∩ K = {I}. Thus, |N K| = p3 so N K = H(Fp ). By Proposition 9.3.9, H(Fp ) ∼ = N oϕ K for an appropriate ϕ : K → Aut(N ). Now referring to Example 9.4.7, we deduce that H(Fp ) ∼ = (Zp ⊕ Zp ) o Zp , where every possible nontrivial ϕ gives an isomorphic semidirect product, i.e., there is only one nondirect semidirect product of this form. Exercise: 14 Section 9.4 Question: In this exercise, we classify groups G of order 56 = 23 · 7. Let P be a Sylow 2-subgroup and let Q be a Sylow 7-subgroup. a) Prove that G has a normal Sylow 2-subgroup or a normal Sylow 7-subgroup. Deduce that G is P oϕ Q or Q oϕ P for an appropriate ϕ. b) List all (three) abelian groups of order 56. (In the remainder of the exercise, assume G is nonabelian.) c) Suppose that Q is normal. Prove that there is: (i) one group when P ∼ = Z23 ; (ii) two nonisomorphic groups ∼ ∼ when P = Z4 ⊕ Z2 ; (iii) one group when P = Z8 ; (iv) three non-isomorphic groups when P ∼ = D4 ; and (v) two nonisomorphic groups when P ∼ = Q8 . d) Suppose that Q is not a normal subgroup. Prove that the only nontrivial homomorphism ϕ : Q → Aut(P ) occurs when P ∼ = Z23 . Prove that there is only one nonisomorphic semidirect product Z23 o Z7 . Give a presentation of this group. Solution: [In the original text, there is a typographical error in (a) and (c).] Let G be a group of order 56 = 23 · 7. Let P be a Sylow 2-subgroup and let Q be a Sylow 7-subgroup. a) By Sylow’s Theorem, we deduce that n7 is either 1 or 8 and that n2 can be either 1 or 7. If n7 = 8, then the group G contains 6 × 8 = 48 elements of order 7. But then G contains exactly 8 remaining elements, which would be the elements of a single Sylow 2-subgroup. Thus, G cannot also have n2 > 1. Thus G either contains a normal Sylow 7-subgroup or a normal Sylow 2-subgroup. Thus P ∩ Q = {1} and P Q = G so by Proposition 9.3.9, G has the form P o Q or Q o P .

490

CHAPTER 9. CLASSIFICATION OF GROUPS

b) The abelian groups of order 56 are Z56 ,

Z28 ⊕ Z2 ,

and Z14 ⊕ Z2 ⊕ Z2 .

From now on assume that G is nonabelian. c) Suppose that Q = hxi ∼ = Z7 is a normal subgroup. We point out that Aut(Q) = Aut(Z7 ) ∼ = U (7) ∼ = Z6 . In particular, there is only two elements in U (7) whose orders are a power of 2, namely 1 and −1. Now there are five possibilities for P , a group of order 8. These lead to the following cases. Case 1: P ∼ = Z23 . Write P as P = hy, z, wi. If ϕ : P → Aut(Z7 ) is trivial, then we obtain an abelian group, which was already counted. So we consider non trivial ϕ : P → Aut(Q) so the image of one of the generators must be −1, say ϕ(y) = −1. Suppose that ϕ(z) = −1, then ϕ(yz) = (−1)(−1) = 1 but we could have used {y, yz, w} as a generating set instead of {y, z, w}. So without loss of generality, we can suppose that ϕ(z) = 1. The same reasoning applies with the generator w, so we can suppose without loss of generality that ϕ(w) = 1 as well. Then the resulting semidirect product is Z7 o Z23 = hx, y, z, w | x7 = y 2 = z 2 = w2 = 1, yz = zy, yw = wy, zw = wz yxy = x−1 , zxz = x, wxw = xi ∼ = D 7 ⊕ Z2 ⊕ Z2 . Case 2: P ∼ = Z4 ⊕ Z2 . Write P as P = hy, zi, with |y| = 4 and |z| = 2, and zy = yz. A trivial homomorphism would give an abelian group so we do not consider it. There are three nontrivial homomorphisms, given by (ϕ(y), ϕ(z)) being (−1, 1), (−1, −1) or (1, −1). We note that ϕ(yz) = ϕ(y)ϕ(z) and that P could just as easily be generated by {yz, z} instead of {y, z}, which means that the latter two pairs will produce isomorphic semidirect products. Without loss of generality, we only consider the situations where (ϕ(y), ϕ(z)) is (−1, 1) or (1, −1). This leads to two nonisomorphic semidirect products Z7 o1 (Z4 ⊕ Z2 ) = hx, y, z | Z7 o2 (Z4 ⊕ Z2 ) = hx, y, z |

x7 = y 4 = z 2 = 1, yz = zy, yxy −1 = x−1 , zxz −1 = xi x7 = y 4 = z 2 = 1, yz = zy, yxy −1 = x, zxz −1 = x−1 i ∼ = D7 ⊕ Z4 .

Case 3: P =∼ = Z8 . Write P as P = hyi. The only nontrivial homomorphism ϕ : P → Aut(Q), has ϕ(y) = −1 and we get the group Z7 o Z8 = hx, y | x7 = y 8 = 1 yxy −1 = x−1 i. Case 4: P ∼ = D4 . Write P as P = hr, si, with r and s satisfying the relations of D4 . There are four different homomorphisms ϕ based on whether it maps r and s to 1 or −1. We have not considered the situation with the trivial homomorphism yet, so we must include it here. Note that if ϕ(r) = ϕ(s) = −1, then ϕ(rs) = ϕ(r)ϕ(s) = 1. However, P is generated by the set {rs, s} just as well as by the set {r, s}. Hence, without loss of generality, we do not need to consider the situation where ϕ(r) = ϕ(s) = −1, because it will produce an isomorphic semidirect product to the situation when ϕ(r) = 1 and ϕ(s) = −1. These possibilities lead to the following nonisomorphic groups = hx, r, s | x7 = r4 = s2 = 1, srs = r−1 , rxr−1 = x, sxs−1 = xi ∼ = Z7 ⊕ D4 Z7 o1 D4 = hx, r, s | x7 = r4 = s2 = 1, srs = r−1 , rxr−1 = x−1 , sxs−1 = xi Z7 o2 D4 = hx, r, s | x7 = r4 = s2 = 1, srs = r−1 , rxr−1 = x, sxs−1 = x−1 i ∼ D28 = hb, s | b28 = s2 = 1, sbs−1 = b−1 i =

(where b = xr).

Case 5: P ∼ = Q8 . Write P as P = hi, ji, with i and j satisfying the relations of Q8 . There are four homomorphisms ϕ : Q8 → U (7), based on whether ϕ maps i and j to 1 or −1. We have not considered the situation with the trivial homomorphism yet, so we must include it here. However, P can be generated by any two of the set {i, j, ij} and if ϕ is not the trivial homomorphism then exactly two of ϕ(i), ϕ(j) and ϕ(ij) must be −1. Hence, without loss of generality in terms of constructing nonisomorphic semidirect products We only need to consider when (ϕ(i), ϕ(j)) = (1, 1) or (−1, 1). These lead to only two nonisomorphic groups = hx, i, j | x7 = i4 = j 4 = 1, i2 = j 2 , ij = j 3 i, ixi−1 = x, jxj −1 = xi ∼ = Z7 ⊕ Q8 Z7 o Q8 = hx, i, j | x7 = i4 = j 4 = 1, i2 = j 2 , ij = j 3 i, ixi−1 = x−1 , jxj −1 = x−1 i.

9.5. NILPOTENT GROUPS

491

d) Suppose that Q is not a normal subgroup but that instead P is normal. Still set Q = hxi, though this is no longer the unique subgoup of order 7. Note that a nontrivial homomorphism ϕ : Q → Aut(P ) is such that ϕ(x) has order 7. In particular, P must be such that Aut(P ) is divisible by 7. We note that | Aut(Z23 )| = | GL3 (F3 )| = 168 | Aut(Z4 ⊕ Z2 )| = |D4 | = 8 | Aut(Z8 )| = |U (8)| = 4 | Aut(D4 )| = 8 | Aut(Q8 )| = |S4 | = 24

see Exercise 3.8.24.

Hence, the only nontrivial homomorphism from Z7 → Aut(P ), where P is a group of order 8 occurs when P = Z23 . After a little checking, we find that the matrix   0 1 1 0 0 1 1 0 0 has order 7 in GL3 (F2 ). This gives us the following nondirect semidirect product (Z23 ) o Z7 = hx, u, v, w | x7 = u2 = v 2 = w2 = 1, uv = vu, uw = wu, vw = wv, xux−1 = w, xvx−1 = u, xwx−1 = uvi By Exercise 9.3.3 and Sylow’s Theorem (stating the Sylow subgroups are conjugate to each other) we deduce that every nondirect semidirect product of Z23 by Z7 is isomorphic to this one. We have found 13 nonisomorphic groups of order 56.

9.5 – Nilpotent Groups Exercise: 1 Section 9.5 Question: Prove that Zi (G) is a characteristic subgroup of G for all i. Solution: We prove this by induction. Z0 (G) = {1} is obviously a characteristic subgroup of G. Also let ψ ∈ Aut(G), let x ∈ Z1 (G) = Z(G) and consider gψ(x)g −1 for an arbitrary element g ∈ G. Since ψ is surjective, there exists h ∈ G such that g = ψ(h). Then gψ(x)g −1 = ψ(h)ψ(x)ψ(h)−1 = ψ(hxh−1 ) = ψ(x), since x ∈ Z(G). Thus, ψ(x) ∈ Z(G). Hence, Z(G) is a characteristic subgroup of G. Now suppose that Zi (G) is a characteristic subgroup of G for some nonnegative integer i. The group Zi+1 (G) is the subgroup of G such that Zi+1 (G)/Zi (G) ∼ = Z(G/Zi (G)). Let ψ ∈ Aut(G) and let x ∈ Zi+1 (G). This means that xZi (G) ∈ Z(G/Zi (G)). Consider the element xZi (G) in G/Zi (G). Since Zi (G) is a characteristic subgroup of G, ψ(xZi (G)) = ψ({xy | y ∈ Zi (G)}) = {ψ(x)ψ(y) | y ∈ Zi (G)} = {ψ(x)y 0 | y 0 ∈ Zi (G)} = ψ(x)Zi (G). Thus, ψ ∈ Aut(G) restricts to a function ψ̄ : G/Zi (G) → G/Zi (G). Furthermore, this function is obviously a homomorphism and it is also invertible since ψ̄ −1 = ψ −1 . Hence, ψ̄ ∈ Aut(G/Zi (G)). But then, using the same reasoning as we did to show that Z(G) is a characteristic subgroup of G, we see that ψ(x)Zi (G) = ψ̄(xZi (G)) =∈ Z(G/Zi (G)). Thus, ψ(x) ∈ Zi+1 (G). This establishes that Zi (G) is a characteristic subgroup. By induction on i, all the groups of the upper central series are characteristic subgroups of G. Exercise: 2 Section 9.5 Question: Prove that Dn is nilpotent if and only if n is a power of 2. Solution: By Exercise 3.5.27, Z(Dn ) = {ι} if n is odd and Z(Dn ) = {ι, rn/2 } if n is even. Suppose that n = 2k m with 2 - m. If m = 1, i.e., when n is a power of 2, then the observation of Exercise 3.5.27 shows that the upper central series is {ι} ≤ hrn/2 i ≤ hrn/4 i ≤ · · · ≤ hrn/n i ≤ Dn .

492

CHAPTER 9. CLASSIFICATION OF GROUPS i

In particular, Zi (Dn ) = hrn/2 i for 0 ≤ i ≤ k, and Zi (Dn ) = Dn for i ≥ k + 1. Thus, if n = 2k , then Dn is nilpotent with nilpotence class k + 1. i If m 6= 1, then similarly Zi (Dn ) = hrn/2 i for 0 ≤ i ≤ k. In this case, Zk (Dn ) = hrm i. Then Dn /Zk (Dn ) = hr̄, s̄ | r̄m = s̄2 = 1, r̄s̄ = s̄r̄m−1 i ∼ = Dm . Thus, Z(Dn /Zk (Dn )) = {ῑ}, so Zi (Dn ) = Zk (Dn ) = hrm i for i ≥ k. Hence, there is no index c such that Zc (Dn ) = Dn . Consequently, if n is not a power of 2, then Dn is not nilpotent. Exercise: 3 Section 9.5 Question: Prove Proposition 9.5.10. Solution: We repeatedly use the result of Exercise 8.4.14, which states that every p-group has a nontrivial center. We prove the result by induction on k. Groups of order p must be Zp and, by Exercise 8.4.15 groups of order p2 must be isomorphic to Zp2 or Zp ⊕ Zp , Since these are abelian, they are nilpotent and, when k ≥ 2, of nilpotence class k − 1. Now suppose that for some integer n all groups of order pk are nilpotent with k < n. Let G be any group of order pn . Then G has a nontrivial center Z(G) and G = G/Z(G) is a group of order pm with m < n. By the induction hypothesis, G is nilpotent with nilpotence class of at most m − 1. This means that it has an upper central series that terminates {1̄} = Z0 (G) ≤ Z1 (G) ≤ Z2 (G) ≤ · · · (G) ≤ Zs = G where s ≤ m − 1. By the Fourth Isomorphism Theorem, each Zi (G) corresponds uniquely to a subgroup Hi of G that contains Z(G), and we obtain a chain {1} = Z0 (G) ≤ Z(G) = Z1 (G) ≤ H1 ≤ H2 ≤ · · · ≤ Hs = G.

(9.1)

Also, Hi−1 E Hi and by the Third Isomorphism Theorem and then the definition of the upper central series, Hi /Hi−1 ∼ = Zi (G)/Zi−1 (G) ∼ = Z(G/Zi−1 (G)) ∼ = Z(G/Hi−1 ). Also, H0 = Z(G) so (9.1) is the upper central series of G. Furthermore, it has s + 1 nontrivial terms in it. Since s + 1 ≤ m < n, then G is nilpotent of nilpotence class at most n − 1. By induction, the result of Proposition 9.5.10 holds for all p-groups. Exercise: 4 Section 9.5 Question: Prove part (1) of Theorem 9.5.11. Solution: This means we should prove (1)=⇒(2). Let G be a nilpotent group. Let H ≤ G be a subgroup. Obviously, H ≤ NG (H) and in fact H E NG (H). Let {1} = Z0 (G) ≤ Z1 (G) ≤ · · · ≤ Zc (G) = G be the upper central series of G that terminates at G since G is nilpotent. There exists an i with 1 ≤ i ≤ c such that Zi−1 (G) ≤ H but Zi (G) H. Then in G/Zi−1 (G) there corresponds a unique proper subgroup H such that H/Zi−1 (G) = H. Since Zi (G) H, then H does not contain Zi (G)/Zi−1 (G), which is the center of G/Zi−1 (G). Consequently, for all x ∈ H and for all z ∈ Zi (G) = Zi (G)/Zi−1 (G), we have x z = z x. Then, in the group G, for all n1 , n2 , n3 ∈ Zi−1 (G), for all x ∈ H and all z ∈ Zi (G), −1 (zn1 )(xn2 )(zn3 )−1 = z(n1 xn2 n−1 ∈ H. 3 )z

In particular, zxz −1 ∈ H, which implies that Zi (G) ≤ NG (H). Since Zi (G) H, then H is a proper subgroup of NG (H). Exercise: 5 Section 9.5 Question: Prove that if G1 and G2 are nilpotent groups, then G1 ⊕ G2 is also nilpotent. Solution: Let H and K be two groups. Then Z(H ⊕ K) = {(x, y) ∈ H ⊕ K | (x, y)(h, k) = (h, k)(x, y) for all (h, k) ∈ H ⊕ K} = {(x, y) ∈ H ⊕ K | xh = hx and yk = ky for all (h, k) ∈ H ⊕ K} = Z(H) ⊕ Z(K).

9.5. NILPOTENT GROUPS

493

We also remind that by the First Isomorphism Theorem if N1 E G1 and N2 E G2 , then N1 ⊕ N2 E G1 ⊕ G2 and (G1 ⊕ G2 )/(N1 ⊕ N2 ) ∼ = (G1 /N1 ) ⊕ (G2 /N2 ). Let {1} = Z0 (G1 ) ≤ Z1 (G1 ) ≤ Z2 (G1 ) ≤ · · · ≤ Zs (G1 ) = G1 {1} = Z0 (G2 ) ≤ Z1 (G2 ) ≤ Z2 (G2 ) ≤ · · · ≤ Zt (G2 ) = G2 be upper central series of G1 and G2 that terminate at G1 and G2 respectively. Without loss of generality, suppose that s ≥ t. Note that Z0 (G1 ) ⊕ Z0 (G2 ) = {1} ⊕ {1} = {(1, 1)} is the trivial subgroup of G1 ⊕ G2 so this is Z0 (G1 ⊕ G2 ). Now for all i with 1 ≤ i ≤ t, we have Zi (G1 ⊕ G2 ) is the subgroup of G1 ⊕ G2 such that Zi (G1 ⊕ G2 )/Zi−1 (G1 ⊕ G2 ) ∼ = Z((G1 ⊕ G2 )/Zi−1 (G1 ⊕ G2 )) ∼ = Z((G1 ⊕ G2 )/(Zi−1 (G1 ) ⊕ Zi−1 (G2 ))) ∼ = Z((G1 /Zi−1 (G1 )) ⊕ (G2 /Zi−1 (G2 ))) ∼ = Z(G1 /Zi−1 (G1 )) ⊕ Z(G2 /Zi−1 (G2 )) ∼ = Zi (G1 )/Zi−1 (G1 ) ⊕ Zi (G2 )/Zi−1 (G2 ) ∼ = (Zi (G1 ) ⊕ Zi (G2 ))/(Zi−1 (G1 ) ⊕ Zi−1 (G2 )). Thus, Zi (G1 ) ⊕ Zi (G2 ) with 0 ≤ i ≤ s are the terms of the upper central series of G1 ⊕ G2 , which terminates (stabilizes) at s. Exercise: 6 Section 9.5 Question: Prove that subgroups and quotient groups of nilpotent groups are again nilpotent. Deduce also that homomorphic images of a nilpotent group is again a nilpotent group. Solution: Let G be a nilpotent group with an upper central series of {1} = Z0 (G) ≤ Z1 (G) ≤ · · · ≤ Zc (G) = G. Let H be a subgroup of G. Consider the chain of subgroups {1} = Z0 (G) ∩ H ≤ Z1 (G) ∩ H ≤ · · · ≤ Zc (G) ∩ H = H. Now H ∩ Z(G) is not necessarily equal to Z(H) but we do know that H ∩ Z(G) ≤ Z(H). Suppose that H ∩ Zi (G) ≤ Zi (H) for some index i. Let x ∈ H ∩ Zi+1 (G). Then x ∈ H and xZi (G) commutes with everything in G/Zi (G). Thus x−1 g −1 xg ∈ Zi (G) for all x ∈ H and all g ∈ Zi (G). In particular, for all h ∈ H, we have x−1 h−1 xh ∈ Zi (G) ∩ H ≤ Zi (H) so x−1 h−1 xh = 1 in H/Zi (H). By induction on i, wWe deduce that H ∩ Zi (G) ≤ Zi (H) for all i. Most importantly, H ∩ Zc (G) = H ≤ Zi (H), and so this chain of subgroups terminates at H. Let N be a proper normal subgroup of G. First, consider the chain of subgroups N = Z0 (G)N ≤ Z1 (G)N ≤ · · · ≤ Zc (G)N = G. This is a resulting chains of subgroups between N and G. Now consider {1̄} = Z0 (G)N/N ≤ Z1 (G)N/N ≤ · · · ≤ Zc (G)N/N = G/N. We claim that Zi (G)N/N ≤ Zi (G/N ) for each i. This is obviously true for i = 0. Suppose it is true for some i. Let x̄ ∈ Zi+1 (G)N/N . Note that all the cosets of Zi+1 (G)N/N have a representative in Zi+1 (G) so we can assume that x ∈ Zi+1 (G). Then gx = xg in G/Zi (G) for all g ∈ G, so g −1 x−1 gx ∈ Zi (G). Consequently, as elements of G/N , we have g −1 x−1 gxN ∈ Zi (G)N/N . So xZi (G/N ) commutes with every element gZi (G/N ) ∈ (G/N )/Zi (G/N ). Thus, Zi+1 (G)N/N ≤ Zi+1 (G/N ). By induction, this result holds for all i. Most importantly, Zc (G)N/N = G/N = Zc (G/N ) and the nilpotence class of G/N is less than or equal to c. The final result follows from the First Isomorphism Theorem, since Im ϕ ∼ = G/ Ker ϕ so the homomorphic image of G is a quotient group of G. Exercise: 7 Section 9.5 Question: Prove that if G/Z(G) is nilpotent, then G is nilpotent.

494

CHAPTER 9. CLASSIFICATION OF GROUPS

Solution: Suppose that G/Z(G) = G is nilpotent. Then there exists an upper central series {1} = Z0 (G) ≤ Z1 (G) ≤ · · · ≤ Zc (G) = G. By the Fourth Isomorphism Theorem, there exists a chain of subgroups in G {1} ≤ Z(G) ≤ H1 ≤ H2 ≤ · · · ≤ Hc = G,

(9.2)

where Hi /Z(G) = Zi (G). However, we note that Z(G) in (9.2) serves as Z1 (G). Also, by the Third Isomorphism Theorem, Hi /Hi−1 ∼ = Z(G/Zi−1 (G)) = Z(G/Hi−1 ). = Zi (G)/Zi−1 (G) ∼ Thus, we see that Hi = Zi+1 (G) in the upper central series of G. Also, since Hc = G, this series terminates at G. The nilpotence class of G is one more than the nilpotence class of G/Z(G). Exercise: 8 Section 9.5 Question: Prove that if |G| = n2 with n = p1 p2 · · · pr with all pi distinct such that pi - p2j − 1 for all pairs of i 6= j, then G is abelian. Solution: Suppose that r = 2. Let G be a group of order p21 p22 such that pi - p2j − 1 for all pairs of i 6= j. By Sylow’s Theorem, np1 ≡ 1 (mod p1 ) and np1 | p22 . Thus, there exists k with 1 + kp1 = 1, p2 or p22 . Then there exists k such that kp1 is equal to 0, p2 −1, or p22 −1. By the hypothesis, we deduce that k = 0. Hence np1 (G) = 1. Similarly, np2 (G) = 1. By Theorem 9.5.11, G is a a direct sum of its Sylow subgroups. Furthermore, since the Sylow pi -subgroup has order p2i , then it is abelian. The direct sum of two abelian groups is abelian. Exercise: 9 Section 9.5 Question: Prove that a maximal subgroup of a nilpotent group has a prime index. Solution: Let M be a maximal subgroup of a nilpotent group. Then by part (2) of Theorem 9.5.11, since M is a proper subgroup of its normalizer NG (M ), then NG (M ) = G. Thus M E G. Then we can take the quotient G/M . Since M is maximal, then G/M has no nontrivial proper subgroups. Thus, G/M ∼ = Zp , so in particular M has prime index. Exercise: 10 Section 9.5 Question: Let G be a nilpotent group. Prove that x, y ∈ G commute whenever gcd(|x|, |y|) = 1. Solution: By Theorem 9.5.11, G is a direct sum of its Sylow subgroups, G ∼ = P1 ⊕ P2 ⊕ · · · ⊕ Ps , where αs 1 α2 . Write an element of G as x = (x , x , . . . , x ), with x ∈ P . We know that xi is an element |G| = pα p · · · p 1 2 s i i s 1 2 whose order is a power of pi . Furthermore, |x| = lcm(|x1 |, |x2 |, . . . , |xs |) = |x1 | |x2 | · · · |xs | since each of the given orders is relatively prime to any other. Suppose that x, y ∈ G are elements such that gcd(|x|, |y|) = 1. Then for all i ∈ {1, 2, . . . , s}, we have |xi | = 1 or |yi |. Let I be the subset {1, 2, . . . , s} such that |xi | > 1. Then x is in the subgroup ( Pi if x ∈ I H = H1 ⊕ H2 ⊕ · · · ⊕ Hs whereHi = {1} if i ∈ /I and y is in the subgroup K = K1 ⊕ K2 ⊕ · · · ⊕ Ks

( {1} whereKi = Pi

if x ∈ I if i ∈ / I.

Note that by the Direct Sum Decomposition Theorem, G ∼ = H ⊕ K so every element in H commutes with every element in K. Thus, x and y commute. Exercise: 11 Section 9.5 Question: Let F be a finite field and let Un (F ) be the subgroup of GLn (F ) of upper triangular matrices with 1s on the diagonal. Prove Un (F ) is a nilpotent group of nilpotence class n − 1.

9.5. NILPOTENT GROUPS

495

Solution: We point out that every upper triangular matrix with 1s on the diagonal is invertible. Hence, |Un (F )| = |F |n(n−1)/2 However, every finite field is such that |F | = pk for some integer k so Un (F ) is a p-group. By Proposition 9.5.10, Un (F ) is a nilpotent group of nilpotence class at most kn(n − 1)/2 − 1. We need to show that its nilpotence class is n − 1. This amounts to describing the upper central series. Let Z` be the set of matrices Z` = {A ∈ Un (F ) | aij = 0 for all i, j with 0 < j − i < n − `}. We claim that {1} ≤ Z1 ≤ Z2 ≤ · · · Zn−1 = Un (F ) is the upper central series of Un (F ). It is not hard to check that for arbitrary A, B ∈ Un (F ), the ijth entry of AB − BA is j−1 X

aik bkj − bik akj .

(9.3)

k=i+1

Note that the ijth entry of AB − BA is 0 for j ≤ i + 1. Since we wish to find the form of a matrix A that commutes with every possible B, we require all of the entries of AB − BA to be 0 for arbitrary B. However, in the summations in (9.3), only the coefficient a1n does not appear. Otherwise, all the coefficients aij with j > i must be 0. We deduce that, Z(Un (F )) = Z1 , the set of matrices in Un (F ) that have as a free variable only the 1nth position. In Un (F )/Z1 , matrices are considered equivalent up to a change in the 1nth entry. Again referring to (9.3), if we ignore the 1n term in AB − BA, then the coefficients of aij with j − i ≥ n − 1 − ` do not appear. With the arbitrary choices for B, we see that the remaining entries of A must be 0, in order for A to commute with arbitrary B. Thus, Z2 (Un (F )) = {A ∈ Un (F ) | aij = 0 for all i, j with 0 < j − i < n − 2}. The same reasoning continues and mutatis mutandis and establishes the claim. We note that when ` = n − 1, we have Z` = Un (F ). Hence, the nilpotence class of Un (F ) is n − 1. Exercise: 12 Section 9.5 Question: Consider the Heisenberg group H(Fp ). (See Exercises 3.2.25 and 9.4.13.) Since H(Fp ) is a p-group it has a trivial center. Find it. Solution: In Exercise 9.4.13, we prove that H(Fp ) ∼ = (Zp ⊕ Zp ) o Zp , where the normal subgroup isomorphic to Zp ⊕ Zp is the set of matrices of the form   1 a b 0 1 a . 0 0 1 This is not the center of H(Fp ) but the center of a group is contained in it. In particular, it is not hard to show that the set of matrices *1 0 b  +   0 1 0 K= b ∈ Fp 0 0 1 commute with every matrix in H(Fp ). Hence, K is the center.

10 | Modules and Algebra 10.1 – Boolean Algebras Exercise: 1 Section 10.1 Question: Use rules of Boolean algebra to simplify the following expression as much as possible: (x ∧ y) ∨ (x ∧ (y ∨ x))

Solution: We expand and collect the expression to make it simpler. (x ∧ y) ∨ (x ∧ (y ∨ x)) = (x ∧ y) ∨ ((x ∧ y) ∨ (x ∧ x))

distributivity

= (x ∧ y) ∨ ((x ∧ y) ∨ 0) = (x ∧ y) ∨ (x ∧ y)

identity

= (x ∨ x) ∧ y

distributivity

=1∧y =y

identity

Exercise: Boolean Algebras 2 Question: Use rules of Boolean algebra to simplify the following expression as much as possible: ((a ∨ b) ∧ (a ∨ b)) ∨ ((a ∨ b) ∧ b)

Solution: We simplify as much as possible ((a ∨ b) ∧ (a ∨ b)) ∨ ((a ∨ b) ∧ b) = ((a ∧ a) ∨ (a ∧ b) ∨ (b ∧ a) ∨ (b ∧ b)) ∨ ((a ∧ b) ∨ (b ∧ b)) = (0 ∨ (a ∧ b) ∨ (b ∧ a) ∨ 0) ∨ ((a ∧ b) ∨ 0) = (a ∧ b) ∨ (b ∧ a) ∨ (a ∧ b) = (a ∧ b) ∨ (b ∧ (a ∨ a)) = (a ∧ b) ∨ (b ∧ 1) = (a ∧ b) ∨ b

Exercise: 3 Section 10.1 Question: Prove that x̄y ∨ ȳz ∨ z̄x = ȳx ∨ z̄y ∨ x̄z for all x, y, z in a Boolean algebra B. Solution: We prove the desired equality: x̄y ∨ ȳz ∨ z̄x = x̄y1 ∨ ȳz1 ∨ z̄x1 = x̄y(z ∨ z̄) ∨ ȳz(x ∨ x̄) ∨ z̄x(y ∨ ȳ) = x̄yz ∨ x̄yz̄ ∨ xȳz ∨ x̄ȳz ∨ xyz̄ ∨ xȳz̄ = xȳ(z ∨ z̄) ∨ yz̄(x̄ ∨ x) ∨ z x̄(y ∨ ȳ) = ȳx ∨ z̄y ∨ x̄z

Exercise: 4 Section 10.1 Question: Consider the expression in a Boolean algebra (x → y) → ¬(y → z), where we are borrowing symbols from propositional (Boolean) logic. 497

498

CHAPTER 10. MODULES AND ALGEBRA

a) Rewrite only using ∧, ∨, and x for negation. b) If we regard this expression as a Boolean function F (x, y, z), evaluate F (1, 0, 0). Solution: We consider the expression (x → y) → ¬(y → z). a) F (x, y, z) = x̄ ∨ y ∨ ȳ ∨ z = (x ∧ ȳ) ∨ (y ∧ z̄) b) F (1, 0, 0) = (1 ∧ 0̄) ∨ (0 ∧ 0̄) = (1 ∧ 1) ∨ (0 ∧ 1) = 1 ∨ 0 = 1. Exercise: 5 Section 10.1 Question: Prove that the Boolean ring associated to the basic Boolean algebra B = {0, 1} is the ring (Z/2Z, +, ×). Solution: The addition in the Boolean ring associated to the basic Boolean algebra satisfies is defined as def

x + y = (x ∧ ȳ) ∨ (x̄ ∧ y). Thus, 0 + 0 = (0 ∧ 1) ∨ (1 ∧ 0) = 0 ∨ 0 = 0 1 + 0 = (1 ∧ 1) ∨ (0 ∧ 0) = 1 ∨ 0 = 1 0 + 1 = (0 ∧ 0) ∨ (1 ∧ 1) = 0 ∨ 1 = 1 1 + 1 = (1 ∧ 0) ∨ (0 ∧ 1) = 0 ∨ 0 = 0. Consider the function between rings f : (B, +, ∧) → (Z/2Z, +, ×) defined by f (0) = 0̄ and f (1) = 1̄. We’ve now shown that f (x + y) = f (x) + f (y). It is easy to check f (x ∧ y) = f (x) × f (y). Hence, f is a ring isomorphism.

Exercise: 6 Section 10.1 Question: On the interval of real numbers [0, 1], define the binary operators x ∧ y = min(x, y) and x ∨ y = max(x, y), and the function x = 1 − x, using 0 as 0 and 1 as the 1. Prove that ([0, 1], ∧, ∨, ) satisfies all the axioms of a Boolean algebra except the complement laws. Prove also that ([0, 1], ∧, ∨, ) satisfies all the supplementary laws in Proposition 10.1.2. Solution: We study which of the axioms and supplementary laws the quadruple ([0, 1], ∧, ∨, ) satisfies. In what follows, x, y, z are arbitrary elements of [0, 1]. Identity: x ∧ 1 = min(x, 1) = x and x ∨ 0 = max(x, 0) = x. Hence 1 is the identity of ∧ and 0 is the identity of ∨. Complement: min(0.2, 0.8) = 0.2 6= 0 so x ∧ x̄ 6= 0 and similarly max(0.8, 0.2) = 0.8 6= 1 so x ∨ x̄ 6= 1. Complement law does not hold. Associativity: It is easy to see that (x∧y)∧z = min(min(x, y), z) = min(x, y, z) = min(x, min(y, z)) = x∧(y∧z). Also (x ∨ y) ∨ z = max(max(x, y), z) = max(x, y, z) = max(x, max(y, z)) = x ∨ (y ∨ z). This law holds. Commutativity: x ∧ y = min(x, y) = min(y, x) = y ∧ x, and similarly for ∨. This law holds. Distributivity: We have   x if z ≤ x ≤ y (x ∧ y) ∨ z = max(min(x, y), z) = y if z ≤ y ≤ x   z otherwise and also

  x (x ∨ z) ∧ (y ∨ z) = min(max(x, z), max(y, z)) = y   z

if z ≤ x ≤ y if z ≤ y ≤ x otherwise

10.1. BOOLEAN ALGEBRAS

499

so (x ∧ y) ∨ z = (x ∨ z) ∧ (y ∨ z). Also,   x (x ∨ y) ∧ z = min(max(x, y), z) = y   z and also

if y ≤ x ≤ z if x ≤ y ≤ z otherwise

  x if y ≤ x ≤ z (x ∧ z) ∨ (y ∧ z) = max(min(x, z), min(y, z)) = y if x ≤ y ≤ z   z otherwise

so (x ∨ y) ∧ z = (x ∧ z) ∨ (y ∧ z). This law holds. Basic Negations: 0̄ = 1 − 0 = 1 and 1̄ = 1 − 1 = 0. This law holds. Dominance: 0 ∧ x = min(0, x) = 0 and 1 ∨ x = max(1, x) = 1. This law holds. Double Negative: x̄ = 1 − (1 − x) = 1 − 1 + x = x. This law holds. Idempotent: x ∧ x = min(x, x) = x and x ∨ x = max(x, x) = x. This law holds. DeMorgan: We have ( x ∧ y = 1 − min(x, y) =

1 − x if x ≤ y = 1 − y if y ≤ x

(

1−x 1−y

if 1 − x ≥ 1 − y = max(1 − x, 1 − y) = x̄ ∨ ȳ. if 1 − y ≥ 1 − x

(

if 1 − x ≥ 1 − y = min(1 − x, 1 − y) = x̄ ∧ ȳ. if 1 − y ≥ 1 − x

Also, ( 1−x x ∨ y = 1 − max(x, y) = 1−y

if x ≥ y = if y ≥ x

1−x 1−y

This law holds. Absorption: We have ( min(x, x) if y ≤ x x ∧ (x ∨ y) = min(x, max(x, y)) = = x. min(x, y) if x ≤ y Also, ( max(x, y) if y ≤ x x ∨ (x ∧ y) = max(x, min(x, y)) = = x. max(x, x) if x ≤ y This law holds.

Exercise: 7 Section 10.1 Question: Use the axioms of a Boolean algebra to prove that the following propositional forms from Boolean logic are tautologies. a) p → (p ∨ q) b) ((p → q) ∧ ¬q) → ¬p c) ((p → q) ∧ (q → r)) → (p → r) Solution: Recall that p → q ≡ p̄ ∨ q. a) For p → (p ∨ q) we have p → (p ∨ q) ≡ p̄ ∨ (p ∨ q) ≡ (p̄ ∨ p) ∨ q

by associativity

≡T∨q

by complement law

≡T

by dominance.

500

CHAPTER 10. MODULES AND ALGEBRA

b) For ((p → q) ∧ ¬q) → ¬p we have ((p → q) ∧ ¬q) → ¬p ≡ ((p̄ ∨ q) ∧ q̄) ∨ p̄ ≡ ((p̄ ∨ q) ∨ q) ∨ p̄

by DeMorgan and double negation

≡ ((p ∧ q̄) ∨ q) ∨ p̄

by DeMorgan and double negation

≡ ((p ∨ q) ∧ (q̄ ∨ q)) ∨ p̄

by distributivity

≡ ((p ∨ q) ∧ T) ∨ p̄

by complement law

≡ (p ∨ q) ∨ p̄

by identity

≡ q ∨ (p ∨ p̄)

by associativity and commutativity

≡q∨T

by complement law

≡T

by dominance.

c) For ((p → q) ∧ (q → r)) → (p → r) we have ((p → q) ∧ (q → r)) → (p → r) ≡ ((p̄ ∨ q) ∧ (q̄ ∨ r)) ∨ (p̄ ∨ r) ≡ ((p̄ ∨ q) ∨ (q̄ ∨ r)) ∨ (p̄ ∨ r)

by DeMorgan

≡ ((p ∧ q̄) ∨ (q ∧ r̄)) ∨ (p̄ ∨ r)

by DeMorgan and double negation

≡ (p̄ ∨ (p ∧ q̄)) ∨ ((q ∧ r̄) ∨ r)

by associativity and commutativity

≡ ((p̄ ∨ p) ∧ (p̄ ∨ q̄)) ∨ ((q ∨ r) ∧ (r̄ ∨ r))

by distributivity

≡ (T ∧ (p̄ ∨ q̄)) ∨ ((q ∨ r) ∧ T)

by complement law

≡ (p̄ ∨ q̄) ∨ (q ∨ r)

by identity

≡ (p̄ ∨ r) ∨ (q̄ ∨ q)

by associativity and commutativity

≡ (p̄ ∨ r) ∨ T

by complement law

≡T

by dominance.

Exercise: 8 Section 10.1 Question: Prove that for all x, y in a Boolean algebra B, x ∧ y = y if and only if x ∨ y = x. Solution: Let x, y ∈ B, where B is a Boolean algebra. Suppose first that x∧y = y. Then x∨(x∧y) = x∨y and by the absorption law, this implies that x = x ∨ y. Conversely, suppose that x ∨ y = x. Then (x ∨ y) ∧ y = x ∧ y. Then using commutativity and the absorption law, we have x ∧ y = y ∧ (y ∨ x) = y. Hence x ∧ y = y if and only if x ∨ y = x. Exercise: 9 Section 10.1 Question: Let (B, ∧, ∨, ) be a Boolean algebra. Define the relation 4 on B by x 4 y when x ∧ y = x. a) Prove that 4 defines a partial order on B. b) Show this partial order 4 on the Boolean algebra P(S), where S is a set, is precisely the containment partial order ⊆. Solution: Let (B, ∧, ∨, ) be a Boolean algebra. Define the relation 4 on B by x 4 y when x ∧ y = x. a) Let x ∈ B. Then x ∧ x = x by the idempotent law. Hence x 4 x, so 4 is reflexive. Let x, y ∈ B. Suppose that x 4 y and y 4 x. Then x ∧ y = x and y ∧ x = y. However, since ∧ is commutative, we deduce that x = y. Hence, 4 is anti-symmetric. Let x, y, z ∈ B and suppose that x 4 y and y 4 z. Then x ∧ y = x and y ∧ z = y. Then, x ∧ z = (x ∧ y) ∧ z = x ∧ (y ∧ z) = x ∧ y = x Thus x 4 z and so 4 is transitive. We have shown that 4 is a partial order on B. b) Consider the Boolean algebra (P(S), ∩, ∪, ), where S is a set. Then A ∩ B = A if and only if every element of A is in B, which means that A ⊆ B. Hence, the relation 4 defined generally above gives the subset relation on P(S).

10.1. BOOLEAN ALGEBRAS

501

Exercise: 10 Section 10.1 Question: Let S be a set and let A be a fixed subset of S. Consider the Boolean algebras of P(S) and P(A). Prove that the function ϕ : P(S) → P(A) defined by ϕ(X) = X ∩ A is a Boolean algebra homomorphism. Solution: We consider the function ϕ : P(S) → P(A) defined by ϕ(X) = X ∩ A. For all X, Y ∈ P(S), using idempotence, commutativity and associativity, we have ϕ(X ∩ Y ) = (X ∩ Y ) ∩ A = (X ∩ Y ) ∩ (A ∩ A) = (X ∩ A) ∩ (Y ∩ A) = ϕ(X) ∩ ϕ(Y ). Also, using just distributivity, ϕ(X ∪ Y ) = (X ∪ Y ) ∩ A = (X ∩ A) ∪ (Y ∩ A) = ϕ(X) ∪ ϕ(Y ). Finally, ϕ(X) = X ∩ A = A − X = A − (A ∩ X) = ϕ(X). Thus, ϕ is a Boolean algebra homomorphism. Exercise: 11 Section 10.1 Question: Let (B, ∧, ∨, ) be a Boolean algebra. a) Show that (B, ∨, ∧, ) is also a Boolean algebra. b) Prove that the function ϕ : B → B defined by ϕ(x) = x is an isomorphism from (B, ∧, ∨, ) to (B, ∨, ∧, ). Solution: Let (B, ∧, ∨, ) be a Boolean algebra. a) To see that (B, ∨, ∧, ) is also a Boolean algebra, it suffices to observe that all of the laws are given in pairs that a symmetric in the role of ∧ and ∨. Note that the identities change roles in this symmetry. b) The DeMorgan law and the double negation law imply that the function ϕ : B → B defined by ϕ(x) = x is a Boolean algebra homomorphism from (B, ∧, ∨, ) to (B, ∨, ∧, ). It is also an isomorphism because it has an inverse, namely itself. Exercise: 12 Section 10.1 Question: Let f : S → T be a function between sets. Prove that the function ϕ : P(T ) → P(S) defined by ϕ(X) = f −1 (X) is a Boolean algebra homomorphism. Solution: Let f : S → T be a function between sets. Let A, B ⊆ T . In Exercise 1.1.26, the reader proved that f −1 (A ∩ B) = f −1 (A) ∩ f −1 (B) and also that f −1 (A ∪ B) = f −1 (A) ∪ f −1 (B). These establish the ∧ and ∨ axioms of a homomorphism. For the complement axiom of a homomorphism, we first point out that f −1 (A) = {x ∈ S | f (x) ∈ A} = {x ∈ S | f (x) ∈ / A} Then, if x ∈ / f −1 (A), then f (x) ∈ / A. In other words, f −1 (A) ⊆ f −1 (A). On the other hand, if x ∈ f −1 (A), then f (x) ∈ / A. However, for all x ∈ S, either f (x) ∈ A or f (x) ∈ / A. So x ∈ f −1 (A). Hence, f −1 (A) ⊆ f −1 (A), and we have established that these are equal. These observations show that the function ϕ : P(T ) → P(S) defined by ϕ(X) = f −1 (X) is a Boolean algebra homomorphism. Exercise: 13 Section 10.1 Question: Let (B1 , ∧1 , ∨1 , ) and (B2 , ∧2 , ∨2 , ) be two Boolean algebras and let ϕ : B1 → B2 be a Boolean algebra homomorphism. Define the kernel and image of ϕ Ker ϕ = {x ∈ B1 | ϕ(x) = 0}

and

Im ϕ = {y ∈ B2 | ∃x ∈ B1 , y = ϕ(x)}.

a) Prove that Ker ϕ is not necessarily a Boolean subalgebra of B1 . b) Prove that Im ϕ is a Boolean subalgebra of B2 . Solution: Let (B1 , ∧1 , ∨1 , ) and (B2 , ∧2 , ∨2 , ) be two Boolean algebras and let ϕ : B1 → B2 be a Boolean algebra homomorphism. a) Let S be a set and let A be a nontrivial proper subset. Recall the Boolean algebra homomorphism ϕ : P(S) → P(A) described in Exercise 10.1.10. In the Boolean algebra P(S), the 1 is S and the 0 is ∅. Note that Ker ϕ = {X ∈ P(S) | X ∩ A = ∅}. However, this set of subsets does not include S and Ker ϕ cannot be close under ∪ because the complement law would imply that S is in any subalgebra.

502

CHAPTER 10. MODULES AND ALGEBRA

b) Let x, y ∈ Im ϕ with x = ϕ(a) and y = ϕ(b). Then x ∧ y = ϕ(a) ∧ ϕ(b) = ϕ(a ∧ b) ∈ Im ϕ x ∨ y = ϕ(a) ∨ ϕ(b) = ϕ(a ∨ b) ∈ Im ϕ. Furthermore, x̄ = ϕ(a) = ϕ(ā) ∈ Im ϕ. Thus, Im ϕ is closed under ∧, ∨, and . Thus, Im ϕ is a subalgebra of B2 . Exercise: 14 Section 10.1 Question: Prove Theorem 10.1.13, Solution: Let (B, +, ∧) be a Boolean ring with additive identity 0 and multiplicative identity of 1. We defined ∨ and as x ∨ y = (x ∧ y) + (x + y) and x = 1 + x. To show that (B, ∨, ∧, ) is a Boolean algebra, we need to verify all the rules of a Boolean algebra. Identity: x ∧ 1 = x by definition of multiplicative identity. x ∨ 0 = (x ∧ 0) + x + 0 = 0 + x + 0 = x, by the property of rings that a0 = 0a = 0 for all elements a in a ring. Complement: x ∧ x̄ = x ∧ (1 + x) = x + x ∧ x = x + x = 0, by properties of a Boolean ring. Also x ∨ x̄ = x ∧ x̄ + x + x̄ = 0 + x + 1 + x = 1, again because x + x = 0. Associativity: The multiplication ∧ is associative. We only need to check that ∨ is associative: x ∨ (y ∨ z) = (x ∧ ((y ∧ z) + (y + z))) + x + ((y ∧ z) + (y + z)) = x ∧ (y ∧ z) + (x ∧ y) + (x ∧ z) + (y ∧ z) + x + y + z = (((x ∧ y) + (x + y)) ∧ z) + ((x ∧ y) + (x + y) + z = (x ∧ y) ∧ z. Commutativity: In any Boolean ring, (x + y) ∧ (x + y) = x + y =⇒ x ∧ x + x ∧ y + y ∧ x + y ∧ y = x + y =⇒ x + x ∧ y + y ∧ x + y = x + y =⇒ x ∧ y + y ∧ x = 0 =⇒ x ∧ y = y ∧ x because r + r = 0 for all elements in a Boolean ring. Then x ∨ y = (x ∧ y) + (x + y) = (y ∧ x) + (y + x) = y ∨ x. Both operations are commutative. Distributivity: Let x, y, z ∈ B. (x ∨ z) ∧ (y ∨ z) = (x ∧ z + x + z) ∧ (y ∧ z + y + z) =x∧z∧y∧z+x∧y∧z+z∧y∧z+x∧z∧y+x∧y+z∧y +z∧y∧z+z∧y+z∧z =x∧y∧z+x∧y∧z+y∧z+x∧y∧z+x∧y+y∧z+y∧z+y∧z+z =x∧y∧z+x∧y+z = (x ∧ y) ∧ z + (x ∧ y + z) = (x ∧ y) ∨ z Also, (x ∧ z) ∨ (y ∧ z) = (x ∧ z) ∧ (y ∧ z) + (x ∧ z) + (y ∧ z) =x∧y∧z+x∧z+y∧z = (x ∧ y + x + y) ∧ z = (x ∨ y) ∧ z. These show that ∧ is distributive over ∨ and that ∨ is distributive over ∧.

10.1. BOOLEAN ALGEBRAS

503

We have proved that (B, ∨, ∧, ) is a Boolean algebra. (The supplementary laws of Proposition 10.1.2 follow.)

Exercise: 15 Section 10.1 Question: Let f ∈ Fn . Prove that f (x1 , x2 , . . . , xn ) = (f (1, x2 , . . . , xn ) ∧ x1 ) ∨ (f (0, x2 , . . . , xn ) ∧ x1 ) for all (x1 , x2 , . . . , xn ) ∈ {0, 1}n . Solution: [Note: there was an error in the original exercise statement.] Let f ∈ Fn . Let g(x1 , x2 , . . . , xn ) = (f (1, x2 , . . . , xn ) ∧ x1 ) ∨ (f (0, x2 , . . . , xn ) ∧ x1 ). Then g(1, x2 , . . . , xn ) = (f (1, x2 , . . . , xn ) ∧ 1) ∨ (f (0, x2 , . . . , xn ) ∧ 0) = f (1, x2 , . . . , xn ) ∨ 0 = f (1, x2 , . . . , xn ) and g(0, x2 , . . . , xn ) = (f (1, x2 , . . . , xn ) ∧ 0) ∨ (f (0, x2 , . . . , xn ) ∧ 1) = 0 ∨ f (0, x2 , . . . , xn ) = f (0, x2 , . . . , xn ). Hence, for all values (x1 , x2 , . . . , xn ) ∈ {0, 1}n , we have f (x1 , x2 , . . . , xn ) = g(x1 , x2 , . . . , xn ). Exercise: 16 Section 10.1 Question: Disjunctive Normal Form.Use the previous exercise to prove that every function f ∈ Fn can be written in the form f (x1 , x2 , . . . , xn ) = p1 ∨ p2 ∨ · · · ∨ pr where each pi is a function involving only ∧ (multiplication) of the inputs xi or the negations xi . Solution: We prove the desired result by induction on n. Let n = 1. When n = 1, there are precisely 4 Boolean functions, namely 0, x1 , x1 , 1 = x1 ∨ x1 . All of these functions are written in the desired form, with the convention that 0 is in the desired form but with r = 0. Suppose that the result holds for Boolean functions of n variables. Let f be a Boolean function of n + 1 variables. Then by the previous exercise, f (x1 , x2 , . . . , xn+1 ) = (f (1, x2 , . . . , xn+1 ) ∧ x1 ) ∨ (f (0, x2 , . . . , xn+1 ) ∧ x1 ) By the induction hypothesis, f (1, x2 , . . . , xn+1 ) can be written as p1 ∨ p2 ∨ · · · ∨ pr and f (0, x2 , . . . , xn+1 ) can be written as q1 ∨ q2 ∨ · · · ∨ qs , where pi and qj involve the variables x2 , . . . , xn+1 , multiplications, and on variables. Then f (x1 , x2 , . . . , xn+1 ) = x1 p1 ∨ x1 p2 ∨ · · · ∨ x1 pr ∨ x1 q1 ∨ x1 q2 ∨ · · · ∨ x1 qs , which gives and expression of f in the desired form, namely x1 p and x1 q involve a product of n + 1 terms, of either a variable or its negation. By induction, every Boolean function can be written in this disjunctive normal form. Exercise: 17 Section 10.1 Question: Find the disjunctive normal form (see Exercise 10.1.16) of the function f ∈ F4 that returns: a) 1 exactly when exactly one of the inputs xi = 1; b) 1 exactly when x1 or x2 are 1 and when x3 or x4 are 1. Solution: Disjunctive normal forms of functions f ∈ F4 . a) Suppose that f returns a 1 when exactly one of the inputs is a 1. The desired function is f (x1 , x2 , x3 , x4 ) = x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 . We can see this is correct because, for example, the term x1 x2 x3 x4 gives a 1 only for (x1 , x2 , x3 , x4 ) = (1, 0, 0, 0) and similarly for the other three. Performing a ∨ on the four terms means that f is equal to 1 only if (x1 , x2 , x3 , x4 ) is one of the four possibilities (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), or (0, 0, 0, 1).

504

CHAPTER 10. MODULES AND ALGEBRA

b) Suppose that f returns a 1 exactly when x1 or x2 are 1 and when x3 or x4 are 1. Consider the function g ∈ F2 that gives a 1 exactly when x1 or x2 is 1. The disjunctive normal form for this function is g(x1 , x2 ) = x1 x2 ∨ x1 x2 ∨ x1 x2 . Then f (x1 , x2 , x3 , x4 ) = g(x1 , x2 ) ∧ g(x3 , x4 ) = x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 ∨ x1 x2 x3 x4 .

Exercise: 18 Section 10.1 Question: Let B be a Boolean algebra. Define the operator x ↓ y = x ∨ y. This is also called the NOR operator since it takes the OR and then negates it. a) Show that x can be obtained using just the ↓ operator (and perhaps combinations thereof). b) Show that x ∧ y can be obtained using just the ↓ operator (and perhaps combinations thereof). c) Show that x ∨ y can be obtained using just the ↓ operator (and perhaps combinations thereof). d) Deduce that every operation in a Boolean algebra can be deduced by a single operation. Solution: We define the operator x ↓ y = x ∨ y. a) x ↓ x = x ∨ x = x by idempotence. b) For the ∧ operator, (x ↓ x) ↓ (y ↓ y) = x ↓ y =x∨y =x∧y

by part a) definition deMorgan’s law

c) For the ∨ operator, (x ↓ y) ↓ (x ↓ y) = x ↓ y

by part a)

=x∨y

definition

=x∨y

double negative

Exercise: 19 Section 10.1 Question: Draw the logic gate diagrams corresponding to each of the following Boolean functions. a) x̄ ∨ y b) (x ∨ y) ∧ x Solution: Logic gates diagrams. a) x̄ ∨ y x

y b) (x ∨ y)x x

y

10.2. VECTOR SPACES

505

Exercise: 20 Section 10.1 Question: Draw the logic gate diagrams corresponding to each of the following Boolean functions. a) xyz ∨ x̄ȳz̄ b) (x̄ ∨ z)(y ∨ z̄) Solution: Logic gates diagrams. a) xyz ∨ x̄ȳz̄ x y z x y z Notice that in this diagram, we let the same variable appear in two different input wires. This is sometimes done to simplify the diagram though the full diagram should only have one input wire for each input variable and make branches come of that one input wire. b) (x̄ ∨ z)(y ∨ z̄) x z y

10.2 – Vector Spaces Exercise: 1 Section 10.2 Question: Show that if T : V → W is a linear transformation between vector spaces, then T (0V ) = 0W , where 0V and 0W are the additive identity elements in V and W , respectively. Solution: Let T : V → W be a linear transformation between vector spaces. Then since 0v + 0v = 0V , T (0V ) + T (0V ) = T (0V ) =⇒ T (0V ) + T (0V ) − T (0V ) = T (0V ) − T (0V ) =⇒ T (0V ) = 0W .

Exercise: 2 Section 10.2 Question: Let T : V → W be a linear transformation between vector spaces. Show that if a set of vectors {T (v1 ), T (v2 ), . . . , T (vn )} in Im T is linearly independent, then the set of vectors {v1 , v2 , . . . , vn } in V is linearly independent. Solution: Let T : V → W be a linear transformation between vector spaces and let {v1 , v2 , . . . , vn } be a set of vectors in V such that {T (v1 ), T (v2 ), . . . , T (vn )} in Im T is linearly independent. Suppose that c1 v1 + c2 v2 + · · · + cn vn = 0V . Then since T is a linear transformation (and using the result of Exercise 10.2.1), T (c1 v1 + c2 v2 + · · · + cn vn ) = T (0V ) =⇒ c1 T (v1 ) + c2 T (v2 ) + · · · + cn T (vn ) = 0W . Since {T (v1 ), T (v2 ), . . . , T (vn )} is linearly independent, then c1 = c2 = · · · = cn = 0 and we deduce that the set {v1 , v2 , . . . , vn } is linearly independent. Exercise: 3 Section 10.2 Question: Let U, V, W be vector spaces and let L1 : U → V and L2 : V → W be linear transformations. Prove

506

CHAPTER 10. MODULES AND ALGEBRA

that the composition L2 ◦ L1 : U → W is a linear transformation. If L1 and L2 are isomorphisms, is it also true that L2 ◦ L1 is an isomorphism? Solution: Let u1 , u2 ∈ U . Then (L2 ◦ L1 )(u1 + u2 ) = L2 (L1 (u1 ) + L1 (u2 ))

because L1 is a linear transformation

= L2 (L1 (u1 )) + L2 (L1 (u2 )))

because L2 is a linear transformation

= (L2 ◦ L1 )(u1 ) + (L2 ◦ L1 )(u2 ). Also, if c ∈ F and u ∈ U , then (L2 ◦ L1 )(cu) = L2 (cL1 (u))

because L1 is a linear transformation

= cL2 (L1 (u) )

because L2 is a linear transformation

= c(L2 ◦ L1 )(u). This shows that L2 ◦ L1 : U → W is a linear transformation. −1 Suppose that L1 and L2 are bijections, then the function L−1 1 ◦ L2 : W → U is such that −1 (L−1 1 ◦ L2 ) ◦ (L2 ◦ L1 ) = idU

and

−1 (L2 ◦ L1 ) ◦ (L−1 1 ◦ L2 ) = idW .

−1 Thus, L2 ◦ L1 has L−1 1 ◦ L2 as an inverse function. Hence, L2 ◦ L1 is also an isomorphism.

Exercise: 4 Section 10.2 Question: Prove Proposition 10.2.4. Solution: Let V and W be vector spaces over a field F and define the operations on V × W as described in the statement of the proposition. The componentwise addition is the same as the direct sum of of the groups (V, +) and (W, +), so the resulting addition of + on V × W does make V × W into an abelian group with identity (0V , 0W ). For all c ∈ F and (v1 , w1 ), (v2 , w2 ) ∈ V × W , c((v1 , w1 )+(v2 , w2 )) = c(v1 +v2 , w1 +w2 ) = (cv1 +cv2 , cw1 +cw2 ) = (cv1 , cw1 )+(cv2 , cw2 ) = c(v1 , w1 )+c(v2 , w2 ). For all c, d ∈ F and for all (v, w) ∈ V × W , (c + d)(v, w) = ((c + d)v, (c + d)w) = (cv + dv, cw + dw) = (cv, cw) + (dv, dw) = c(v, w) + d(v, w). For all c, d ∈ F and for all (v, w) ∈ V × W , (cd)(v, w) = (cdv, cdw) = c(dv, dw) = c(d(v, w)). Finally, for all (v, w) ∈ V × W , 1(v, w) = (1v, 1w) = (v, w). This finishes proving the 4 additional axioms of a vector space. Hence, the set V ×W equipped with the described operations is a vector space. Exercise: 5 Section 10.2 Question: Let V and W be finite-dimensional vector spaces over a field F . Prove that dim(V ⊕ W ) = (dim V ) + (dim W ). Solution: In order to prove that dim(V ⊕ W ) = (dim V ) + (dim W ), we must make reference to a basis. Let B = {v1 , v2 , . . . , vm } be a basis of V and let B 0 = {w1 , w2 , . . . , wn } be a basis of W . Note that dim V = m and dim W = n. We claim that B 00 = {(v1 , 0), (v2 , 0), . . . , (vm , 0), (0, w1 ), (0, w2 ), . . . , (0, wn )} is a basis of V ⊕ W . Let (v, w) ∈ V ⊕ W . Since V is spanned by B and W is spanned by B 0 then v = c1 v1 + c2 v2 + · · · + cm vm

for some c1 , c2 , . . . , cm ∈ F

w = d1 w1 + d2 w2 + · · · + dn wn

for some d1 , d2 , . . . , dn ∈ F.

10.2. VECTOR SPACES

507

Thus (v, w) = (v, 0) + (0, w) = (c1 v1 + c2 v2 + · · · + cm vm , 0) + (0, d1 w1 + d2 w2 + · · · + dn wn ) = c1 (v1 , 0) + · · · + cm (vm , 0) + d1 (0, w1 ) + · · · + dn (0, wn ), and so B 00 spans V ⊕ W . On the other hand, suppose that c1 (v1 , 0) + · · · + cm (vm , 0) + d1 (0, w1 ) + · · · + dn (0, wn ) = (0V , 0W ) for some set of constants. Then (0V , 0W ) = (c1 v1 +c2 v2 +· · ·+cm vm , 0)+(0, d1 w1 +d2 w2 +· · ·+dn wn ) = (c1 v1 +c2 v2 +· · ·+cm vm , d1 w1 +d2 w2 +· · ·+dn wn ) and so

( c1 v1 + c2 v2 + · · · + cm vm = 0V d1 w1 + d2 w2 + · · · + dn wn = 0W .

Since B is linearly independent, c1 = c2 = · · · = cm = 0, and since B 0 is linearly independent, d1 = d2 = · · · = dn = 0. Hence, B 00 is also linearly independent. We have shown that B 00 is a basis of V ⊕ W . Thus dim(V ⊕ W ) = |B 00 | = |B| + |B 0 | = m + n = (dim V ) + (dim W ).

Exercise: 6 Section 10.2 Question: Let W be a subspace of a vector space V . Prove that dim W ≤ dim V . Solution: Let S = {v1 , v2 , . . . , vm } be a basis of W . (We are implicitly invoking Corollary 10.2.17 to even talk about dimensions.) Obviously, V = Span(V ). By Theorem 10.2.16, V has a basis B such that S ⊆ B. Since this set inclusion defines an injection of S into B, then as cardinalities, |S| ≤ |B|. Thus, dim W ≤ dim V . (We point out that the given proof does not require that V and W be finite dimensional. If a vector space is infinite dimensional, then its dimension is as always the cardinality of a basis, regardless of whether the basis is finite. Thus, the dimension can be any cardinal number, including infinite cardinalities. Recall that if A and B are sets, we say that |A| ≤ |B| if there exists an injective function f : A → B. That is what we proved here.) Exercise: 7 Section 10.2 Question: Let V be a vector space over a field F and let U and W be subspaces of V . Prove that U ∪ W is not necessarily a subspace of V . Solution: As an example, let = R and let V = R2 . Set U = Span 10 and W = Span 01 . Then c 0 , | c, d ∈ F . U ∪W = 0 d But U ∪ W is not closed under addition. For example, 10 + 01 = 11 , which is not an element of U ∪ W . Exercise: 8 Section 10.2 Question: Sum of Subspaces. Let V be a vector space and let U and W be subspaces of V . Define the sum of the subspaces, denoted U + W as the set U + W = {v ∈ V | v = u + w for some u ∈ U and w ∈ W }. Show that U + W is a subspace of V . Solution: Let x, y ∈ U + W . Then x = u1 + w1 and y = u2 + w2 for some u1 , u2 ∈ U and w1 , w2 ∈ W . Then x + y = (u1 + w1 ) + (u2 + w2 ) = (u1 + u2 ) + (w1 + w2 ) so x + y ∈ U + W because u1 + u2 ∈ U and w1 + w2 ∈ W . Furthermore, if c ∈ F , then cx = c(u1 + w1 ) = (cu1 ) + (cw1 )

508

CHAPTER 10. MODULES AND ALGEBRA

so cx ∈ U + W because cu1 ∈ U and cw1 ∈ W . Exercise: 9 Section 10.2 Question: Let V be a vector space over a field F and let U and W be subspaces of V . Prove that U ∩ W is a subspace of V . Solution: Let v1 , v2 ∈ U ∩ W . Since v1 , v2 ∈ U , then v1 + v2 ∈ U . Since v1 , v2 ∈ W , then v1 + v2 ∈ W . So v1 + v2 ∈ U ∩ W . Let c ∈ F and v ∈ U ∩ W . Since v ∈ U , then cv ∈ U . Since v ∈ W , then cv ∈ W . So cv ∈ U ∩ W . Since V ∩ W is closed under addition and scalar multiplication, it is a subspace of V . Exercise: 10 Section 10.2 Question: Let V be a vector space over a field F and let U and W be subspaces of V . Prove that dim(U +W ) = dim U + dim W − dim(U ∩ W ). [See Exercise 10.2.8.] Solution: Let B be a basis of U ∩ W . By Theorem 10.2.16, there exists a basis BU of U that contains B and there exists a basis BW of W that contains B. Define B 0 = BU ∪ BW . We claim that B 0 is a basis of U + W . Let v ∈ U + W . Then v = u + w for some u ∈ U and w ∈ W . Then u = c1 u1 + c2 u2 + · · · + cm um with ci ∈ F and ui ∈ BU and also w = d1 w1 + d2 w2 + · · · + dn wn with di ∈ F and wi ∈ BW . Hence, v = c1 u1 + c2 u2 + · · · + cm um + d1 w1 + d2 w2 + · · · + dn wn and like basis vectors can be combined in this expression. This presents v as a (finite) linear combination of elements in B 0 . This shows that B 0 spans U + W . Suppose that c1 v1 + c2 v2 + · · · + cs vs = 0 with vi ∈ B 0 for all 1 ≤ i ≤ s. Relabeling if necessary, suppose that vi ∈ BU for 1 ≤ i ≤ t and that vi ∈ BW − BU for t + 1 ≤ i ≤ s where t is some integer with 0 ≤ t ≤ s. Then 0 = (c1 v1 + c2 v2 + · · · + ct vt ) + (ct+1 vt+1 + ct+2 vt+2 + · · · + cs vs ) and each of the two parentheses groups is in U and W respectively. Thus c1 v1 + c2 v2 + · · · + ct vt = v0 and ct+1 vt+1 + ct+2 vt+2 + · · · + cs vs = −v0 , for some v0 ∈ U ∩ W . However, since vt+1 , . . . , vs ∈ BW − BU , we must have v0 = 0. Consequently, c1 v1 + c2 v2 + · · · + ct vt = 0 and so c1 = c2 = · · · = ct = 0, since v1 , v2 , . . . , vt are in the basis BU but also ct+1 = · · · = cs = 0 because vt+1 , . . . , vs are in the basis |mathcalBW . Thus, B 0 is linearly independent. We have proved the claim. By the Inclusion-Exclusion Principle, |B 0 | = |BU | + |BW | − |B|. This translates into the formula dim(U + W ) = dim U + dim W − dim(U ∩ W ).

Exercise: 11 Section 10.2 Question: Subspace Complement. Let U be a vector space and V a subspace of U . Another subspace W of U is called a complement to V if V ∩ W = {0} and if every vector u ∈ U can be written as u = v + w for some v ∈ V and some w ∈ W . a) Show that if W is a complement to V , then every vector u ∈ U can be written as u = v + w in a unique way. b) Show that if W is a complement to V , then dim U = dim V + dim W . Solution: Let U be a vector space and V a subspace of U . a) Let u ∈ U . Suppose that u = v1 + w1 = v2 + w2 . Then v1 − v2 = w2 − w1 . But then v1 − v2 and w1 − w2 are vectors in V ∩ W . But since V ∩ W = {0}, we deduce that v1 = v2 and w1 = w2 . Hence, u can be written as a sum v + w with v ∈ V and w ∈ W in a unique way. b) Since every vector u ∈ U can be written as a sum v + w with v ∈ V and w ∈ W , then U = V + W . By Exercise 10.2.10, we have dim U = dim V + dim W − dim(V ∩ W ) = dim V + dim W.

10.2. VECTOR SPACES

509

Exercise: 12 Section 10.2 Question: Let F be a field, let K be a finite field extensions of F and let V be a vector space over K. Prove that V is a vector space over F and that dimF V = [K : F ](dimK V ). Solution: Let B = {v1 , v2 , . . . , vn } be a basis of V as a vector space of K. Also let {α1 , α2 , . . . , αs } ⊆ K be a basis of K over F as an F -vector space. Note that dimK V = n and [K : F ] = s. Consider the set B 0 = {αi vj | 1 ≤ i ≤ s and 1 ≤ j ≤ n}. Obviously, the set B 0 spans V as an F -vector space. Furthermore, suppose that X cij αi vj = 0 1≤i≤s 1≤j≤n

in V with cij ∈ F . Then n s X X j=1

! cij αi

vj = 0.

i=1

Ps But, for each j, the summation dj = i=1 cij αi is an element of K. Since {v1 , v2 , . . . , vn } is a basis of V over K, then d1 v1 + d2 v2 + · · · + dn vn = 0 implies d1 = d2 = · · · = dn = 0. But then, for each j, c1j α1 + c2j α2 + · · · + csj αs = 0. Since, cij ∈ F and since {α1 , α2 , . . . , αs } is a basis of K over F , then we deduce that cij = 0 for all i, j, with 1 ≤ i ≤ s and 1 ≤ j ≤ n. We deduce that B 0 spans V over F and is linearly independent, hence is a basis. Thus dimF V = |B 0 | = sn = [K : F ](dimK V ).

Exercise: 13 Section 10.2 Question: Let V be a vector space over a field F and define End(V ) = HomF (V, V ), as the space of endomorphisms on V (linear transformations of V into itself). Prove that (End(V ), +, ◦) is a ring, where ◦ is function composition. Solution: By Proposition 10.2.27, End(V ) is a vector space with addition of linear transformations and scalar multiplication. In particular, (End(V ), +) is an abelian group with the trivial linear transformation as the identity. As a function composition, if f, g, h ∈ End(V ), then f ◦ (g ◦ h) = (f ◦ g) ◦ h and so ◦ is associative. For distributivity, note that for all v ∈ V , we have (f ◦ (g + h))(v) = f (g(v) + h(v)) = f (g(v)) + f (h(v)) = (f ◦ g)(v) + (f ◦ h)(v) = (f ◦ g + f ◦ h)(v), so f ◦ (g + h) = f ◦ g + f ◦ h. Also, ((f + g) ◦ h)(v) = (f + g)(h(v)) = f (h(v)) + g(h(v)) = (f ◦ h)(v) + (g ◦ h)(v) = (f ◦ h + g ◦ h)(v), so (f + g) ◦ h = f ◦ h + g ◦ h. We have established the axioms of a ring. In addition, the identity function idV serves as the multiplicative identity. Exercise: 14 Section 10.2 Question: Let T : V → W be a linear transformation between vector spaces. Let U be a subspace of V . Prove that T (U ) = {T (u) | u ∈ U } is a subspace of W . [This generalizes the fact that Im T is a subspace of W since Im T = T (V ).] Solution: Let w1 , w2 ∈ T (U ). Then there exist u1 , u2 ∈ U such that T (u1 ) = w1 and T (u2 ) = w2 . Then w1 + w2 = T (u1 ) + T (u2 ) = T (u1 + u2 ). Hence, w1 + w2 ∈ T (U ), so T (U ) is closed under addition. Now let c ∈ F and w ∈ T (U ). There exists u ∈ U such that w = T (u). Then cw = cT (u) = T (cu). Thus, cw ∈ T (U ), so T (U ) is closed under scalar multiplication. This establishes that T (U ) is a subspace of W .

510

CHAPTER 10. MODULES AND ALGEBRA

Exercise: 15 Section 10.2 Question: Let T : V → W be a linear transformation between vector spaces. Let U be a subspace of W . Define the set T −1 (U ) = {v ∈ V | T (v) ∈ U }. Prove that T −1 (U ) is a subspace of V . [This generalizes the fact that Ker T is a subspace of V because Ker T = T −1 ({0}).] Solution: Let v1 , v2 ∈ T −1 (U ). Then T (v1 + v2 ) = T (v1 ) + T (v2 ) ∈ U since U is closed under addition. Thus v1 + v2 ∈ T −1 (U ). Let c ∈ F and v ∈ T −1 (U ). Then T (cv) = cT (v) ∈ U , since U is closed under scalar multiplication. We have shown that T −1 (U ) is a subspace of V . Exercise: 16 Section 10.2 Question: Consider the vector space V = C 0 ([a, b], R) as in Example 10.2.29. Let g ∈ V . a) Prove that the function Ψg : V → R defined by Z b Ψg (f ) = f (t)g(t) dt a ∗

is an element of V . b) Prove that the mapping g 7→ Ψg is an injective linear transformation from V to V ∗ . c) Let c ∈ [a, b]. Prove that the evaluation evc ∈ V ∗ is not equal to Ψg for any function g ∈ V . Solution: Consider V = C 0 ([a, b], R) as a vector space over R. a) Let f1 , f2 ∈ V . Then Z b Z b Z b Z b Ψg (f1 +f2 ) = (f1 (t)+f2 (t))g(t) dt = f1 (t)g(t)+f2 (t)g(t) dt = f1 (t)g(t) dt+ f2 (t)g(t) dt = Ψg (f1 )+Ψg (f2 ). a

a

a

a

Let c ∈ R and let f ∈ V . Then Z b Ψg (cf ) =

Z b cf (t)g(t) dt = c

a

f (t)g(t) dt = cΨg (f ). a

This proves that for all g ∈ V , the function Ψg ∈ Hom(V, R) = V ∗ . b) Consider the function Ψ : V → V ∗ given by Ψ(g) = Ψg . For all f ∈ V and for all g1 , g2 ∈ V , we have Z b Z b Z b Z b Ψg1 +g2 (f ) = f (t)(g1 (t)+g2 (t)) dt = f (t)g1 (t)+f (t)g2 (t) dt = f (t)g1 (t) dt+ f (t)g2 (t) dt = Ψg1 (f )+Ψg2 (f ). a

a

a

a

∗

Thus, Ψg1 +g2 = Ψg1 + Ψg2 as elements in V . Also, for all f ∈ V , for all c ∈ F , and all g ∈ V , Z b Z b Ψcg (f ) = cf (t)g(t) dt = c f (t)g(t) dt = cΨg (f ). a

a

∗

Thus, Ψcg = cΨg as elements in V . This establishes that the function Ψ is in fact a linear transformation. Now suppose that Ψg1 = Ψg2 . This means to for all f ∈ V , Z b Z b Z b f (t)g1 (t) dt = f (t)g2 (t) dt ⇐⇒ f (t)(g1 (t) − g2 (t)) dt = 0. a

a

a

Assume that g1 6= g2 . Then g1 −g2 is not the zero function. Let t0 ∈ [a, b] such that g1 (t0 )−g2 (t0 ) = y0 6= 0. Without loss of generality, suppose that y0 > 0. (Otherwise, interchange the role of g1 and g2 .) Using the definition of continuity, there exists a δ > 0 such that |t − t0 | < 4δ implies |(g1 (t) − g2 (t)) − y0 | ≤ 21 y0 . In particular, this implies that for t ∈ [t0 − 2δ, t0 + 2δ], we have y0 − 21 y0 ≤ g1 (t) − g2 (t) ≤ y0 + 21 y0 and thus that g1 (t) − g2 (t) is positive over the whole interval [t0 − 2δ, t0 + 2δ]. Let f (t) be the function   0 if a ≤ t ≤ t0 − 2δ    1   (t − t + 2δ) if t0 − 2δ ≤ t ≤ t0 − δ 0 δ f (t) = 1 if t0 − δ ≤ t ≤ t0 + δ   1  − δ (t − t0 − 2δ) if t0 + δ ≤ t ≤ t0 + 2δ    0 if t0 + 2δ ≤ t ≤ b.

10.3. INTRODUCTION TO MODULES

511

Then, Z b

Z t0 +2δ f (t)(g1 (t) − g2 (t)) dt =

f (t)(g1 (t) − g2 (t)) dt

because of f

f (t)(g1 (t) − g2 (t)) dt

since the integrand is positive

t0 −2δ Z t0 +δ

a

≥ t0 −δ Z t0 +δ

g1 (t) − g2 (t) dt

≥ t0 −δ

≥

1 y0 2δ = δy0 > 0. 2

This contradicts Ψg1 (f ) = Ψg2 (f ). We have established that Ψ : V → V ∗ is an injective function. c) Let c ∈ [a, b]. It is easy to see that the evaluation transformation evc defined by evc (f ) = f (c) is a linear transformation V → R, so evc ∈ V ∗ . Assume that evc = Ψg for some continuous function g ∈ V . Then Z b f (t)g(t) dt = f (c) a

for all continuous functions f . So for all a1 < a2 with a2 < c, this implies that Z a2 f (t)g(t) dt = 0 a1

and by the (Weighted) Mean Value Theorem (for Integral), this implies that there exists t1 with a1 < t1 < a2 such that Z a 2

f (t) dt = 0.

g(t1 ) a1

Since this is true for all continuous f (t), with an appropriate f (t), we deduce that g(t1 ) = 0 for some t1 ∈ (a1 , a2 ). However, since a1 and a2 are chosen arbitrarily, we deduce that g(t1 ) = 0 for all t1 < c. Similarly, we deduce that g(t1 ) = 0 for all t1 > c. The only continuous function g ∈ V with these requirements on g is the 0 function. However, Ψ0 (f ) = 0 for all f . Consequently, there is no function g such that Ψg = evc and therefore the function Ψ : V → V ∗ is injective but not surjective.

10.3 – Introduction to Modules Exercise: 1 Section 10.3 Question: Let R be a ring and let M be a left R-module. Prove that the action of R on M also induces a group action of U (R) on M . Solution: Consider the group of units U (R) in R. The action of R on the module M restricts to a scalar multiplication of U (R) acting on M . The axioms (3) and (4) in Definition 10.3.1 give the requirements of a group action of U (R) on M . Exercise: 2 Section 10.3 Question: Let R be a ring and let M be a left R-module. Suppose that rm = 0 for some r ∈ R and some nonzero m ∈ M . Prove that r does not have a left inverse in R. Solution: Suppose that rm = 0 for some r ∈ R and some nonzero m ∈ M . Assume that r has a left inverse in R, so that sr = 1. Then s(rm) = s0 = 0 by Proposition 10.3.2(2). But also then 0 = (sr)m = 1m = m. Since we started with a nonzero m, then we arrive at a contradiction. Exercise: 3 Section 10.3 Question: Let M be a left R-module and let S be a subring of R. Show that M is a left S-module as well.

512

CHAPTER 10. MODULES AND ALGEBRA

Solution: Let M be a left R-module and let S be a subring of R. Restricting the scalars to the subring S does give a well-defined function S × M → M . Item (1) of Definition 10.3.1 holds trivially. Item (2) and (3) of Definition 10.3.1 holds because S is closed under addition and multiplication respectively. Item (4) also holds trivially since it only applies when S contains the identity 1. Exercise: 4 Section 10.3 Question: Let M be a left R-module and let ϕ : S → R be a unital ring homomorphism (i.e., a homomorphism satisfying ϕ(1S ) = 1R , if S has an identity). Show that M is a left S-module if for all s ∈ S and all m ∈ M we define sm = ϕ(s)m. Solution: Let M be a left R-module and let ϕ : S → R be a unital ring homomorphism. Define s · m as ϕ(s)m, where the latter involves the action of R on M . • For all s ∈ S and for all m, n ∈ M , we have s · (m + n) = ϕ(s)(m + n) = ϕ(s)m + ϕ(s)n, be axiom (1). Then s · (m + n) = s · m + s · n. • For all s, s0 ∈ S and for all m ∈ M , we have (s + s0 ) · m = ϕ(s + s0 )m = (ϕ(s) + ϕ(s0 ))m = ϕ(s)m + ϕ(s0 )m by axiom (2). So (s + s0 ) · m = s · m + s0 · m. • For all s, s0 ∈ S and for all m ∈ M , we have (ss0 ) · m = ϕ(ss0 )m = (ϕ(s)ϕ(s0 ))m = ϕ(s)(ϕ(s0 )m) by axioms (3). So (ss0 ) · m = s · (s0 · m). • Suppose that S contains a multiplicative identity. Then 1S · m = ϕ(1S )m = 1R m = m. We have shown all axioms to support that s · m = ϕ(s)m defined M as an S module. Exercise: 5 Section 10.3 Question: Show that there is no nontrivial action of Z[i] on Z that gives Z the structure of a Z[i]-module. Solution: Suppose that there exists an action of Z[i] on Z as desired. Consider the element i · 1. It must be some integer n. Suppose that n is a positive integer. Then n times

n times

n times }| { z z }| { z }| { i · (i · 1) = i · (1 + 1 + · · · + 1) = i · 1 + i · 1 + · · · + i · 1 = n + n + · · · + n = n2 .

But i · (i · 1) = (i2 ) · 1 = (−1) · 1. But 1 · 1 = 1 and 0 · 1 = 0, so 0 = 0 · 1 = (1 + (−1)) · 1 = 1 · 1 + (−1) · 1. Thus, (−1) · 1 = −1. We deduce that n2 = −1, which is a contradiction. Suppose that n is a negative integer. Note that 0 = i · 0 = i · (1 + (−1)) = i · 1 + i · (−1), so i · (−1) = −(i · 1) = −n. Then |n| times

n times

|n| times

z }| { z }| { }| { z i · (i · 1) = i · ((−1) + (−1) + · · · + (−1)) = i · 1 + i · 1 + · · · + i · 1 = |n| + |n| + · · · + |n| = n2 . This again leads to a contradiction. The only possibility is i · 1 = 0. Since i · 1 = 0, then since i4 = 1 in Z[i], using axiom (3) of modules, we deduce that 1 · 1 = 0. Now by axiom (3), α · 1 = 0 for all α ∈ Z[i]. Applying axiom (1), we deduce that α · n = 0 for all positive integers n and then we can deduce also that α · m = 0 for all α ∈ Z[i] and all m ∈ Z. The only action of Z[i] on Z is trivial. Exercise: 6 Section 10.3 Question: Prove Proposition 10.3.7. Solution: Let M and N be two left R-modules. Define the summation and and the scalar multiplication on M × N as in the statement of Proposition 10.3.7. First, we point out that the componentwise addition defines the usual addition on the direct sum of (M, +) and (N, +) as abelian groups. Thus (M × N, +) is another abelian group. We know check the four remaining R-module axioms. • For all (m1 , n1 ), (m2 , n2 ) ∈ M × N and for all r ∈ R, we have r((m1 , n1 ) + (m2 , n2 )) = r(m1 + m2 , n1 + n2 ) = (r(m1 + m2 ), r(n1 + n2 )) = (rm1 + rm2 , rn1 + rn2 ) = (rm1 , rn1 ) + (rm2 , rn2 ) = r(m1 , n1 ) + r(m2 , n2 ).

10.3. INTRODUCTION TO MODULES

513

• For all (m, n) ∈ M × N and for all r, s ∈ R, we have (r + s)(m, n) = ((r + s)m, (r + s)n) = (rm + sm, rn + sn) = (rm, rn) + (sm, sn) = r(m, n) + s(m, n). • For all (m, n) ∈ M × N and for r, s ∈ R, we have (rs)(m, n) = ((rs)m, (rs)n) = (r(sm), r(sn)) = r(sm, sn) = r(s(m, n)). • If R has a multiplicative identity, then 1(m, n) = (1m, 1n) = (m, n). We have established all the axioms of a left R-module.

Exercise: 7 Section 10.3 Question: Prove the details in the second half of Example 10.3.14. Conclude that a representation of G into a F -vector space V makes V into an F [G]-module. Solution: Let ρ : G → GL(V ) be a group representation. We prove that it defines a left F [G]-module via the left action   X X  ag g  · v = ag ρ(g)(v). g∈G

g∈G

First, we point out that V is a vector space, so in particular, (V, +) is an abelian group. Now we prove the remainder of the module axioms. • For all α ∈ F [G], and all v, w ∈ V , if we write α = α · (m + n) =

X

P

g∈G ag g then we have

ag ρ(g)(v + w)

g∈G

=

X

ag (ρ(g)(v) + ρ(g)(w))

because ρ(g) is a linear transformation

g∈G

=

X

ag ρ(g)(v) +

g∈G

X

ag ρ(g)(w)

separating the summations

g∈G

= α · v + α · w. • For all α, β ∈ F [G], and all v ∈ V , if we write α =

P

g∈G ag g and β =

 (α + β) · v = 

g∈G bg g then

 X

ag g +

g∈G

X

bg g  · v

g∈G

 =

P

 X

(ag + bg )g  · v

g∈G

=

X

(ag + bg )ρ(g)(v)

g∈G

=

X

ag ρ(g)(v) +

g∈G

= α · v + β · v.

X g∈G

bg ρ(g)(v)

514

CHAPTER 10. MODULES AND ALGEBRA • For all α, β ∈ F [G], and all v ∈ V , if we write α =  (αβ) · v = 

 X

ag g  

g∈G

P

g∈G ag g and β =

 X

P

g∈G bg g then



bg g  · v

X

g∈G

xy=g

=

g∈G

!  X

ax by

g · v

! =

X

X

g∈G

xy=g

X

X

g∈G

xy=g

X

X

g∈G

xy=g

ax by

ρ(g)(v) !

=

ax by ρ(xy)(v) !

= =

XX

ax by ρ(x)(ρ(y)(v))

ax bx−1 g ρ(x)(ρ(x−1 g)(v))

g∈G x∈G

 =

X x∈G

ax 

 X

bx−1 g ρ(x)(ρ(x

−1

g)(v))

g∈G

 =

X

ax ρ(x) 

x∈G

 X

g∈G

 = =

X

bx−1 g ρ(x−1 g)(v)

ax ρ(x) 

 X

x∈G

y∈G

X

ax ρ(x)(β · v)

by ρ(y)(v)

x∈G

= α · (β · v). • The ring F [G] has an identity 1G and for all v ∈ V we have 1G · v = ρ(1G )(v) = idV (v) = v.

Exercise: 8 Section 10.3 Question: Let R be a ring with an identity 1 6= 0. Prove that Rn is a free left R-module. Solution: [This question should not have been asked until Section 10.5.] Exercise: 9 Section 10.3 Question: Let F be a field and let R be the ring Mn (F ) of n × n matrices with coefficients in F . a) Prove that the F -vector space Mn×m (F ) is a left R-module. b) Prove that the F -vector space Mm×n (F ) is a right R-module. Solution: Let F be a field and let R be the ring Mn (F ) of n × n matrices with coefficients in F . a) The set Mn×m (F ) of n × m matrices with coefficients in F is an abelian group. Furthermore, for all X, Y ∈ Mn×m (F ) and for all A, B ∈ Mn×n the following identities hold A(X + Y ) = AX + AY (A + B)X = AX + BX (AB)(X) = A(BX) In X = X. This shows that Mn×m (F ) is a left Mn (F )-module.

10.3. INTRODUCTION TO MODULES

515

b) The set Mm×n (F ) of m × n matrices with coefficients in F is an abelian group. Furthermore, for all X, Y ∈ Mm×n (F ) and for all A, B ∈ Mn×n the following identities hold (X + Y )A = XA + Y A X(A + B) = XA + XB (X)(AB) = (XA)B X = XIn . This shows that Mm×n (F ) is a right Mn (F )-module. Exercise: 10 Section 10.3 Question: Let R be a commutative ring and let M be a left R-module. Show that defining a scalar rightdef multiplication by mr = rm gives M the structure of a right R-module. Solution: The pair (M, +) is obviously an abelian group. We the given definition of right scalar multiplication, for all r, s ∈ R and all m1 , m2 ∈ M , we have (m1 + m2 )r = r(m1 + m2 ) = rm1 + rm2 = m1 r + m2 r m(r + s) = (r + s)m = rm + sm = mr + ms. These follow from the definition and the axioms of a left R-module. Furthermore, m(rs) = (rs)m = (sr)m = s(rm) = s(mr) = (mr)s where the second equality holds because of the commutativity in R. Finally, if R has an identity 1R , then m1r = 1R m = m. Exercise: 11 Section 10.3 Question: Let R and S be rings. Let M be a left R-module and let N be a left S-module. Prove that M × N is a left R ⊕ S-module when equipped with the following addition and scalar multiplication: def

(m1 , n1 ) + (m2 , n2 ) = (m1 + m2 , n1 + n2 ) def

(r, s) · (m, n) = (rm, sn).

Solution: The set M ×N equipped with addition is the abelian group M ⊕N . We now check the four remaining axioms of a left module. • For all (r, s) ∈ R ⊕ S and for all (m1 , n1 ), (m2 , n2 ) ∈ M × N , we have (r, s) · ((m1 , n1 ) + (m2 , n2 )) = (r, s) · (m1 + m2 , n1 + n2 ) = (r(m1 + m2 ), s(n1 + n2 )) = (rm1 + rm2 , sn1 + sn2 ) = (rm1 , sn1 ) + (rm2 , sn2 ) = (r, s) · (m1 , n1 ) + (r, s) · (m2 , n2 ). • For all (m, n) ∈ M × N and for all (r1 , s1 ), (r2 , s2 ) ∈ R ⊕ S, we have ((r1 , s1 ) + (r2 , s2 )) · (m, n) = (r1 + r2 , s1 + s2 ) · (m, n) = ((r1 + r2 )m, (s1 + s2 )n) = (r1 m + r2 m, s1 n + s2 n) = (r1 m, s1 m) + (r2 m, s2 n) = (r1 , s1 ) · (m, n) + (r2 , s2 ) · (m, n). • For all (m, n) ∈ M × N and for (r1 , s1 ), (r2 , s2 ) ∈ R ⊕ S, we have ((r1 , s1 )(r2 , s2 ))(m, n) = (r1 r2 , s1 s2 ) · (m, n) = ((r1 r2 )m, (s1 s2 )n) = (r1 (r2 m), s1 (s2 n)) = (r1 , s2 ) · (r2 m, s2 m) = (r1 , s2 ) · ((r2 , s2 ) · (m, n)).

516

CHAPTER 10. MODULES AND ALGEBRA • If R and S have a multiplicative identity, then (1R , 1S ) is the multiplicative identity of R ⊕ S and (1R , 1S ) · (m, n) = (1R m, 1S n) = (m, n).

Exercise: 12 Section 10.3 Question: Let F be a field and consider the polynomial ring F [x, y]. a) Prove that an F [x, y]-module consists of vector space V over F along with commuting linear transformations T1 and T2 . b) Explicitly describe the action of a polynomial p(x, y) on a vector in V . Solution: Let F be a field and consider the polynomial ring F [x, y]. a) If we consider the action of the constant polynomials c ∈ F [x, y] on an F [x, y], as in the case of F -modules, we deduce that an F [x, y]-module must consist first of a vector space V , where the constant polynomials are the scalars. As in Example 10.3.13, the module axioms (1) and (3) imply that x must act on V as a linear transformation T1 : V → V . Similarly, the variable y must also act as a linear transformation T2 : V → V . However, since xy = yx ∈ F [x, y], then for all v ∈ V , we have T1 (T2 (v)) = x · (y · v) = (xy) · v = (yx) · v = y · (x · v) = T2 (T1 (v)). Thus, T1 and T2 must be commuting matrices. With these requirements, the remaining axioms for a left ring module hold. b) To be explicit, suppose that a polynomial in F [x, y] has the form p(x, y) =

n m X X

ai,j xi y j

i=0 j=0

where m is the highest power occurring on x and n is the highest power occurring on y. Then for all v ∈ V ,   n m X n m X n m X n m X X X X X i j i j i j  ai,j T1i (T2j (v)). ai,j x · (y · v) = ai,j (x y ) · v = ai,j x y ·v = p(x, y) · v = i=0 j=0

i=0 j=0

i=0 j=0

i=0 j=0

Exercise: 13 Section 10.3 Question: Let R be a commutative ring and let M be a left R-module. Suppose that D is a multiplicatively closed subset of R not containing 0. Define the equivalence relation ∼ on D × M by (d1 , m1 ) ∼ (d2 , m2 ) ⇐⇒ r(d2 m1 − d1 m2 ) = 0 for some r ∈ R. Prove that D−1 M is a left D−1 R-module. Solution: Just as with rings of fractions, we define modules of fractions as the set of equivalence classes under the ∼ relation defined here. We first need to define the addition on D−1 M and the scalar multiplication of D−1 R on D−1 M . (This is implicit in the question of the exercise.) The addition, as with addition in the ring of fractions D−1 R is def

(d1 , m1 ) + (d2 , m2 ) = (d1 d2 , d2 m1 + d1 m2 ) and the scalar multiplication is like usual multiplication, namely, def

(d1 , r) · (d2 , m) = (d1 d2 , rm). Before we proceed though, we need to check that these definitions are well defined. In other words, if there are other representatives for the same ∼-equivalence class, then the resulting operations must be equivalent. We first check the addition. Suppose that (d1 , m1 ) ∼ (d01 , m01 ) and that (d2 , m2 ) ∼ (d02 , m02 ). This means that there exist r1 , r2 ∈ R such that r1 (d01 m1 − d1 m01 ) = 0

and

r2 (d02 m2 − d2 m02 ) = 0.

10.3. INTRODUCTION TO MODULES

517

We must consider the difference d01 d02 (d2 m1 + d1 m2 ) − d1 d2 (d02 m01 + d01 m02 ) = d2 d02 (d01 m1 − d1 m01 ) + d1 d01 (d02 m2 − d2 m02 ). Multiplying this difference by r1 r2 , we get r1 r2 (d01 d02 (d2 m1 + d1 m2 ) − d1 d2 (d02 m01 + d01 m02 )) = d2 d02 r2 r1 (d01 m1 − d1 m01 ) + d1 d01 r1 r2 (d02 m2 − d2 m02 ) = 0. Thus, (d1 d2 , d2 m1 + d1 m2 ) ∼ (d01 d02 , d02 m01 + d01 m02 ), so the given operation of addition on equivalence classes is indeed well-defined. We now check that scalar multiplication is well-defined. Suppose that (d1 , r) ∼ (d01 , r0 ) and that (d2 , m) ∼ (d02 , m0 ). This means that there exist s, t ∈ R such that s(r0 d1 − rd01 ) = 0

and

t(d02 m − d2 m0 ) = 0.

We consider the difference d01 d02 rm − d1 d2 r0 m0 . Multiplying by st gives us st(d01 d02 rm − d1 d2 r0 m0 ) = st(d01 d02 rm − d01 d2 rm0 + rd01 d2 m0 − d1 d2 r0 m0 ) = std01 r(d02 m − d2 m0 ) + std2 (d01 r − d1 r0 )m0 = sd01 r(t(d02 m − d2 m0 )) + td2 (s((d01 r − d1 r0 ))m0 = 0 + 0 = 0. Thus, (d1 d2 , rm) ∼ (d01 d02 , r0 m0 ), so the given operation of scalar multiplication on equivalence classes is indeed well-defined. We now proceed to check the axioms of a left module. • Associativity of addition: Let (d1 , m1 ), (d2 , m2 ), (d3 , m3 ) ∈ D−1 M as equivalence classes. Then (d1 , m1 ) + ((d2 , m2 ) + (d3 , m3 )) = (d1 , m1 ) + (d2 d3 , d3 m2 + d2 m3 ) = (d1 d2 d3 , d2 d3 m1 + d1 d3 m2 + d1 d2 m3 ) = (d1 d2 , d2 m1 + d1 m2 ) + (d3 , m3 ) = ((d1 , m1 ) + (d2 , m2 )) + (d3 , m3 ). • Commutativity of addition: Let (d1 , m1 ), (d2 , m2 ) ∈ D−1 M as equivalence classes. Then (d1 , m1 ) + (d2 , m2 ) = (d1 d2 , d2 m1 + d1 m2 ) = (d2 d1 , d1 m2 + d2 m1 ) = (d2 , m2 ) + (d1 , m1 ). • Identity of addition: Let (d, m) ∈ D−1 M and consider any element (d0 , 0) ∈ D−1 M . Then (d, m) + (d0 , 0) = (dd0 , d0 m + d0) = (dd0 , d0 m) = (d, m). We point out that (d0 , 0) = (d00 , 0) for any d0 , d00 ∈ D. These are the additive zero. • Inverse for addition: Let (d, m) ∈ D−1 M and consider the elements (d, −m) ∈ D−1 M . Then (d, m) + (d, −m) = (d2 , dm + d(−m)) = (d2 , dm + (−dm)) = (d2 , 0). This shows that (d, −m) is the additive inverse of (d, m). • Let (d, r) ∈ D−1 R and let (d1 , m1 ), (d2 , m2 ) ∈ D−1 M . Then (d, r)((d1 , m1 ) + (d2 , m2 )) = (d, r)(d1 d2 , d2 m1 + d1 m2 ) = (dd1 d2 , rd2 m1 + rd1 m2 ) = (d2 d1 d2 , dd2 rm1 + dd1 rm2 ) = (dd1 , rm1 ) + (dd2 , rm2 ) = (d, r)(d1 , m1 ) + (d, r)(d2 , m2 ). • Let (d1 , r1 ), (d2 , r2 ) ∈ D−1 R and let (d, m) ∈ D−1 M . Then ((d1 , r1 ) + (d2 , r2 ))(d, m) = (d1 d2 , d2 r1 + d1 r2 )(d, m) = (d1 d2 d, d2 r1 m + d1 r2 m) = (d1 d2 d2 , d2 dr1 m + d1 dr2 m) = (d1 d, r1 m) + (d2 d, r2 m) = (d1 , r1 )(d, m) + (d2 , r2 )(d, m).

518

CHAPTER 10. MODULES AND ALGEBRA • Let (d1 , r1 ), (d2 , r2 ) ∈ D−1 R and let (d, m) ∈ D−1 M . Then ((d1 , r1 )(d2 , r2 ))(d, m) = (d1 d2 , r1 r2 )(d, m) = (d1 d2 d, r1 r2 m) = (d1 , r1 )(d2 d, r2 m) = (d1 , r1 )((d2 , r2 )(d, m)). • Finally, the ring D−1 R always has a multiplicative identity, namely (d0 , d0 ) for any d0 ∈ D. Then for any (d, m) ∈ D−1 M , we have (d0 , d0 )(d, m) = (d0 d, d0 m) = (d, m).

We have established all the required axioms to show that D−1 M is a D−1 R module. Exercise: 14 Section 10.3 Question: Consider the Z-module M = Zn . Decide with a proof which of the following subsets are submodules of M . a) {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn = 0} b) {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn = 5} c) {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn ∈ 5Z} d) {(x1 , x2 , . . . , xn ) ∈ M | x21 + x22 + · · · + x2n ∈ 5Z} e) {(x1 , x2 , . . . , xn ) ∈ M | xi ∈ iZ} o n 2 +···+nxn = 5 ∪ {(0, 0, . . . , 0)} f) (x1 , x2 , . . . , xn ) ∈ M | xx1 +2x +x +···+x 1 2 n Solution: Consider the Z-module M = Zn . a) Let N = {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn = 0}. The set N is nonempty since 0 ∈ N . Let x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) be in Zn and let r ∈ Z. Then for x + ry = (x1 + ry1 , x2 + ry2 , . . . , xn + ryn ) we have (x1 + ry1 ) + (x2 + ry2 ) + · · · + (xn + ryn ) = (x1 + x2 + · · · + xn ) + r(y1 + y2 + · · · + yn ) = 0, so x + ry ∈ N . By the One-Step Submodule Criterion, N is a submodule of Zn . b) Let N = {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn = 5}. This set is not closed under the addition. If x ∈ N and y ∈ N , then (x1 + y1 ) + (x2 + y2 ) + · · · + (xn + yn ) = (x1 + x2 + · · · + xn ) + (y1 + y2 + · · · + yn ) = 5 + 5 = 10 so x + y ∈ / N. c) Let N = {(x1 , x2 , . . . , xn ) ∈ M | x1 + x2 + · · · + xn ∈ 5Z}. The set N is nonempty since 0 ∈ N . Let x = (x1 , x2 , . . . , xn ) and y = (y1 , y2 , . . . , yn ) be in Zn and let r ∈ Z. Then for x + ry, we have (x1 + ry1 ) + (x2 + ry2 ) + · · · + (xn + ryn ) = (x1 + x2 + · · · + xn ) + r(y1 + y2 + · · · + yn ). But (x1 +x2 +· · ·+xn ) is a multiple of 5 by assumption and so is r(y1 +y2 +· · ·+yn ) because (y1 +y2 +· · ·+yn ) is. By the One-Step Submodule Criterion, N is a submodule of Zn . d) Let N = {(x1 , x2 , . . . , xn ) ∈ M | x21 + x22 + · · · + x2n ∈ 5Z}. This is not a submodule. For example (2, 1, 0, . . . , 0) ∈ N and (3, 1, 0, . . . , 0) ∈ N but the sum of these two (5, 2, 0, . . . , 0) has a sum of squares of 29, so is not in N . The subset N is not closed under addition. e) Let N = {(x1 , x2 , . . . , xn ) ∈ M | xi ∈ iZ}. This subset is nonempty since 0 = (0, 0, . . . , 0) ∈ N . Let x, y ∈ N and let r ∈ Z. Then xi ∈ iZ and yi ∈ iZ for i = 1, 2, . . . , n. But then ryi is also a multiple of i and hence so is xi + ryi . Thus, x + ry ∈ N . By the One-Step Submodule Criterion, N is a submodule of Zn . o n 2 +···+nxn = 5 ∪ {(0, 0, . . . , 0)}. In the definition, we obviously asf) Let N = (x1 , x2 , . . . , xn ) ∈ M | xx1 +2x +x +···+x 1 2 n sume that we cannot divide by 0, but the element 0 = (0, 0, . . . , 0) is explicitly included. This subset N is not closed under addition. Let x = (0, 2, −3, 0, . . . , 0) and y = (0, 0, 0, 0, 1, 0, . . . , 0). Then 0 + 2 · 2 + 3(−3) + 0 −5 = =5 2−3 −1

and

0 + 0 + 0 + 0 + 5(1) = 5. 1

Hence, x, y ∈ N . But x + y = (0, 2, −3, 0, 1, 0, . . . , 0) and the sum of the entries is not zero. Hence, the fraction is not defined so we cannot say that this element is in N . Hence N is not a submodule of M = Zn .

10.3. INTRODUCTION TO MODULES

519

Exercise: 15 Section 10.3 Question: Let M be a left R-module and let N1 and N2 be submodules. Define the addition of submodules as N1 + N2 = {m1 + m2 | m1 ∈ N1 , m2 ∈ N2 }. Prove that N1 + N2 is a submodule of M . Solution: Note that 0 ∈ N1 + N2 as 0 = 0 + 0. Hence, N1 + N2 is nonempty. Let x, y ∈ N1 + N2 and let r ∈ R. Then x = x1 + x2 and y = y1 + y2 for some x1 , y1 ∈ N1 and some x2 , y2 ∈ N2 . Thus, x + ry = (x1 + x2 ) + r(y1 + y2 ) = (x1 + ry1 ) + (x2 + ry2 ). Since N1 and N2 are submodules of M , then x1 + ry1 ∈ N1 and also x2 + ry2 ∈ N2 . Hence, x + ry ∈ N1 + N2 . By the One-Step Submodule Criterion, N1 + N2 is a submodule of M . Exercise: 16 Section 10.3 Question: Let M be a left R-module and let N1 and N2 be submodules. a) Prove that if N1 and N2 are submodules of M , then N1 ∩ N2 is a submodule of M . b) Prove that the intersection (not necessarily finite) of a collection of submodules of M is again a submodule. Solution: Let M be a left R-module and let N1 and N2 be submodules. a) Let N1 and N2 be submodules of M . Obviously, 0 ∈ N1 ∩ N2 . Now let x, y ∈ N1 ∩ N2 and let r ∈ R. Then x + ry ∈ N1 since N1 is a submodule of M and x + ry ∈ N2 since N2 is a submodule of M . Thus, x + ry ∈ N1 ∩ N2 . By the One-Step Submodule Criterion, N1 ∩ N2 is a submodule of M . T b) Let {Ni }i∈I be a collection of submodules of M . The additive 0 element is in Ni for all I ∈ I so N = i∈I is nonempty. Now let x, y ∈ N and let r ∈ R. Then x + ry ∈ Ni for all I ∈ I since Ni is a submodule of M . Thus, x + ry ∈ N . By the One-Step Submodule Criterion, the intersection N is a submodule of M . Exercise: 17 Section 10.3 Question: Let M be a left R-module and let I be an ideal contained in Ann(M ). Prove that M is a left R/I-module if for all r ∈ R and all m ∈ M we define (r + I)m = rm. Solution: Let us write r̄ for the element r + I in the quotient ring R/I. The set (M, +) is still an abelian group. Concerning the scalar multiplication, we first need to show that it is well defined. Let r1 , r2 be two representatives of the same coset in R/I. This means that (r1 − r2 ) ∈ I so (r2 − r1 )m = 0 for all m ∈ M . Hence, r1 m − r2 m = 0 so r1 m = r2 m. This shows that the scalar multiplication is well-defined. We prove the remaining axioms of a left ring module • Let r̄ ∈ R/I and let m1 , m2 ∈ M . Then r̄(m1 + m2 ) = r(m1 + m2 ) = rm1 + rm2 = r̄m1 + r̄m2 . • Let r̄, s̄ ∈ R/I and let m ∈ M . Then (r̄ + s̄) = (r + s)m = (r + s)m = rm + sm = r̄m + s̄m. • Let r̄, s̄ ∈ R/I and let m ∈ M . Then (r̄ s̄) = (rs)m = (rs)m = r(sm) = r̄(s̄m). • If R contains an identity 1, then 1̄ is an identity in R/I and for all m ∈ M we have 1̄m = 1m = m. This shows that M is an R/I module. Exercise: 18 Section 10.3 S∞ Question: Let N1 ⊆ N2 ⊆ · · · be a chain of submodules of an R-module M . Prove that i=1 Ni is a submodule of M . Solution: The result is true whether M is a left or a right module. We prove it for a left R-module. The S∞element 0 ∈ Ni for all integers i ≥ 1, but more importantly it is in one of the submodules of M . Thus N = i=1 Ni is nonempty. Now let x, y ∈ N and let r ∈ R. Since x, y ∈ N , then x ∈ Ni for some i ≥ 1 and y ∈ Nj for some j ≥ 1. Let k = max(i, j). Then x, y ∈ Nk because the submodules are in a chain of inclusion. Since Nk is a submodule of M , then x + ry ∈ Nk and thus x + ry ∈ N . By the One-Step Submodule Criterion, N is a submodule of M . Exercise: 19 Section 10.3 Question: Consider the Z-module M = Z45 ⊕ Z12 . Find the annihilator of M .

520

CHAPTER 10. MODULES AND ALGEBRA

Solution: Recall that the collection of abelian groups is precisely the collection of Z-modules. We write that abelian group Z45 ⊕ Z12 in addition notation as M = Z/45Z ⊕ Z/12Z. We note that an integer r satisfies r(1, 0) = (0, 0) if 45 | r and that r satisfies r(0, 1) = (0, 0) whenever 12 | r. Thus r annihilates both (1, 0) and (0, 1) if lcm(45, 12) = 180 divides r. Furthermore, it is easy to see that any element in 180Z annihilates every element in M . Hence, Ann(M ) = 180Z. Exercise: 20 Section 10.3 Question: Let R be a ring and let M be a left R-module. Prove that if R is an integral domain, then the set of torsion elements Tor(M ), defined in (10.6), is a submodule of M . Give an example of a module M with R not an integral domain where Tor(M ) is not a submodule of M . Solution: [The textbook has a typo in equation (10.6). The definition of torsion is corrected here below.] Let R be an integral domain, let M be a left R-module and consider the set of torsion elements Tor(M ) = {m ∈ M | ∃r ∈ R − {0}, rm = 0}. Obviously Tor(M ) is nonempty since it contains the 0 element. Now let x, y ∈ Tor(M ) and let r ∈ R. Then there exists r1 , r2 ∈ R such that r1 x = 0 and r2 y = 0. Then since R is an integral domain, r1 r2 6= 0 and r1 r2 (x + ry) = r2 (r1 x) + r1 r(r2 y) = 0 + 0 = 0. By the One-Step Subgroup Criterion, Tor(M ) is a submodule. However, consider the situation where R = Z ⊕ Z acting on itself as a module by left multiplication. Then x = (1, 0) and y = (0, 1) are torsion elements because (0, 1)(1, 0) = (0, 0). However, x + y = (1, 1) is not a torsion element: (r, s)(1, 1) = (0, 0) if and only if r = 0 and s = 0, so (r, s) = (0, 0). Exercise: 21 Section 10.3 Question: Let R be an integral domain and let M1 and M2 be R-modules. Prove that Tor(M1 ⊕ M2 ) = Tor(M1 ) ⊕ Tor(M2 ). Solution: By definition, Tor(M1 ⊕ M2 ) = {(m1 , m2 ) ∈ M1 ⊕ M2 | ∃r ∈ R − {0}, r(m1 , m2 ) = (0, 0)}. This shows that m1 ∈ Tor(M1 ) and m2 ∈ Tor(M2 ). Thus, Tor(M1 ⊕ M2 ) ⊆ Tor(M1 ) ⊕ Tor(M2 ). Conversely, let (m1 , m2 ) ∈ Tor(M1 ) ⊕ Tor(M2 ). Then there exists r1 ∈ R − {0} such that r1 m1 = 0 and r2 ∈ R − {0} such that r2 m2 = 0. Since R is an integral domain, then r1 r1 = r2 r1 6= 0. Then r1 r2 (m1 , m2 ) = (r2 (r1 m1 ), r1 (r2 m2 )) = (0, 0). Hence, Tor(M1 ) ⊕ Tor(M2 ) ⊆ Tor(M1 ⊕ M2 ). This shows that these sets are equal. Exercise: 22 Section 10.3 Question: Consider the ring R = Z[x]. a) Let M = F25 and define the scalar multiplication by Z on M as the usual one with integers and where x acts on F25 by multiplying the vectors by 0 1 . 0 0 Prove that this scalar multiplication equips M with the structure of an R-module. b) Determine the annihilator Ann(M ). Solution: Consider the ring R = Z[x]. a) It is clear that (M, +) is an abelian group. Consequently, it is a Z-module. We notice that the element x satisfies axiom (1) for ring modules. Using axioms (1-3) we define how polynomials in R act on M so that p(x) · (v1 + v2 ) = p(x) · v1 + p(x) · v2 (p(x) + q(x)) · v = p(x) · v + q(x) · v (p(x)q(x)) · v = p(x) · (q(x) · v). The last axiom is perhaps not as obvious as the others. If p(x) = am xm + · · · + a1 x + a0 and q(x) = bn xn + · · · + b1 x + b0 , then p(x)q(x) = cm+n xm+n + · · · + c1 x + c0

where ck =

k X i=0

ai bk−i .

10.3. INTRODUCTION TO MODULES

521

Then for all v ∈ M , (p(x)q(x)) · v =

=

m+n X k=0 m+n X

=

i=0

k

0 ck 0

1 0

0 ai 0

i

k=0 m X

ck x k · v

1 0

v  ·

n X j=0

bj

0 0

 j 1 · v 0

= p(x) · (q(x) · v). b) The set Ann(M ) is the ideal in R defined as Ann(M ) = {r ∈ R | rm = 0 for all m ∈ M }. We note that the element 5 is in the annihilator of M . Furthermore, x2 acts on M as the 0 matrix. Hence, x2 ∈ Ann(M ). We deduce that the ideal I = (5, x2 ) is in Ann(M ). There are only 10 elements in R/(5, x2 ), namely polynomials that are congruent modulo I to a + bx where a, b ∈ {0, 1, 2, 3, 4}. However, 0̄ b̄ (a + bx) · = . 1̄ ā Thus, among these polynomials, on the 0 polynomial annihilates this vector. Hence, we deduce that Ann(M ) ⊆ (5, x2 ) and thus that Ann(M ) = (5, x2 ). Exercise: 23 Section 10.3 Question: Let M be a left R-module. Let I be a right ideal of R. Prove that I ⊆ Ann(Ann(I)). Give an example to show that this containment might be strict. Solution: Consider the submodule N = Ann(I) and the ideal Ann(N ). By definition Ann(I) = {m ∈ M | am = 0 for all a ∈ I}. Now let b ∈ I. Every element n ∈ N satisfies bn = 0 so b ∈ Ann(N ). Thus I ⊆ Ann(Ann(I)). a Consider the previous exercise. Suppose that I = (x) as an ideal in R = Z[x]. Then Ann((x)) = { ∈ 0 F25 | a ∈ F5 }. But, Ann(Ann((x)) = (5, x). This gives an example of strict containment. Exercise: 24 Section 10.3 Question: Let M be a left R-module. Let N be a submodule of M . Prove that N ⊆ Ann(Ann(N )). Give an example to show that this containment might be strict. Solution: Let I = Ann(N ). By definition I = {a ∈ R | an = 0 for all n ∈ N }. So if n ∈ N , then an = 0 for all a ∈ I. In other words, n ∈ Ann(I) = Ann(Ann(N )). Thus, ⊆ Ann(Ann(N )). Let R = R and let M be the vector space M = R2 . Consider the submodule N = Span 10 . We have Ann(N ) = (0), the 0-ideal in R. However, Ann((0)) = M . Thus N is strictly contained in Ann(Ann(N )). Exercise: 25 Section 10.3 Question: Consider the R[x]-module V = R2 , equipped with the linear transformation on V corresponding to projection onto the y = x line. Prove that the only submodules of V are Span 11 and Span −1 . 1 Solution: The only nontrivial, proper submodules of V are Span 11 and Span −1 . A module always contains 1 {0} and V as submodules. Because a submodule of V closed under scalar multiplication (and closed under addition) is a vector subspace. A submodule must also be closed under multiplication by polynomial p(x) ∈ R[x]. However, in this case, if a subspace is closed under the action of x, then it is closed under the action of any polynomial. Since V is 2 dimensional, a nontrivial proper submodule will be any 1-dimensional eigenspace of the linear transformation corresponding to projection onto the y = x line. Any linear transformation on a 2dimensional vector space has at most 2 eigenspaces. We note that Span 11 is an eigenspace of eigenvalue since any vector on y = x is unchanged by the projection, and that Span −1 is the subspace perpendicular to the 1 y = x line so this subspace is an eigenspace of eigenvalue 0. Exercise: 26 Section 10.3 Question: Consider the C[x]-module V = C2 , equipped with the linear transformation T : V → V given by z1 0 1 z1 T = . z2 −1 0 z2

522

CHAPTER 10. MODULES AND ALGEBRA

a) Prove that V has exactly four submodules. b) Explain why this differs from the first specific example in Example 10.3.19. Solution: a) Obviously, the C[x]-module V = C2 as defined in the problem has the submodules of V and {~0}. Since a submodule of V is closed under addition and scalar multiplication, it is a vector subspace. Furthermore, it must be closed under the action of the linear transformation T as well. The only nontrivial proper subspaces of V are 1 dimensional (over C) subspaces. If N = Span(~v ) is a one-dimensional subspace that is closed under T , then T (v) ∈ N , so T (~v ) = λ~v . In other words, N must be an eigenspace. In this example, the eigenvalues of T are λ = ±i. The eigenspace of λ = i is Span 1i and the eigenspace 1 of λ = −i is Span −i . We conclude that the C[x]-module V = C2 has exactly four subspaces. Exercise: 27 Section 10.3 Question: Consider the R[x]-module V = R3 , equipped with the linear transformation T : V → V given by      x 0 2 0 x T y  = −1 3 0 y  . z 0 0 3 z Determine all the R[x]-submodules of V . Solution: We note that eigenspaces of T are submodules of the R[x]-module V = R3 . Elementary linear algebra shows that T has three distinct eigenvalues, namely 3, 2, and 1 with respective eigenspaces       1 2 0 E3 = Span 0 , E2 = Span 1 , E1 = Span 1 . 0 0 1 Call ~e3 , ~e2 , and ~e1 the eigenvectors given in the above expression. Since a submodule W of V is closed under scalar multiplication and under addition of vectors, then it is a vector subspace. The subspace W must also satisfy T (w) ~ ∈ W for all w ~ ∈ W . As always, {0} and V are submodules. Also, the eigenspaces are submodules. For the two-dimensional submodules, we have W = Span(w ~ 1, w ~ 2 ) with T (w ~ i ) ∈ W . Now since {~e1 , ~e2 , ~e3 } is a basis of V , then we can write ( w ~1 w ~2

= c1~e1 + c2~e2 + c3~e3 = d1~e1 + d2~e2 + d3~e3 .

By doing Gauss-Jordan elimination on these coefficients, we arrive at three possible cases. Case 1: c1 = d1 = 0: In this case W = Span(~e2 , ~e3 ). Case 2: c1 = 1, d1 = d2 = 0, d3 = 1 and c3 = 0: In this case, vw2 = ~e3 is an eigenvector and T (w ~ 1) = c1~e1 + 2c2~e2 , which forces c2 = 0 (since c1 = 1) in order for T (w ~ 1 ) to be in W . Thus W = Span(~e1 , ~e3 ). Case 3: c1 = 1, c2 = 0, d1 = 0, d2 = 1: Then w ~ 1 = ~e1 + c3~e3 and w ~ 2 = ~e2 + d3~e3 . Then since W is closed under T , then we have T (w ~ 1) − w ~ 1 = 2c3~e3 ∈ W and also T (w ~ 2 ) − 2w ~ 2 = d3~e3 . If either c3 6= 0 or d3 6= 0, then W contains all three eigenvectors. Hence, we conclude that we must have c3 = d3 = 0, in which cases W = Span(~e1 , ~e2 ). We have shown the V has 8 submodules: {~0}, E1 , E2 , E3 , E1 + E2 , E1 + E3 , E2 + E3 , and V . Exercise: 28 Section 10.3 Question: Consider the R[x]-module V = R3 , equipped with the linear transformation T : V → V given by      x 0 1 0 x T y  = 0 0 1 y  . z 0 0 0 z Determine all the R[x]-submodules of V .

10.3. INTRODUCTION TO MODULES

523

Solution: As in the previous exercises, submodules of V must be subspaces of V that are preserved under the linear transformation T . We observe that T has a single eigenvalue of λ = 0 and also a one-dimensional eigenspace, namely   1 E0 = Span 0 = Span(~ı). 0 This is the only one-dimensional submodule. We also observe that         0 1 0 0 T 0 = 1 and T 1 = 0 1 0 0 0 Suppose that a submodule W contains a vector ~v = a~ı + b~ + c~k with c 6= 0. Then 1c T 2 (~v ) = ~i ∈ W . Then W also contains w ~ = b~ + c~k and also T (w) ~ = b~ı + c~. Consequently, W also contains 1c (T (w) ~ − b~ı) = ~. Then W also 1 ~k. Thus, the only submodule containing a vector with a nonzero component in the ~k direction ( w ~ − a~ ı − b~  ) = c is the whole vector space. Thus, the only two-dimensional submodule of V is Span(~ı, ~). Exercise: 29 Section 10.3 Question: Let (M, +) be an abelian group. Recall that End(M ) is the set of endomorphisms of M . For any f, g ∈ End(M ), define the addition f + g and product f · g by def

(f + g)(x) = f (x) + g(x)

and

def

(f · g)(x) = f (g(x)).

Also define the 0 on End(M ) as the 0 function and the 1 in End(M ) as the identity function on M . a) Prove that (End(M ), +, ·) is a ring with additive identity 0 and multiplicative identity 1. b) Prove that any structure that makes M into a left R-module defines a ring homomorphism from R into (End(M ), +, ·). Solution: We consider an abelian group (M, +) and consider the addition + and multiplication ◦ on the set of homomorphisms of M into itself, denoted End(M ). a) We first show that + is a binary operation on End(M ). Let f, g : M → M be homomorphisms. Then for all x, y ∈ M , (f + g)(x + y) = f (x + y) + g(x + y) = f (x) + f (y) + g(x) + g(y) = (f (x) + g(x)) + (f (y) + g(y)) = (f + g)(x) + (f + g)(y). Thus, f + g is again a homomorphism on M . Associativity, commutativity and identity hold easily on End(M ). The additive inverse of a function f (x) is −f (x) for all x ∈ M . Thus, End(M ) is a group. Function composition is associative so the multiplication · is associative. Finally, for all f, g, h ∈ End(M ) and for all x ∈ M , we have (f · (g + h))(x) = f ((g + h)(x)) = f (g(x) + h(x)) = f (g(x)) + f (h(x)) because f is a homomorphism. Hence, f · (g + h) = f · g + f · h. Furthermore, ((f + g) · h)(x) = (f + g)(h(x)) = f (h(x)) + g(h(x)) so we also have f · h + g · h. The identity function 1 serves as the multiplicative identity. b) Let R be any ring and let M be a left R-module (which must by definition be an abelian group). Consider the function ψ : R → End(M ) by ψ(r) = (m → rm). In other words, ψ(r) is the function that maps m to rm. We note that ψ(r) ∈ End(M ) because of the first axiom of modules, namely that r(m1 + m2 ) = rm1 + rm2 , which means that ψ(r)(m1 + m2 ) = ψ(r)(m1 ) + ψ(r)(m2 ). The function ψ(r) is a ring homomorphism because of axioms (2) and (3) of modules imply that ψ(r1 + r2 ) = ψ(r1 ) + ψ(r2 ) and ψ(r1 r2 ) = ψ(r1 )ψ(r2 ). Furthermore, ψ is a unital ring homomorphism in that if R has an identity, then ψ(1R ) = idM , which is the multiplicative identity in End(M ). Exercise: 30 Section 10.3 Question: Let R be a ring with identity. Let M be the abelian group (R, +).

524

CHAPTER 10. MODULES AND ALGEBRA

a) Prove that the left-multiplication of R on itself makes M into a left R-module. b) Use Exercise 10.3.29 to show that R is isomorphic to a subring of End(M ). [The result of this exercise shows that every ring is isomorphic to a subgroup of some endomorphism ring of an abelian group.] Solution: Let R be a ring with identity. a) If we only consider the addition on R, then (R, +) is an abelian group. If we consider the left-multiplication of R on itself, then axiom (1) and (2) follow from distributivity of multiplication over +, (3) follows from associativity of multiplication in R, and (4) is the definition of the identity in the ring. Thus R is a left R-module. b) By Exercise 10.3.29, we know that the action of left-multiplication defines a homomorphism from ψ : R → End(R) via ψ(r)(x) = rx. Now ψ(r)(1) = r for all r in R so ψ(r1 ) = ψ(r2 ) implies that r1 = r2 . Thus, ψ is injective. By the First Isomorphism Theorem, R is isomorphic to a subring of End(R). Exercise: 31 Section 10.3 Question: Section 6.7 introduced the notion of an algebraic integer and in particular the concept of algebraic closure. (See Definition 6.7.7.) Let R be a subring of a commutative ring S and suppose that S has an identity 1 that is also in R. Prove that the following three statements are equivalent: a) s ∈ S is integral over R; b) R[s] is a finitely generated R-module; c) s is an element in some subring T , where R ⊆ T ⊆ S, and such that T is a finitely generated R-module. Solution: Recall that an element s ∈ S is called integral over R if there exists a monic polynomial f (x) ∈ R[x] such that f (s) = 0. (1) =⇒ (2): Let s be integral and let f (x) = xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial such that f (s) = 0. Then sn = −an−1 xn−1 − · · · − a1 x − a0 Suppose that for some k ≥ n, the element sk can be written as an R-linear combination of {sn−1 , . . . , s, 1}. Say sk = cn−1 sn−1 + · · · + c1 s + c0 . Then sk+1 = cn−1 sn + · · · + c1 s2 + c0 s = cn−2 sn−1 + · · · + c1 s2 + c0 s + cn−1 (−an−1 xn−1 − · · · − a1 x − a0 ). Hence, sk+1 can be written as an R-linear combination of {sn−1 , . . . , s, 1}. By induction, this result holds for all positive integers k. Consequently, every polynomial in R[s] is an R-linear combination of {sn−1 , . . . , s, 1}. This means that R[s] is a finitely generated R-submodule. (2) =⇒ (3): This is obvious, since we can use the ring T = R[s]. (3) =⇒ (1): We note that T is an R[s]-module by left multiplication in S. Since S contains an identity 1 that is also in S, then for all a ∈ R[s], we have a · 1 = 0 if and only if a = 0. Let {t1 , t2 , . . . , tm } be a set of elements that generate T over R. Then for each i, there exists aij ∈ R such that  (a11 − s)t1 + a12 t2 + ··· + a1m tm = 0    m  X a21 t1 + (a22 − s)t2 + · · · + a2m tm = 0 sti = aij tj ⇐⇒ .. .. .. ..  . . . .  j=1   am1 t1 + am2 t2 + · · · + (amm − s)tm = 0. Consider the matrix A − sI, where A = (aij ). This matrix A − sI ∈ Mm (T ). If T were a field, we would use linear algebra and say that det(A − sI) = 0 because   t1  t2    (A − sI)  .  = ~0  ..  tm and ~t 6= ~0. However, we do not have linear algebra at our disposal. Instead, recall that the classical adjoint B ∗ of a matrix B, defined by b∗ij = (−1)i+j Bji where Bji is the determinant of the matrix B with row j and

10.3. INTRODUCTION TO MODULES

525

column i removed, is such that B ∗ B = det(B)I. This property holds still holds where the coefficients are in any commutative ring. Thus,     t1 t1  t2   t2      (−1)m (A − sI)∗ (A − sI)  .  = ~0 ⇐⇒ (−1)m det(A − sI)  .  = ~0, . .  .   .  tm

tm

so each ti is annihilated by det(sI − A). Now if 1 = r1 t1 + r2 t2 + · · · + rm tm , then we see that det(sI − A)1 = r1 det(sI − A)t1 + r2 det(sI − A)t2 + · · · + rm det(sI − A)tm = 0. However, we had pointed out that a · 1 = 0 if and only if a = 0. Thus, we deduce that det(sI − A) = 0 and this gives a monic equation in s with coefficients in R. Exercise: 32 Section 10.3 Question: Use Exercise 10.3.31 to prove the following facts about integrality in ring extensions. Let R and S be as in Exercise 10.3.31. a) If s and s0 are in S and integral over R, then s ± s0 and ss0 are also integral over R. b) The integral closure of R is a subring of S that contains R. Solution: Let R be a subring of a commutative ring S and suppose that S has an identity 1 that is also in R. a) Let s, s0 ∈ S be integral over R. By (3) of the previous exercise, s is in a subring T of S that is finitely generated by {t1 , t2 , . . . , tm } as an R-module and s0 is in a subring T 0 of S that is finitely generated by {t01 , t02 , . . . , t0n } as an R-module. Consider the submodule M of S generated as an R-module by {ti t0j | 1 ≤ i ≤ m, 1 ≤ j ≤ n}. Then it is not hard to check that M is a subring of S (because tk t` is an R-linear combination of the ti and because t0k t0` is an R-linear combination of the t0j , which leads to M being closed under multiplication). Furthermore, M contains both T and T 0 because 1 is a linear combination of the ti with 1 = r1 t1 + r2 t2 + · · · + rm tm with ri ∈ R so t0j = r1 t0j t1 + r2 t0j t2 + · · · + rm t0j tm and a similar reason holds to show that ti ∈ M . Thus M is a finitely generated R-module, subring of S, that contains s + s0 , s − s0 , and ss0 . By (3) of the previous exercise, we deduce that these three elements are integral over R. b) By the previous part, the integral closure of R in S is closed under subtraction and multiplication. Thus, the integral closure is a subring of S. Furthermore, every element r ∈ R is the root of the monic polynomial x − r. Hence, the integral closure contains R as well. Exercise: 33 Section 10.3 Question: Let R be a commutative ring and consider the set Mn (R) of n × n matrices with coefficients in R. Define the function [, ] : Mn (R) × Mn (R) → Mn (R) by [A, B] = AB − BA. Prove that [, ] is an R-bilinear product on the R-module Mn (R). Show that the bilinear function [, ] is not associative on Mn (R) and deduce that (Mn (R), [, ]) is an R-algebra that is not a ring. Solution: Define the function [, ] : Mn (R) × Mn (R) → Mn (R) by [A, B] = AB − BA. Then for all A1 , A2 , B1 , B2 ∈ Mn (R) and for all c1 c2 , d1 , d2 ∈ R we have [c1 A1 + c2 A2 , B] = (c1 A1 + c2 A2 )B − B(c1 A1 + c2 A2 ) = c1 (A1 B − BA1 ) + c( A2 B − BA2 ) = c1 [A1 , B] + c2 [A2 , B], [A, d1 B1 + d2 B2 ] = A(d1 B1 + d2 B2 ) − (d1 B1 + d2 B2 )A = d1 (AB1 − B1 A) + d2 (AB2 − B2 A) = d1 [A, B1 ] + d2 [A, B2 ]. Thus [, ] is an R-bilinear product on the R-module Mn (R). Furthermore, [A, [B, C]] = A(BC − CB) − (BC − CB)A = ABC − ACB − BCA + CBA [[A, B], C] = (AB − BA)C − C(AB − BA) = ABC − BAC − CAB + CBA and we notice that these are not the same. Their difference is −ACB − BCA + BAC + CAB, which is not necessarily 0. Consequently, this product is not associative on Mn (R).

526

CHAPTER 10. MODULES AND ALGEBRA

Exercise: 34 Section 10.3 Question: Let R be a ring with identity 1R . Suppose that A is a ring with identity 1A that is also a left R-module satisfying r · (ab) = (r · a)b = a(r · b) for all r ∈ R and all a, b ∈ A. a) Prove that the function f : R → A defined by f (r) = r · 1A is a ring homomorphism mapping 1R to 1A . b) Prove that f (R) is contained in the center C(A). c) Deduce that A has the structure of a unital associative R-algebra. Solution: a) For all r1 , r2 ∈ R, we have f (r1 + r2 ) = (r1 + r2 ) · 1A = r1 · 1A + r2 · 1A = f (r1 ) + f (r2 ) f (r1 r2 ) = (r1 r2 ) · 1A = r1 · (r2 · 12A ) = (r1 · 1A )(r2 · 1A ) = f (r1 )f (r2 ). Thus f is a ring homomorphism. b) Let r ∈ R and let a ∈ A. Then f (r) = r · 1A . By the properties of the module action, af (r) = a(r · 1A ) = r · a = r · (1A a) = (r · 1)a = f (r)a. Thus f (r) is in the center of the ring A, so f (R) is a subring of C(A). c) We see that A is an R-algebra with the product of A as the product. Since the multiplication in A is associative and since A contains an identity 1A , then A is a unital associative R-algebra.

Exercise: 35 Section 10.3 Question: Let (V, [, ]) be an algebra over a field F (an vector space over F equipped with a bilinear function). k Suppose that V is finite dimensional and has an ordered basis B = (e1 , e2 , . . . , en ). Let γij ∈ F for 1 ≤ i, j, k ≤ n 3 be a collection of n constants defined by [ei , ej ] =

n X

k γij ek .

k=1

(Note that k is a superscript index and not a power.) Prove that if 

 a1  a2    [v]B =  .   .. 

and

  b1  b2    [w]B =  .  ,  .. 

an

bn

then   c1  c2    [v, w] B =  .   ..  cn

where

ck =

n X n X

k γij ai bj .

i=1 j=1

k [The collection of constants γij is called the structure constants of the F -algebra (V, [, ]).]

10.4. HOMOMORPHISMS AND QUOTIENT MODULES

527

Solution: Since [, ] is a bilinear function on V we have   n n X X [v, w] =  ai ei , bj ej  i=1

=

n X

j=1

 ai ei ,

i=1

=

n X

 bj ej 

j=1

n X n X

ai bj [ei , ej ] i=1 j=1 n X n X n X k = ai bj γij ek i=1 j=1 k=1 =

n X



n X n X

 k=1

 k ai bj γij ek .

i=1 j=1

This proves the desired result.

10.4 – Homomorphisms and Quotient Modules Exercise: 1 Section 10.4 Question: Suppose that R has an identity 1 6= 0. Prove that a function ϕ : M → N between left R-modules is an R-module homomorphism if and only if ϕ(rm1 + m2 ) = rϕ(m1 ) + ϕ(m2 ) for all m1 , m2 ∈ M and all r ∈ R. Solution: Suppose that ϕ : M → N is a left R-module homomorphism. Then for all r ∈ R and for all m1 , m2 ∈ M , ϕ(rm1 + m2 ) = ϕ(rm1 ) + ϕ(m2 ) = rϕ(m1 ) + ϕ(m2 ). Conversely, suppose that ϕ : M → N is a function such that ϕ(rm1 + m2 ) = rϕ(m1 ) + ϕ(m2 ) for all m1 , m2 ∈ M and all r ∈ R. Then setting r = 1R we deduce that ϕ(m1 + m2 ) = ϕ(m1 ) + ϕ(m2 ) for all m1 , m2 . Now let a ∈ R and let m ∈ M ; setting r = a − 1R and m1 = m2 = m, we have aϕ(m) = (a − 1)ϕ(m) + ϕ(m) = ϕ((a − 1)m + m) = ϕ(am). Thus, we’ve shown the two defining properties of a module homomorphism. Thus, ϕ is an R-module homomorphism. Exercise: 2 Section 10.4 Question: Let ϕ : M → N be an R-module homomorphism. Prove that ϕ is injective if and only if Ker ϕ = {0}. Solution: Suppose that ϕ : M → N is an injective R-module homomorphism. We know that ϕ(0) = 0 but since it is injective Ker ϕ = {m ∈ M | ϕ(m) = 0} = {0}. Now suppose that Ker ϕ = {0} and suppose that ϕ(m1 ) = ϕ(m2 ). Then ϕ(m1 ) − ϕ(m2 ) = 0, which implies that ϕ(m1 − m2 ) = 0. Since, Ker ϕ = {0}, we deduce that m1 − m2 = 0, so m1 = m2 . his shows that ϕ is injective. Exercise: 3 Section 10.4 Question: Let M1 , M2 , . . . , Mk be a collection of R-modules. Define the ith projection function πi : M1 ⊕ M2 ⊕ · · · ⊕ Mk → Mi by πi (m1 , m2 , . . . , mk ) = mi . Prove that πi is an R-module homomorphism. Solution: Let m = (m1 , m2 , . . . , mk ) and m̄ = (m̄1 , m̄2 , . . . , m̄k ) be two elements in M1 ⊕ M2 ⊕ · · · ⊕ Mk . Then πi (m + m̄) = πi (m1 + m̄1 , m2 + m̄2 , . . . , mk + m̄k ) = mi + m̄i = ϕ(m) + ϕ(m̄).

528

CHAPTER 10. MODULES AND ALGEBRA

Furthermore for all r ∈ R and all m ∈ M1 ⊕ M2 ⊕ · · · ⊕ Mk , we have πi (rm) = πi (rm1 , rm2 , . . . , rmk ) = rmi = rπi (m). Hence, the projection onto the ith component is an R-module homomorphism. Exercise: 4 Section 10.4 Question: Let R be a ring and suppose that constants c1 , c2 , . . . , cn are in the center of R. Prove that the function ϕ : Rn → R defined by ϕ(r1 , r2 , . . . , rn ) = c1 r1 + c2 r2 + · · · + cn rn is an R-module homomorphism. Solution: Let r = (r1 , r2 , . . . , rn ) and s = (s1 , s2 , . . . , sn ) be two elements in the module Rn . Then ϕ(r + s) = ϕ(r1 + s2 , r2 + s2 , . . . , rn + sn ) = c1 (r1 + s2 ) + c2 (r2 + s2 ) + · · · + cn (rn + sn ) = c1 r1 + c2 r2 + · · · + cn rn + c1 s1 + c2 s2 + · · · + cn sn = ϕ(r) + ϕ(s). Also, let a ∈ R. Then ϕ(ar) = ϕ(ar1 , ar2 , . . . , arn ) = c1 ar1 + c2 ar2 + · · · + cn arn = a(c1 r1 + c2 r2 + · · · + cn rn ), because ri ∈ C(R). Then, ϕ(ar) = aϕ(r). These two statements establish that ϕ is an R module homomorphism.

Exercise: 5 Section 10.4 Question: Let R be a ring and let ϕ : M → N be an R-module homomorphism between two left R-modules. a) Prove that if U is a submodule of M , then ϕ(U ) is a submodule of N . b) Prove that if V is a submodule of N , then ϕ−1 (V ) is a submodule of M . Solution: Let R be a ring and let ϕ : M → N be an R-module homomorphism a) Let n1 , n2 ∈ ϕ(U ). Then there exist m1 , m2 ∈ U with ϕ(m1 ) = n1 and ϕ(m2 ) = n2 . Then n1 + n2 = ϕ(m1 + m2 ) so this is in ϕ(U ). Hence, ϕ(U ) is closed under addition. Also, let r ∈ R and n ∈ ϕ(U ). Then there exists m ∈ U such that ϕ(m) = n. But then rm ∈ U so ϕ(rm) = rϕ(m) ∈ ϕ(U ). This shoes that ϕ(U ) is a submodule of N . b) Let U be a submodule of N . Let m1 , m2 ∈ ϕ−1 (U ). This means that ϕ(m1 ) ∈ U and ϕ(m2 ) ∈ U . Since is closed under addition, we deduce that ϕ(m1 + m2 ) = ϕ(m1 ) + ϕ(m2 ) ∈ U since U is closed under addition. Hence, m1 + m2 ∈ ϕ−1 (U ). Also, for all r ∈ R and all m ∈ ϕ−1 (U ), then ϕ(m) ∈ U . Then ϕ(rm) = rϕ(m) ∈ U since U is closed under the scalar multiplication. Exercise: 6 Section 10.3 2 2 ∼ Question: Let R = Q[x] and consider the √module M = Q[x]/(x − 2). We know that as a ring Q[x]/(x − 2) = √ Q[ 2] so we write elements in M as a + b 2. √ a) Determine x · (a + b 2). √ b) Determine xn · (a + b 2). √ c) Calculate (x3 − x2 + 3x + 1) · (a + b 2). √ Solution: 2]. When we write the elements in M = Q[x]/(x2 −2) √ This exercise is defining an √ action of √ Q[x] on Q[ √ as a + b 2, we are equating x with 2 (or − 2). So a + b 2 is in fact a + bx. √ √ √ a) x · (a + b 2) = x(a + bx) = ax + bx2 = 2b + ax, so we write x · (a + b 2) = 2b + a 2. √ b) xn · (a + b 2) has a different value for whether n is odd or even. ( n √ √ 2 2 (a + b 2) if n is even n √ x · (a + b 2) = n−1 2 2 (2b + a 2) if n is odd. c) We calculate √ √ √ √ √ (x3 − x2 + 3x + 1) · (a + b 2) = x3 · (a + b 2) − x2 · (a + b 2) + 3x · (a + b 2) + 1 · (a + b 2) √ √ √ √ = 2(2b + a 2) − 2(a + b 2) + 3(2b + a 2) + (a + b 2) √ = (−a + 10b) + (5a − b) 2.

10.4. HOMOMORPHISMS AND QUOTIENT MODULES

529

Exercise: 7 Section 10.4 Question: Let R be a ring. Let A = (aij ) ∈ Mn×m (R). Supposing that we write elements of Rn and Rm as column vectors, prove that the function f : Rm → Rn defined by      r1 a11 a12 · · · a1m r1  r2   a21 a22 · · · a2m   r2       f . = . .. ..   ..  ..  ..   .. . . .  .  rm

an1

an2

···

anm

rm

is an R-module homomorphism if R is a commutative ring. Prove that f might not be an R-module homomorphism when R is not commutative. Solution: Let m1 , m2 ∈ Rm with     s1 r1  s2   r2      m1 =  .  and m2 =  .   ..   ..  sm rm with ri , si ∈ R. Then  a11  a21  f (m1 + m2 ) =  .  ..

a12 a22 .. .

··· ··· .. .

     r1 s1 a1m  r2   s2  a2m       ..   ..  +  ..  .   .   . 

rm sm an1 an2 · · · anm   a11 (r1 + s1 ) + a12 (r2 + s2 ) + · · · + a1m (rm + sm )  a21 (r1 + s1 ) + a22 (r2 + s2 ) + · · · + a2m (rm + sm )    =  ..   . am1 (r1 + s1 ) + am2 (r2 + s2 ) + · · · + amm (rm + sm )    a11 s1 + a12 s2 + · · · + a1m sm a11 r1 + a12 r2 + · · · + a1m rm  a21 r1 + a22 r2 + · · · + a2m rm   a21 s1 + a22 s2 + · · · + a2m sm      =  + .. ..     . . 

am1 r1 + am2 r2 + · · · + amm rm

= f (m1 ) + f (m2 ).

am1 s1 + am2 s2 + · · · + amm sm

Now let c ∈ R. We have 

a11  a21  f (cm1 ) =  .  ..

a12 a22 .. .

··· ··· .. .

    a1m cr1 a11 cr1 + a12 cr2 + · · · + a1m crm     a2m    cr2   a21 cr1 + a22 cr2 + · · · + a2m crm   ..   ..  =  ..  .  .   .

an1 an2 · · · anm crm am1 cr1 + am2 cr2 + · · · + amm crm   ca11 r1 + ca12 r2 + · · · + ca1m rm  ca21 r1 + ca22 r2 + · · · + ca2m rm    =  ..   . cam1 r1 + cam2 r2 + · · · + camm rm   a11 r1 + a12 r2 + · · · + a1m rm  a21 r1 + a22 r2 + · · · + a2m rm    = c  = cf (m1 ). ..   . am1 r1 + am2 r2 + · · · + amm rm These show that f is an R-module homomorphisms. The latter axiom relied on the fact that c could commute with all the aij . As a counter-example when R is not commutative, consider R = H and consider a situation f : H2 → H2 defined by α 1 i α f = . β j k β

530

CHAPTER 10. MODULES AND ALGEBRA

The first axiom of module homomorphisms still holds but for example let m = f (cm) =

1 j

i k

2+i j

and c = k. Then

j + 2k 1 + j + 2k = −iβ −1 + 2i − jβ

but

1 cf (m) = k j

i k

2+i 2+i+k −1 + j + 2k =k = . jβ 2j − k − iβ 1 − 2i − jβ

In particular, f (cm) 6= cf (m). Exercise: 8 Section 10.4 Question: Let F be a field and let V and W be F [x]-modules with such that x acts on V according to T : V → V and x acts on W according to S : W → W . Prove that V ∼ = W as F [x] modules if and only if V and W are isomorphic as vector spaces and T and S are similar, i.e., S = ϕ ◦ T ◦ ϕ−1 . Solution: Suppose that V ∼ = W . Then there exists a module isomorphism ϕ : V → W . Thus ϕ is first of all a bijective function. Since ϕ(v1 + v2 ) = ϕ(v1 ) + ϕ(v2 ) and ϕ(cv) = cϕ(v) for all v1 , v2 , v ∈ V and c ∈ F , then ϕ is also a linear transformation. Hence, ϕ is a vector space isomorphism. Furthermore, for all v ∈ V we have ϕ(xv) = xϕ(v), ϕ(T (v)) = S(ϕ(v)). Thus, ϕ ◦ T = S ◦ ϕ so ϕ ◦ T ◦ ϕ−1 = S as functions. Conversely, suppose that T and S are similar with ϕ ◦ T ◦ ϕ−1 = S for some linear transformation ϕ : V → W . Implicit in this statement is that ϕ is a vector space isomorphism. Then the similarity condition shows that for all v ∈ V , we have ϕ(xv) = ϕ(T (v)) = S(ϕ(v)) = xϕ(v). We now, need to show that ϕ is an F [x]-module homomorphism. (We already know that ϕ is a bijection, so we only need to prove that module homomorphism to establish module isomorphism.) Note that since ϕ◦T ◦ϕ−1 = S, we alos have S n = (ϕ ◦ T ◦ ϕ−1 )n = ϕ ◦ T n ϕ−1 , so S n ◦ ϕ = ϕ ◦ T n . For all an xn + · · · + a1 x + a0 ∈ F [x] and all v ∈ V , ϕ((an xn + · · · + a1 x + a0 ) · v) = ϕ(an (xn · v) + · · · + a1 (x · v) + a0 v)

module action on V

n

linear transformation

n

= an ϕ(T (v)) + · · · + a1 ϕ(T (v)) + a0 ϕ(v)

action of x on V

= an S n (ϕ(v)) + · · · + a1 S(ϕ(v)) + a0 ϕ(v)

similarity

= an ϕ(x · v) + · · · + a1 ϕ(x · v) + a0 ϕ(v)

n

action of X on W

n

module action on W .

= an x · (ϕ(v)) + · · · + a1 x · ϕ(v) + a0 ϕ(v) = (an x + · · · + a1 x + a0 ) · ϕ(v)

This establishes the scalar multiple property of module homomorphisms. Hence, ϕ is a module isomorphism. Exercise: 9 Section 10.4 Question: Let G be a group. Prove that an R[G]-module homomorphism is a homomorphism of group representations as defined in Definition 8.6.13. Solution: In Definition 8.6.13, we defined a homomorphism of group representations as follows. If V and W are representations of a group G, a G-representation homomorphism is a linear transformation T : V → W such that T (g · v) = g · T (v) for all v ∈ V . (More precisely, V is equipped with a homomorphism ρ1 : G → GL(V ) and W with the homomorphism ρ2 : G → GL(W ) and T (ρ1 (g)v) = ρ2 (g)T (v).) Note that this definition supposes that R is a field F so we will only consider that case. Example 10.3.14, showed the left F [G]-modules are precisely representations of G. Let V and W be two left R[G]-modules and let ϕ : V → W be a module homomorphism. Because ϕ(v1 + v2 ) = ϕ(v1 ) + ϕ(v2 ) and ϕ(cv) = cϕ(v) for all v1 , v2 , v ∈ V and all c ∈ F , then ϕ is a linear transformation. However, the module homomorphism axioms require that ϕ(g · v) = g · ϕ(v) for all g ∈ G, where we identify g ∈ G with the element 1 · g in F [G]. Hence, a module homomorphism is a homomorphism of group G representations. Conversely, let

10.4. HOMOMORPHISMS AND QUOTIENT MODULES

531

T be a group representation homomorphism. It is a by definition a linear transformation. Let α ∈ F [G] and let v ∈ V . Then α = a1 g1 + a2 g2 + · · · + an gn for ai ∈ F and gi ∈ G and T (α · v) = T (a1 (g1 · v) + a2 (g2 · v) + · · · + an (gn · v))

module action on V

= a1 T (g1 · v) + a2 T (g2 · v) + · · · + an T (gn · v)

linear transformation

= a1 g1 · T (v) + a2 g2 · T (v) + · · · + an gn · T (v)

group representation homomorphism

= (a1 g1 + a2 g2 + · · · + an gn ) · T (v)

module action on W .

This shows that a group representation homomorphism is in fact an F [G]-module homomorphism. Exercise: 10 Section 10.4 Question: Prove that HomZ (Z/mZ, Z/nZ) ∼ = Z/dZ, where d = gcd(m, n). Solution: A Z-module homomorphism ϕ : Z/mZ → Z/nZ is a group homomorphism such that ϕ(k·ā) = k·ϕ(ā) for all k ∈ Z. However, this condition is redundant with the requirement that ϕ is a group homomorphism. Module homomorphisms in this case are determined completely by ϕ(1̄). Suppose that ϕ(1̄) = ā with 1 ≤ a ≤ n but ϕ(m · 1̄) = ϕ(0̄) = 0̄ = m · ϕ(1̄) = m · ā. Hence, n | ma or in other words there exists an integer k such that nk = ma. So ma is a multiple of m and n, so ma = c lcm(m, n) for some integer c and hence, n a = c lcm(m,n) = c gcd(m,n) = c nd . In Z/nZ, the element n/d is the smallest element of order d. Consequently, m multiples of n/d give the subgrou isomorphic to Z/dZ in Z/nZ. The statement that HomZ (Z/mZ, Z/nZ) ∼ = Z/dZ as Z-modules means that there is a Z-module isomorphism ψ : Z/dZ → HomZ (Z/mZ, Z/nZ). We construct this isomorphism as ψ(c̄) = ϕc where ϕc is the homomorphism described above that maps 1̄ to c(n/d), i.e., such that ϕc (x) = c(n/d)x. Note that for b̄, c̄ ∈ Z/dZ we have for all x ∈ Z/mZ, ψ(b̄ + c̄)(x) = ϕ(b+c) (x) = (b + c)(n/d)x = b(n/d)x + c(n/d)x = ϕb (x) + ϕb (x) = ψ(b̄)(x) + ψ(c̄)(x). Thus, ψ(b̄ + c̄) = ψ(b̄) + ψ(c̄). Also for all c̄ ∈ Z/dZ and for all k ∈ Z, and all x ∈ Z/mZ, we have ψ(k · c̄)(x) = ψ(kc)(x) = kc(n/d)x = k · c(n/d)x = k · ψ(c̄)(x). Hence, ψ(k · c̄) = k · ψ(c̄). This shows that ψ is a Z-module homomorphism. We have already shown that ψ is a bijection so ψ is an isomorphism. Exercise: 11 Section 10.4 Question: Prove Proposition 10.4.12. Solution: This was the content of Exercise 10.3.29 so we do not repeat it here. Exercise: 12 Section 10.4 Question: Consider the Z-module M = Z2 and let N be the subspace N = {s(3, 1) + t(2, 5) | s, t ∈ Z}. Prove that M/N ∼ = Z13 . See Figure 10.1. Solution: In Figure fig:exe:Z2QuoMod, the vertices indicate the elements in the module M = Z2 . The vectors (3, 1) and (2, 5) are indicated in the figure. By integral translations by these vectors, the elements in the quotient module M/N are cosets of the form (a, b)+N . Because of translations, the vertices of the parallelogram represent the same cosets. Furthermore, every element in M is congruent to one of the vertices inside the parallelogram or the single vertex of the parallelogram. There are precisely 12 points of Z2 inside the parallelogram and there is one vertex of the parallelogram. Hence, M/M is an abelian group of 13 elements. There exist a unique abelian group of order 13, so M/N ∼ = Z/13Z. Exercise: 13 Section 10.4 Question: Consider the Z-module M = Z2 and let N be the subspace N = {s(4, 1) + t(2, 5) | s, t ∈ Z}. Determine the isomorphism type of M/N . Solution: See the explanation for the previous exercise. Using the same diagram as that in Figure 10.1, we see that the parallelogram spanned by (4, 1) and (2, 5) has 16 points in its interior. As in the previous exercise, the four vertices of this parallelogram represent the same element in M/N . Consequently, M/N is an abelian group (because it is a Z-module) with 17 elements. THere exists only one abelian group of order 17, so M/N ∼ = Z/17Z.

532

CHAPTER 10. MODULES AND ALGEBRA

Figure 10.1: Fundamental region in a quotient of Z2 Exercise: 14 Section 10.4 Question: Let M1 , M2 be left R-modules and let Ni be a submodule of Mi for i = 1, 2. Prove that (M1 ⊕ M2 )/(N1 ⊕ N2 ) ∼ = (M1 /N1 ) ⊕ (M2 /N2 ). Solution: Consider the R-module homomorphism ϕ : M1 ⊕ M2 → (M1 /N1 ) ⊕ (M2 /N2 ) defined by ϕ(m1 , m2 ) = (m1 + N1 , m2 + N2 ). The kernel of this homomorphism is Ker ϕ = {(m1 , m2 ) ∈ M1 ⊕ M2 | m1 ∈ N1 and m2 ∈ N2 } = N1 ⊕ N2 . By the first isomorphism theorem for modules, (M1 ⊕ M2 )/(N1 ⊕ N2 ) ∼ = (M1 /N1 ) ⊕ (M2 /N2 ). Exercise: 15 Section 10.4 Question: Consider the cyclic group Z4 = hz | z 4 = 1i and the group ring R[Z4 ]. Let V = R4 and consider the action of R[Z4 ] on V by scalar multiplication for real numbers and z acting on the standard basis vectors by shifting through, them, i.e., z~ei = ~ei+1 for i = 1, 2, 3 and z~e4 = ~e1 . a) Show that the data equips V with the structure of a R[Z4 ]-module. b) Prove that the subspace W = {(x1 , x2 , x3 , x4 ) | x1 − x2 + x3 − x4 = 0} is a submodule. c) Determine the structure of V /W and state the action of R[Z4 ] on it. Solution: Consider the cyclic group Z4 = hz | z 4 = 1i, the group ring R[Z4 ] and the vector space V = R4 along with the described action of R[Z4 ] on V . a) The described action of the element z on the standard basis vectors corresponds to a linear transformation. Consider the function ρ : Z4 → GL(V ) that satisfies   0 0 0 1 1 0 0 0  ρ(z) =  0 1 0 0 0 0 1 0 (expressed with respect to the standard basis). We see that ρ(z)4 = I so by the Extension Theorem on Generators, ρ defines a homomorphism with ρ(z a ) = ρ(z)a . Thus, V , equipped with ρ is a representation of the group Z4 . A group representation of G over the field R is the same as an R[Z4 ] module. b) Let w ~ ∈ W . Then     w1 w4 w2  w1     z·w ~ = ρ(z)(w) ~ =z· w3  = w2  . w4 w3

10.4. HOMOMORPHISMS AND QUOTIENT MODULES

533

Since w1 − w2 + w3 − w4 = 0 then w4 − w1 + w2 − w3 = 0 so z · w ∈ W . Consequently, z a · w ~ ∈ W for all w ~ ∈ W and all a ∈ N and so any R-linear combination of {1, z, z 2 , z 3 } acts on w ~ by mapping back into W . Since W is also closed under addition, W is a submodule of V . c) We can describe W is an alternate way. Consider the function ϕ : V → R defined by ϕ(~x) = x1 −x2 +x3 −x4 . In the previous part, we calculated that ϕ(z · w) ~ = −ϕ(w). ~ Consequently, we think of V 0 = R as a left R[Z4 ]-module by letting the generator z act on a real number x as z ·x = −x. This defines a homomorphism Z4 → GL1 (R) and hence a representation of Z4 and also make R into an R[Z4 ]-module. We defined V 0 as we did because now, since ϕ is already a linear transformation, since ϕ(z · w) ~ = z · ϕ(w), ~ we deduce that ϕ is an R[Z4 ]-module homomorphism with Ker ϕ = W . Since ϕ is surjective, by the First Isomorphism Theorem, V /W ∼ = R, equipped with the R[Z4 ]-module structure as defined above: i.e., with z · x = −x. Exercise: 16 Section 10.4 Question: Let ϕ : M → M be an R-module homomorphism such that ϕ2 = ϕ. (We call ϕ idempotent.) a) Prove that Im ϕ ∩ Ker ϕ = {0}. b) Prove that M = Im ϕ + Ker ϕ. c) Deduce that M = Im ϕ ⊕ Ker ϕ. [An idempotent endomorphism is called a projection onto the submodule U where U = Im ϕ.] Solution: Let ϕ : M → M be an R-module homomorphism such that ϕ2 = ϕ. (We call ϕ idempotent.) a) Let m ∈ Im ϕ ∩ Ker ϕ. Since m ∈ Im ϕ, there exists n ∈ M such that m = ϕ(n). Since m ∈ Ker ϕ, then ϕ(m) = 0. But then ϕ(ϕ(n)) = 0 and since ϕ2 = ϕ, we deduce that m = ϕ(n) = 0. Thus, Im ϕ ∩ Ker ϕ = {0}. b) Let m ∈ M . Call m1 = ϕ(m) and m2 = m − ϕ(m). Obviously m = m1 + m2 . Furthermore, m1 ∈ Im ϕ. Also, since ϕ(m2 ) = ϕ(m − ϕ(m)) = ϕ(m) − ϕ2 (m) = ϕ(m) − ϕ(m) = m. Thus ϕ(m2 ) = 0 so m2 ∈ Ker ϕ. Therefore, every element in M can be written as an element in Im ϕ and an element in Ker ϕ. Hence, M = Im ϕ + Ker ϕ. c) Define a function ψ : M → Im ϕ ⊕ Ker ϕ by ψ(m) = (ϕ(m), m − ϕ(m)). For m1 , m2 ∈ M , ψ(m1 + m2 ) = (ϕ(m1 + m2 ), m1 + m2 − ϕ(m1 + m2 )) = (ϕ(m1 ) + ϕ(m2 ), m1 + m2 − ϕ(m1 ) − ϕ(m2 )) = (ϕ(m1 ), m1 − ϕ(m2 )) + (ϕ(m2 ), m2 − ϕ(m2 )) = ψ(m1 ) + ψ(m2 ) and for all r ∈ R and all m ∈ M , ψ(rm) = (ϕ(r), rm − ϕ(rm)) = (rϕ(m), r(m − ϕ(m))) = r(ϕ(m), m − ϕ(m)) = rψ(m). Thus, ψ is an R-module homomorphism. However, for all ϕ(n) ∈ Im ϕ and all a ∈ Ker ϕ, the element m = ϕ(n) + a satisfies ψ(m) = (ϕ2 (n), ϕ(n) + a − ϕ2 (n)) = (ϕ(n), a). Hence, ψ is surjective. Furthermore, Ker ψ = {m ∈ M | ϕ(m) = 0 and m − ϕ(m) = 0}, from which we easily see that Ker ψ = {0}. Thus, ψ is injective, so M ∼ = Im ϕ + Ker ϕ. Exercise: 17 Section 10.4 Question: Let F be a field, V be a finite-dimensional vector space over F , and U a vector subspace. Prove that V ∼ = U ⊕ V /U . Solution: We use the principle of Exercise 10.4.16. Let B1 = (~v1 , ~v2 , . . . , ~vs ) be an ordered basis of U and complete B1 to an ordered basis B = (~v1 , ~v2 , . . . , ~vn ) for V . Define the function ϕ : V → U by ϕ(c1~v1 + c2~v2 + · · · + cn~vn ) = cs+1~vs+1 + cs+2~vs+2 + · · · + cn~vn . Since every vector ~v ∈ V can be written uniquely as a linear combination of the basis vectors in B, this is a well-defined function. It is not hard to see that it is also a linear transformation. Furthermore, ϕ is a projection function because ϕ2 = ϕ and Ker ϕ = U . by the result of the previous exercise, as a vector space (i.e., an F -module) V ∼ = Ker ϕ ⊕ Im ϕ = U ⊕ Im ϕ.

534

CHAPTER 10. MODULES AND ALGEBRA

By the First Isomorphism Theorem, we also have V ∼ = U ⊕ V / Ker ϕ = U ⊕ V /U.

10.5 – Free Modules and Module Decomposition Exercise: 1 Section 10.5 Question: Prove Proposition 10.5.2. Solution: Let R be a ring and let M be a left R-module. Let S be a subset of M and consider the subset Span(S). Let x, y ∈ Span(S). Then x = a1 s1 + a2 s2 + · · · + am sm

and

y = b1 s01 + b2 s02 + · · · + bn s0n

where ai , bj ∈ R and si , s0j ∈ S. Then x + y = a1 s1 + a2 s2 + · · · + am sm + b1 s01 + b2 s02 + · · · + bn s0n is again a finite R-linear combination of elements in S so x + y ∈ Span(S). Let r ∈ R and again x ∈ Span(S). Then rx = ra1 s1 + ra2 s2 + · · · + ram sm , which is a finite R-linear combination of elements in S so rx ∈ Span(S). Since Span(S) is closed under addition and left R-multiplication, then Span(S) is a submodule of M . Exercise: 2 Section 10.5 Question: Show that if M = M1 ⊕ M2 , then M1 ∼ = M/M2 and M2 ∼ = M/M1 . Solution: Let π1 : M1 ⊕ M2 → M1 and π2 : M1 ⊕ M2 → M2 be the projections defined as π1 (m1 , m2 ) = m1

and

π2 (m1 , m2 ) = m2 .

Since the addition on the direct sum is a componentwise addition, and since the scalar multiplication is a componentwise multiplication, then πi are both homomorphisms. Furthermore, πi is surjective. Also, Ker π1 = {(m1 , m2 ) | m1 = 0} = {(0, m2 ) | m2 ∈ M2 } ∼ = M2 and likewise Ker π2 ∼ = M1 . By the First Isomorphism Theorem applied to π1 , Im π1 ∼ = M/ Ker π1 ⇐⇒ M1 ∼ = M/M2 , and applied to π2 ,

Im π2 ∼ = M/ Ker π2 ⇐⇒ M2 ∼ = M/M1 .

Exercise: 3 Section 10.5 Question: Let R = Z[x] and consider the ideal I = (3, x + 1) in R. Prove that I is not a free R-module. Solution: Let R = Z[x] and consider the ideal I = (3, x + 1) in R. We first prove that I cannot be generated by a single element. Assume that I = (a(x)) for one polynomial in R. Then 3 = a(x)b(x) for some other polynomial b(x), which implies that deg 3 = deg a(x) + deg b(x) =⇒ deg a(x) + deg b(x) = 0. Since a(x) has degree 0, then a(x) must be a constant polynomial, a(x) = c. Furthermore, c must divide 3 in Z so c = −3, −1, 1, or 3. If a(x) = ±1, then I = R = Z[x]. This is a contradiction because I contains polynomials of the form f (x) = 3p(x) + (x + 1)q(x) but this means that f (−1) = 3p(−1) is a multiple of 3. this does not account for all polynomials in Z[x]. Consequently, a(x) = ±1 is a contradiction. If a(x) = ±3, then I would consist of

10.5. FREE MODULES AND MODULE DECOMPOSITION

535

polynomials whose coefficients are multiples of 3. This is not all of I since x + 1 is not such a polynomial. We have shown that I is not a principal ideal. Now {3, x + 1} might not be the only generating set of I but we have shown that I cannot be generated by a single element. Let {r1 , r2 } be any spanning set of I. Then r2 r1 + (−r1 )r2 = 0 with r2 6= 0 and −r1 6= 0. Hence, no spanning set is linearly independent. Thus I is not a free R-module. Exercise: 4 Section 10.5 Question: Let R = Q[x] and consider the module M = Q where the action of R on M is defined by (p(x), m) 7→ p(1)m. a) Prove that this action makes M into an R-module. b) Prove that M is a torsion module and that Ann(M ) = (x − 1). c) Deduce that M is not free. Solution: a) Obviously, (Q, +) is an abelian group. Now let p(x), q(x) ∈ R = Q[x] and let a, b ∈ M = Q. Then p(x) · (a + b) = p(1)(a + b) = p(1)a + p(1)b = p(x) · a + p(x) · b (p(x) + q(x)) · a = (p(1) + q(1))a = p(1)a + q(1)a = p(x) · a + q(x) · a (p(x)q(x) · a = p(1)q(1)a = p(1)(q(x) · a) = p(x) · (q(x) · a) 1 · a = a. This shows all the axioms of a left R-module. b) The element x − 1 ∈ R is such that (x − 1) · a = 0 for all a ∈ M . In particular, every element in M is a torsion element, so M is a torsion module. Furthermore, Ann(M ) = {p(x) ∈ R | p(1)a = 0 for all a ∈ Q} = {p(x) ∈ R | p(1) = 0}. The set of polynomials that have 1 as a root is Ann(M ) = (x − 1). c) Let S be an set that spans M with R = Q[x] coefficients. Then (x − 1)s1 + (x − 1)s2 + · · · + (x − 1)sk = 0, but x − 1 is not the zero element in R. Hence, there is no spanning set that is linearly independent. Hence, M is not a free left Q[x]-module. Exercise: 5 Section 10.5 Question: Let N be a submodule of the R-module M . Prove that if N and M/N are finitely generated, then so is M . Solution: Suppose that N is a submodule of a left R-module M such that N and M/N are finitely generated. Let {n1 , n2 , . . . , nk } be a generating set of N and let {m1 , m2 , . . . , m` } be a generating set of M/N , where mi means the coset mi + N . Note that mi ∈ M . Let m ∈ M be an arbitrary element. Then there exist r1 , r2 , . . . , r` ∈ R such that m = r1 m1 + r2 m2 + · · · + r` m` = r1 m1 + r2 m2 + · · · + r` m` . Since m − (r1 m1 + r2 m2 + · · · + r` m` ) ∈ N , then there exist elements a1 , a2 , . . . , ak ∈ R such that m − (r1 m1 + r2 m2 + · · · + r` m` ) = a1 n1 + a2 n2 + · · · + ak nk . We deduce that the arbitrary element m is a linear combination of the finite set {m1 , m2 , . . . , m` , n1 , n2 , . . . , nk }. Hence, M is finitely generated. Exercise: 6 Section 10.5 Question: Prove that a commutative ring R is a PID if and only if every submodule of R is a free R-module. Solution: Note that a submodule of a ring R is an ideal of R. Let R be a commutative ring. Let R be a PID. The every submodule has the form Ra for some a. Hence, it is free with a basis of a single element.

536

CHAPTER 10. MODULES AND ALGEBRA

Conversely, suppose that R is such that every submodule (ideal) is free. Let I be an ideal of R and let S be a basis of I = (S). Assume that S contains distinct elements r1 , r2 ∈ S. Then r2 r1 + (−r1 )r2 = 0. Since this is a linear combination of elements from a basis, we deduce that r2 = −r1 = 0, but then r1 = r2 and contradicts the assumption that S contains distinct elements. Hence, every basis of every submodule contains a single element. Thus, I is a principal ideal. Since R is commutative, we deduce that R is a principal ideal domain. Exercise: 7 Section 10.5 Question: Let R be a commutative ring. Prove that for any R-module homomorphism ϕ : Rn → Rm , there exists a unique matrix A ∈ Mm×n (R) such that ϕ(r1 , r2 , . . . , rn ) can be written as (using column vectors)   r1  r2    A . .  ..  rn Solution: Write e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), and so forth to en . Then an element m ∈ Rn can be written as m = (r1 , r2 , . . . , rn ) = r1 e1 + r2 e2 + · · · + rn en in a unique way with r1 , r2 , . . . , rn ∈ R. Suppose that 

 a1i  a2i    ϕ(ei ) =  .  .  ..  ami for all 1 ≤ i ≤ n. Since ϕ is a R-module homomorphism, 

     a11 a12 a1n  a21   a22   a2n        ϕ(m) = r1 ϕ(e1 ) + r2 ϕ(e2 ) + · · · + rn ϕ(en ) = r1  .  + r2  .  + · · · + rn  .   ..   ..   ..  am1 





a11 r1 + a12 r2 + · · · + a1n rn a11  a21 r1 + a22 r2 + · · · + a2n rn   a21    =  =  .. ..    . . am1 r1 + am2 r2 + · · · + amn rn am1

am2

amn

a12 a22 .. .

··· ··· .. .

  a1n r1  r2  a2n    ..   ..  . .  . 

am2

···

amn

rn

This matrix is unique since the expression for each ϕ(ei ) as an element in Rm (expressed by coordinates with respect to the standard basis) is unique. Exercise: 8 Section 10.5 Question: Let R be a commutative ring. Prove that an ideal I in R is a free R-module if and only if I is a principal ideal generated by an element that is not a zero divisor. Solution: [This exercise is redundant with Exercise 10.5.6.] Exercise: 9 Section 10.5 Question: Let R be an integral domain and let M be a finitely generated torsion module (i.e., Tor(M ) = M ). Prove that M has a nonzero annihilator. Give an example of a torsion module whose annihilator is the zero ideal. Solution: Let R be an integral domain and let M be a finitely generated torsion module (i.e., Tor(M ) = M ). Suppose that M is generated by a finite set S = {m1 , m2 , . . . , mn }. Let Ik = Ann(mk ) and consider the ideal I

n \ k=1

Ik .

10.5. FREE MODULES AND MODULE DECOMPOSITION

537

Let ak be nonzero elements such that ak ∈ Ik . Since R is an integral domain, a1 a2 6= 0, then a1 a2 a3 6= 0 and so forth until a = a1 a2 · · · an 6= 0. But a ∈ Ik for all k so a is a nonzero element in I and we deduce that I is a nonzero ideal. Let b ∈ I and let m ∈ M . We can write m = c1 m1 + c2 m2 + · · · + cn mn . Thus bm = c1 bm1 + c2 bm2 + · · · + cn bmn = c1 0 + c2 0 + · · · + cn 0 = 0. We conclude that I ⊆ Ann(M ) and since I is a nonzero ideal, Ann(M ) is nonzero as well. If we suppose that R is an integral domain, because of what we showed, if we hope to find a torsion module whose annihilator is the zero ideal, we must involve a torsion module that is not finitely generated. Let R = Z and let ∞ M M= Z/nZ. n=2

As a direct sum of modules, this consists of sequences m = (a2 , a3 , a4 , . . .) with an ∈ Z/nZ where the ai are 0 except for a finite number of indices i ≥ 2. For any m ∈ M , let k be the highest index such that the k entry is nonzero. Then (k!) · m = 0, the 0-sequence. Hence M is a torsion module. Now let ek be the element of M that is 0 in all entries except for 1 in the kth entry. The annihilator of ek is Ann(ek ) = (k). So Ann(M ) must be a subset of (k) for all k ∈ Z. But this implies that Ann(M ) = (0). Exercise: 10 Section 10.5 Question: Let R be an arbitrary ring and let M be a free module over R. Prove that if M has a finite basis, then every basis of M is finite. Solution: The work for this exercise has already been done in the text. It is a matter of the reader identifying it. The first paragraph of the proof of Theorem 10.5.11 applies for an arbitrary ring R and a module M with a finite basis. The given reasoning does not involve commuting elements or the use of an identity 1 6= 0, which were the conditions of the Theorem. Those conditions on R are used to prove that every basis has the same number of elements. Exercise: 11 Section 10.5 Question: Show that the R-module HomR (Rm , Rn ) is free and of rank mn. Solution: We must assume the R is commutative so that Hom(Rm , Rn ) is an R-module. Let {e1 , e2 , . . . , em } be a basis of Rm and let {f1 , f2 , . . . , fn } be a basis of Rn . Define the function ϕij : Rm → n R that maps (r1 , r2 , . . . , rm ) to i position

(0, . . . , 0,

z}|{ rj , 0, . . . , 0).

It is easy to see that this is an R-module homomorphism. We will show that the set {ϕij | 1 ≤ i ≤ n and 1 ≤ j ≤ m} is a basis for HomR (Rm , Rn ). Let ϕ ∈ HomR (Rm , Rn ). For each j with 1 ≤ j ≤ m set ϕ(ej ) =

n X

aij fi

i=1

for some constants aij ∈ R. Then for s = s1 e1 + s2 e2 + · · · + sm em ∈ Rm we have ϕ(s) = ϕ(s1 e1 + s2 e2 + · · · + sm em ) = s1 ϕ(e1 ) + s2 ϕ(e2 ) + · · · + sm ϕ(em ) ! m n X X = sj aij fi j=1

=

n X i=1

i=1

  m X  sj aij  fi . j=1

Since R is commutative, we have ϕ(s) =

n X m X i=1 j=1

aij sj fi =

n X m X i=1 j=1

aij ϕij (s).

538

CHAPTER 10. MODULES AND ALGEBRA

Hence, the set {ϕij | 1 ≤ i ≤ n and 1 ≤ j ≤ m} spans HomR (Rm , Rn ). Now suppose that for some aij ∈ R, the linear combination n X m X

aij ϕij

i=1 j=1

Pn gave the 0-homomorphism. Then for each j, we have ϕ(ej ) = i=1 aij fi = 0. However, since {f1 , f2 , . . . , fn } is a basis of Rn , then for this given j we have aij = 0 for all i with 1 ≤ i ≤ n. Since j is arbitrary, aij = 0 for all indices. This means that {ϕij } is linearly independent. Thus, {ϕij | 1 ≤ i ≤ n and 1 ≤ j ≤ m} is a basis of the free R-module HomR (Rm , Rn ) and that the rank is mn. Exercise: 12 Section 10.5 Question: Let M, N1 , N2 be left R-modules. Prove the following R-module isomorphisms: a) HomR (N1 ⊕ N2 , M ) ∼ = HomR (N1 , M ) ⊕ HomR (N2 , M ). ∼ b) HomR (M, N1 ⊕ N2 ) = HomR (M, N1 ) ⊕ HomR (M, N2 ). Solution: (The ring R must be commutative for HomR (M, N ) to be an R-module.) a) Define the function f : HomR (N1 , M ) ⊕ HomR (N2 , M ) → Hom(N1 ⊕ N2 , M ) as Ψ(ϕ1 , ϕ2 ) = ϕ, where ϕ(n1 , n2 ) = ϕ1 (n1 ) + ϕ2 (n2 ). In other words, f (ϕ1 , ϕ2 )(n1 , n2 ) = ϕ1 (n1 ) + ϕ2 (n2 ). We prove that this is a homomorphism of R-modules. Let (ϕ1 , ϕ2 ), (ψ1 , ψ2 ) ∈ HomR (N1 , M )⊕HomR (N2 , M ). Then for all (n1 , n2 ) ∈ N1 ⊕ N2 , f ((ϕ1 , ϕ2 ) + (ψ1 , ψ2 ))(n1 , n2 ) = f (ϕ1 + ψ1 , ϕ2 + ψ2 )(n1 , n2 ) = ϕ1 (n1 ) + ψ1 (n1 ) + ϕ2 (n2 ) + ψ2 (n2 ) = ϕ1 (n1 ) + ϕ2 (n2 ) + ψ1 (n1 ) + ψ2 (n2 ) = f (ϕ1 , ϕ2 )(n1 , n2 ) + f (ψ1 , ψ2 )(n1 , n2 ). Thus, f ((ϕ1 , ϕ2 ) + (ψ1 , ψ2 )) = f (ϕ1 , ϕ2 ) + f (ψ1 , ψ2 ) in HomR (N1 ⊕ N2 , M ). Furthermore, for r ∈ R, and for all (n1 , n2 ) ∈ N1 ⊕ N2 , f (r(ϕ1 , ϕ2 ))(n1 , n2 ) = f (rϕ1 , rϕ2 )(n1 , n2 ) = rϕ1 (n1 ) + rϕ(n2 ) = r(ϕ1 (n1 ) + ϕ2 (n2 )) = r(f (ϕ1 , ϕ2 )(n1 , n2 )) Thus, f (r(ϕ1 , ϕ2 )) = rf (ϕ1 , ϕ2 ) in HomR (N1 ⊕ N2 , M ). Consequently, f is an R-module homomorphism. The inverse function is f −1 : HomR (N1 ⊕N2 , M ) ∼ = HomR (N1 , M )⊕HomR (N2 , M ) with f −1 (ϕ) = (ϕ1 , ϕ2 ), where ϕ1 (n1 ) = ϕ(n1 , 0) and ϕ2 (n2 ) = ϕ(0, n2 ). We see that this is the inverse function because first ((f −1 ◦f )(ϕ1 , ϕ2 )) = f −1 ((n1 , n2 ) 7→ ϕ1 (n1 )+ϕ2 (n2 )) = (n1 7→ ϕ1 (n1 )+ϕ2 (0), n2 7→ ϕ1 (0)+ϕ2 (n2 )) = (ϕ1 , ϕ2 ) and secondly (f ◦f −1 )(ϕ) = f (n1 7→ ϕ(n1 , 0), n2 7→ ϕ(0, n2 )) = (n1 , n2 ) 7→ ϕ(n1 , 0)+ϕ(0, n2 ) = (n1 , n2 ) 7→ ϕ(n1 , n2 ) = ϕ. Since f is a bijection R-module homomorphism, then f is an isomorphism, establishing the desired isomorphism. b) Call π1 : N1 ⊕ N2 → N1 and π2 : N1 ⊕ N2 → N2 , the projection homomorphisms. (See Exercise 10.4.3.) Define the function g : HomR (M, N1 ⊕ N2 ) → HomR (M, N1 ) ⊕ HomR (M, N2 ) by g(ϕ) = (π1 ◦ ϕ, π2 ◦ ϕ). If ϕ, ψ ∈ HomR (M, N1 ⊕ N2 ), g(ϕ + ψ) = (m → π1 ((ϕ + ψ)(m)), m → π2 ((ϕ + ψ)(m))) = (m → π1 (ϕ(m) + ψ(m)), m → π2 (ϕ(m) + ψ(m))) = (m → π1 (ϕ(m)) + π1 (ψ(m)), m → π2 (ϕ(m)) + π2 (ψ(m))) = (m → π1 (ϕ(m)) + π1 (ψ(m)), m → π2 (ϕ(m)) + π2 (ψ(m))) = (π1 ◦ ϕ + π1 ◦ ψ, π2 ◦ ϕ + π2 ◦ ψ) = (π1 ◦ ϕ, π2 ◦ ϕ) + (π1 ◦ ψ, π2 ◦ ψ) = g(ϕ) + g(ψ).

10.5. FREE MODULES AND MODULE DECOMPOSITION

539

Furthermore, if r ∈ R, then g(rϕ) = (m → π1 ((rϕ)(m)), m → π2 ((rϕ)(m))) = (m → π1 (rϕ(m)), m → π2 (rϕ(m))) = (m → rπ1 (ϕ(m)), m → rπ2 (ϕ(m))) = (rπ1 ◦ ϕ, rπ2 ◦ ϕ) = r(π1 ◦ ϕ, π2 ◦ ϕ) = rg(ϕ). So, g is a homomorphism. The inverse function is g −1 : Hom(M, N1 ) ⊕ Hom(M, N2 ) → Hom(M, N1 ⊕ N2 ) with g −1 (ϕ1 , ϕ2 ) = (m → (ϕ1 (m), ϕ2 (m))). We check this claim. We have (g◦g −1 )(ϕ1 , ϕ2 ) = g(m → (ϕ1 (m), ϕ2 (m))) = (m → π1 (ϕ1 (m), ϕ2 (m)), m → π1 (ϕ1 (m), ϕ2 (m))) = (ϕ1 , ϕ2 ) and (g −1 ◦ g)(ϕ) = g −1 (π1 ◦ ϕ, π2 ◦ ϕ) = (m → (π1 ◦ ϕ(m), π1 ◦ ϕ(m))) = ϕ. Since g is a bijective R-module homomorphism, we establish the desired isomorphism. Exercise: 13 Section 10.5 Question: Let R be an arbitrary ring and let M and M 0 be free left R-modules. Let B be a basis of M and let B 0 be a basis of M 0 . Prove that if |B| = |B 0 | then M ∼ = M 0 . Do not assume that the bases are finite. Solution: Suppose that M and M 0 are free R-modules with bases B and B 0 such that |B| = |B 0 |. Let f : B → B 0 be a bijection between the bases. Every element in M can be written uniquely a finite linear combination m = r1 m1 + r2 m2 + · · · + rn mn such that ri ∈ R and mi ∈ B. Define the function ϕ : M → M 0 as ϕ(m) = r1 f (m1 ) + r2 f (m2 ) + · · · rn f (mn ). We first prove that this is a left R-module homomorphism. Suppose that union of the set of basis elements involved nontrivially in the expressions of m and of m0 is the set {m1 , m2 , . . . , ms } ⊆ B. Then we can write m and m0 uniquely as m = r1 m1 + r2 m2 + · · · + rs ms

and

m0 = r10 m1 + r20 m2 + · · · + rs0 ms ,

where ri is 0 if mi appears nontrivially in m0 but not in m, and vice versa. Then ϕ(m + m0 ) = ϕ((r1 + r10 )m1 + (r2 + r20 )m2 + · · · + (rn + rn0 )mn ) = (r1 + r10 )f (m1 ) + (r2 + r20 )f (m2 ) + · · · + (rs + rs0 )f (ms ) = r1 m1 + r2 m2 + · · · + rs ms + r10 m1 + r20 m2 + · · · + rs0 ms = ϕ(m) + ϕ(m0 ). Now let a ∈ R and m ∈ M . Write m = r1 m1 +r2 m2 +· · ·+rn mn as a linear combination with {m1 , m2 , . . . , mn } ⊆ B. Then ϕ(am) = ϕ(a(r1 m1 + r2 m2 + · · · + rn mn )) = ϕ(ar1 m1 + ar2 m2 + · · · + arn mn ) = ar1 f (m1 ) + ar2 f (m2 ) + · · · + arn f (mn ) = a(r1 f (m1 ) + r2 f (m2 ) + · · · + rn f (mn )) = aϕ(m). Hence, ϕ is a homomorphism as claimed. Furthermore, Im ϕ = Span(f (m) | m ∈ B) = Span(B 0 ) = M 0 so ϕ is surjective. Also, suppose that m ∈ Ker ϕ so that if m = r1 m1 + r2 m2 + · · · + rn mn with mi ∈ B, then 0 = ϕ(m) = r1 f (m1 ) + r2 f (m2 ) + · · · + rn f (mn ). Since f is a bijection, each f (mi ) is a distinct element in the basis B 0 . Hence, r1 = r2 = · · · = rn = 0. Thus m = 0 an we deduce that Ker ϕ = {0} so that ϕ is injective. This establishes that ϕ is a left R-module isomorphism between M and M 0 . Thus M ∼ = M 0.

540

CHAPTER 10. MODULES AND ALGEBRA

Exercise: 14 Section 10.5 Question: Find all the irreducible Z-modules. Solution: Recall that a Z-module is an abelian group. Let M be module that is generated by at least 2 elements. Then if m 6= 0, the submodule Span(m) is a nonzero strict submodule. Hence, every irreducible Zmodule can be generated by one nonzero element. By the Fundamental Theorem of Finitely Generated Abelian Groups, the only abelian groups generated by 1 element are isomorphic to Z or Z/nZ where n ≥ 2 is an integer. However, if n is not prime, then Z/nZ contains a nontrivial proper subgroup. Hence, the irreducible Z-modules are Z and Z/pZ, where p is prime. Exercise: 15 Section 10.5 Question: Show that an R-module M is irreducible if and only if M 6= {0} and M is cyclic with every nonzero elements as a generator. Solution: Let M be an irreducible R-module. Let m be a nonzero element in M . Then Span(m) = Rm is a nontrivial submodule. So if there exists one nonzero element m such that Span(m) 6= M , then M is not irreducible. Thus, M is cyclic such that every nonzero element is a generator. Conversely, suppose that M 6= {0} and M is such that every nonzero element spans M . Then for all subsets S ⊆ M with S 6= ∅ and S 6= {0}, we have Span(S) = M . Thus M has exactly two submodules, {0} and itself. This means that M is irreducible. Exercise: 16 Section 10.5 Question: (Schur’s Lemma) Show that if M1 and M2 are irreducible R-modules, then every homomorphism between them is the 0 homomorphism or is an isomorphism. Deduce that EndR (M ) is a division ring. Solution: Let M1 and M2 be irreducible R-modules. Let ϕ : M1 → M2 be an R-module homomorphism. Since Ker ϕ is a submodule of M1 and M1 is irreducible, then Ker ϕ = {0} or to M1 . If Ker ϕ = M1 , then ϕ is the trivial homomorphism, i.e., maps every element to 0. Suppose then that Ker ϕ 6= M1 . Furthermore, Im ϕ is a submodule of M2 . Since M2 is irreducible, then Im ϕ = {0} or M2 . Note that Im ϕ = {0} precisely when ϕ is the trivial homomorphism, i.e., when Ker ϕ = M1 . So suppose that ϕ is not the trivial homomorphism. By the First Isomorphism Theorem, M1 ∼ = M1 / Ker ϕ ∼ = Im ϕ = M2 , establishing the isomorphism between M1 and M2 . If R is a commutative ring, by Proposition 10.4.12, End(M ) with + and ◦ is an unital associative R-algebra, so in particular it is a ring. If in addition M is an irreducible R-module, then we have just shown that every nonzero element in End(M ) has an inverse under the product (composition), giving End(M ) the structure of a division ring. Exercise: 17 Section 10.5 Question: Let M be an R-module and suppose that M = M1 ⊕ M2 , where M1 and M2 are nonisomorphic irreducible R-modules. Prove that EndR (M ) ∼ = EndR (M1 ) ⊕ EndR (M2 ) as unital associative R-algebras. Solution: Assuming that R is a commutative ring, by Schur’s Lemma, EndR (M ) = HomR (M, M1 ) ⊕ HomR (M, M2 ) = HomR (M1 , M1 ) ⊕ HomR (M1 , M2 ) ⊕ HomR (M2 , M1 ) ⊕ HomR (M2 , M2 ). By Exercise 10.5.16, HomR (M1 , M2 ) and HomR (M2 , M1 ) are the trivial R module {0}. Thus, EndR (M ) ∼ = EndR (M1 ) ⊕ EndR (M2 ) as R-modules. The product (function composition) operates componentwise so it retains the distributivity properties on pairs of elements in EndR (M1 ) ⊕ EndR (M2 ) and the (multiplicative) identity is (idM1 , idM2 ). Hence, EndR (M1 ) ⊕ EndR (M2 ) as a direct sum of R-modules is also a unital associative R-algebra. Exercise: 18 Section 10.5 Question: An element in a ring R is called a central idempotent element if it is idempotent and in the center of R. a) Prove that if e is a central idempotent element and M is a left R-module, then M = eM ⊕ (1 − e)M . b) Let M be a left R-module. Prove that the central idempotent elements in the ring EndR (M ) are the projection functions.

10.5. FREE MODULES AND MODULE DECOMPOSITION

541

c) Clearly explain how part (a) is tantamount to Proposition 10.5.20. Solution: a) Let e be a central idempotent element of a ring R and let M be a left R-module. Since the center C(R) is a subring and (1 − e)2 = 1 − 2e + e2 = 1 − 2e + e = 1 − e, then 1 − e is also a central idempotent element. Consider the set eM . For all m1 , m2 ∈ M we have em1 − em2 = e(m1 − m2 ) and for all r ∈ R and m ∈ M , we have r(em) = e(rm). This shows that eM is a submodule of M . Since 1 − e is also a central idempotent element, (1 − e)M is also a submodule of M . For all m ∈ M , we obviously have m = (e + 1 − e)m = em + (1 − e)m. Hence, M = eM + (1 − e)M . Now suppose that n ∈ eM ∩ (1 − e)M . Then en = n and (1 − e)n = n as well. In particular, n = e(1 − e)n = (e − e2 )n = 0 so eM ∩ (1 − e)M = {0}. By the Direct Sum Decomposition Theorem, M ∼ = eM ⊕ (1 − e)M . b) Proposition 10.5.20 defines a projection function as a function π : M → M1 , where M1 is a submodule of M , such that π 2 = π. This latter condition is precisely the condition of idempotence. c) Now the projection elements are not central in the associative unital ring EndR (M ) (when R is commutative) but they commute with the multiplication by elements in r. In fact, every element in the algebra EndR (M ) commutes with elements in R. Hence, projections on M are precisely idempotent elements in the algebra EndR (M ). Then part (a) is equivalent to Proposition 10.5.20 with this modification: that in the action of the algebra EndR (M ) we consider elements that are idempotent and commute with R.

Exercise: 19 Section 10.5 Question: Show that any direct sum of free left R-modules is again free. Solution: Let M and N be free left R-modules with bases {m1 , m2 , . . . , ms } and {n1 , n2 , . . . , nt } respectively. We claim that B = {(m1 , 0), . . . , (ms , 0), (0, n1 ), . . . , (0, nt )} is a basis of M ⊕ N . Since {m1 , m2 , . . . , ms } is a basis of M , then every element m ∈ M can be written as a linear combination m = a1 m1 + a2 m2 + · · · + as ms with ai ∈ R. Similarly, since {n1 , n2 , . . . , nt } is a basis of N , then every element n ∈ N can be written as a linear combination n = b1 n1 + b2 n2 + · · · + bt nt with bj ∈ R. Thus (m, n) = (m, 0) + (0, n) = (a1 m1 + a2 m2 + · · · + as ms , 0) + (0, b1 n1 + b2 n2 + · · · + bt nt ) = a1 (m1 , 0) + · · · + as (ms , 0) + b1 (0, n1 ) + · · · + bt (0, nt ). This shows that B spans M ⊕ N . Now suppose that a1 (m1 , 0) + · · · + as (ms , 0) + b1 (0, n1 ) + · · · + bt (0, nt ) = (0, 0). Then (

a1 m1 + a2 m2 + · · · + as ms b1 n1 + b2 n2 + · · · + bt nt

=0 =0

from which we deduce that a1 = a2 = · · · = as = 0 and b1 = b2 = · · · = bt = 0. This shows that B is linearly independent. Thus B is a basis of M ⊕ N which establishes that M ⊕ N is a free left R-module.

Exercise: 20 Section 10.5 Question: Prove that if R is a ring such that R ∼ = R2 , then R ∼ = Rk for any positive integer k. Solution: This is a simple proof by induction on k. The hypothesis that R ∼ = R2 is the basis step. Suppose k ∼ that R = R for some k ≥ 2. Then Rk+1 ∼ = R. = Rk ⊕ R ∼ = R ⊕ R = R2 ∼ By induction Rk ∼ = R for all positive integers k.

542

CHAPTER 10. MODULES AND ALGEBRA

10.6 – Finitely Generated Modules over PIDs, I 10.7 – Finitely Generated Modules over PIDs, II Exercise: 1 Section 10.7 Question: Let L be a free module over a PID R. Prove that the torsion submodule Tor(L) is trivial. Solution: Let B be a basis of L. For all m ∈ L, the element m can be expressed uniquely as a (finite) linear combination m = a1 m1 + a2 m2 + · · · + as ms such that mi ∈ B and ai ∈ R. Then rm = 0 implies that ra1 m1 + ra2 m2 + · · · + ras ms = 0, which in turn implies that rai = 0 for all i. Since R is a PID, r = 0 or ai = 0. If we suppose that r 6= 0, then ai = 0 for all i and thus m = 0. Hence, Tor(L) = {m ∈ L | rm = 0 for some r ∈ R − {0}} = {0}. Exercise: 2 Section 10.7 Question: Consider the free Z-module Z3 . Let f1 = (1, 2, 3), f2 = (4, 5, 6), and f3 = (7, 8, 9) and let K = Span(f1 , f2 , f3 ). Find the invariant factors of K along with a K-preferred basis. Use this to determine Z3 /K. Solution: The generating elements f1 , f2 , f3 are expressed with coordinates in terms of the standard basis, e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1). We use row and column operations to find the Smith normal form of the matrix A with f1 , f2 , f3 as the column vectors.  1 A = 2 3  1 −−−−−−−−−−−→  R3 → R3 − 3R1 0 0  1 −−−−−−−−−−−→  C2 → C2 − 4C1 0 0  1 −−−−−−−→  R2 → −R2 0 0

 7 8 9

 1 −−−−−−−−−−−→  R2 → R2 − 2R1 0 3   1 4 7 −−−−−−−−−−−→ −3 −6 R3 → R3 − 2R2 0 0 −6 −12   1 0 7 −−−−−−−−−−−→ −3 −6 C3 → C3 − 7C1 0 0 0 0   1 0 0 −−−−−−−−−−−→  3 6 C3 → C3 − 2C2 0 0 0 0

4 5 6

 7 −6 9  4 7 −3 −6 0 0  0 0 −3 −6 0 0  0 0 3 0 . 0 0 4 −3 6

We now calculate the corresponding S and T matrices by applying the corresponding matrix transformations as described in Section 10.6:  1 S = 0 0

and

0 −1 0

 1 T = 0 0

 0 1 0 0 1 0 −4 1 0

 0 1 0  0 1 −3

0 1 −2

 0 1 0 0 1 0

0 1 0

0 1 0

 −7 1 0  0 1 0

 0 1 0 −2 1 0 0 1 0

0 1 0

  0 1 0 = 2 1 1

  0 1 −2 = 0 1 0

−4 −1 0

0 −1 −2

 0 0 1

 1 −2 . 1

We finally calculate S

−1

 1  = 2 3

 0 0 −1 0 . −2 1

From this information, we deduce that the invariant factors are (d1 , d3 ) = (1, 3), a K-preferred basis is x1 = (1, 2, 3), x2 = (0, −1, −2), and x3 = (0, 0, 1), a with respect to this basis, K has the basis {x1 , 3x2 }. We deduce that Z3 /K ∼ = (Z/1Z)x1 ⊕ (Z/3Z)x2 ⊕ Z = Z ⊕ (Z/3Z).

Exercise: 3 Section 10.7 Question: Consider the free Z-module Z3 . Let f1 = (2, 5, 10), f2 = (3, −4, 3), and f3 = (−7, 2, −9) and let K = Span(f1 , f2 , f3 ). Find the invariant factors of K along with a K-preferred basis. Use this to determine Z3 /K.

10.7. FINITELY GENERATED MODULES OVER PIDS, II

543

Solution: The generating elements f1 , f2 , f3 are expressed with coordinates in terms of the standard basis, e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1). We use row and column operations to find the Smith normal form of the matrix A with f1 , f2 , f3 as the column vectors.    −1 3 −7 2 3 −7 − − − − − − − − − − → −4 2 2  C1 → C1 − C2  9 A =  5 −4 7 3 −9 10 3 −9     1 −3 7 1 −3 7 − − − − − − − − − − − → −−−−−−−→  2  R2 → R2 − 9R1 0 23 −61 R1 → −R1 9 −4 7 3 −9 7 3 −9     1 0 7 1 −3 7 − − − − − − − − − − − → −−−−−−−−−−−→  R3 → R3 − 7R1 0 23 −61 C2 → C2 + 3C1 0 23 −61 0 24 −58 0 24 −58     1 0 0 1 0 0 − − − − − − − − − − → −−−−−−−−−−−→  C3 → C3 − 7C1 0 23 −61 R2 → R2 − R3 0 −1 −3  0 24 −58 0 24 −58     1 0 0 1 0 0 − − − − − − − − − − − − → −−−−−−−→  3  3  R3 → R3 − 24R2 0 1 R2 → −R2 0 1 0 0 −130 0 24 −58     1 0 0 1 0 0 −−−−−−−−−−−→  − − − − − − − → 0  0  C3 → C3 − 3C2 0 1 C3 → −C3 0 1 0 0 −130 0 0 130 

We calculated the S and T matrices as   1 0 0 1 1 0 0 S = 0 0 −24 1 0   −1 0 0 1  = −2 −1 55 24 −23

0 −1 0

 0 1 0 0 1 0

0 1 0

 0 1 −1  0 1 −7

0 1 0

 0 1 0 −9 1 0

0 1 0

 0 −1 0  0 1 0

0 1 0

 0 0 1

and 

1 T = −1 0  1 = −1 0

The inverse of S is

 0 1 3 0  0 1 1 0 0  3 16 −2 −13 . 0 −1

0 1 0

 0 1 0 0 1 0