GE1 Counting with groups

GE1 Counting with groups

Shirleen Stibbe s.stibbe@open.ac.uk

1

Consider a glass brooch made up of 4 triangles, which we've numbered 1 to 4:

2

4 3

1 Group action on the brooch:

Let X = {1, 2, 3, 4} be the set of triangles in the brooch.

The symmetry group, G, of the brooch is D4 = {e, r, r2, r3, s, rs, r2s, r3s} We let G act on X in the natural way.

Reminders: 

Orbit Stabiliser Theorem: |Orb(x)| × |Stab(x)| = |G|



For g ∈ G and x ∈ X, (g, x) is an inert pair if and only if g ∧ x = x, i.e. g ∈ Stab(x) and x ∈ Fix(g)

1

2

3

4

1

2

3

4

r

2

3

4

1

r

2

3

4

1

2

r

3

4

1

2

3

s

3

2

1

4

rs

2

1

4

3

rs

2

1

4

3

2

3

4

3

2

1

G X e

rs

In the Group Action table, the cell (g, x) is shaded if (g, x) is an inert pair.

In the diagram, the inert pairs are linked by lines.

1

e

r

2

2

r

3

r

3

s

4

rs

2

rs

3

rs

The number of inert pairs = the number of shaded cells = the number of lines in the diagram above.

Counting inert pairs:

For each g ∈ G,  in the table: the number of shaded cells in the row labelled g is |Fix(g)|  in the diagram: the number of lines attached to g is |Fix(g)| So the total number of inert pairs =

∑ Fix(g) g∈G

For each x ∈ X,  in the table: the number of shaded cells in the column labelled x is |Stab(x)|  in the diagram: the number of lines attached to x is |Stab(x)| So the total number of inert pairs =

∑ Stab(x) x∈X

This illustrates the Counting Lemma:

http://www.shirleenstibbe.co.uk/m336

1 G

∑ Stab( x)

x∈ X

=

1 G

∑ Fix(g ) g∈G

1

2 Group Action on the set of colourings of the brooch 1 2

We suppose now that each triangle of our glass brooch may be coloured either red (r) or yellow (y), and we count the number of essentially different colourings.

4 3

Symmetry group D4 :

{e,

r,

r 2,

Equivalent cycles G :

{e,

(1234),

(13)(24),

r 3,

s,

r2s,

rs,

r3s}

(1432), (2)(4)(13), (14)(23), (1)(3)(24), (12)(34)}

Notes:   

since there are two choices for each of the 4 triangles, the total number of colourings is 24 = 16 two colourings are equivalent if they are in the same orbit under the action of the symmetry group (since one can be transformed into the other under the group action) so the number of essentially different colourings is equal to the number of orbits of the group action

1 G

1 The Counting Lemma: Number of orbits =

|Fix(e)| |Fix(r)| = |Fix(r3)| |Fix(r2)| |Fix(s)| = |Fix( r2s)| |Fix(rs)| = |Fix( r3s)| By the counting lemma:

= 24 = 21 = 22 = 23 = 22

∑ Fix( g ) , |D | = 8 4

g∈G

[e fixes all colourings] [all triangles must be the same colour] [opposite pairs must be the same colour] [1 opposite pair must be the same, 2 triangles coloured independently] [adjacent pairs must be the same colour]

number of orbits = 1/8[24 + 2 × 21 + 22 + 2 × 23 + 2 × 22] = 6 2 The Cycle Index Theorem: g

cycle type

cycle symbol

number

e r, r3 r2, rs, r3s s, r2s

(1)(2)(3)(4) (1234) (12)(34) (1)(2)(34)

x 14 x 41 x 22 x 12x 21

1 2 3 2

Cycle index:

1

/8[x14 + 2x41 + 3x22 + 2x12x21]

Number of orbits with m colours:

1

/8[m4 + 2m + 3m2 + 2m3]

Number of orbits with 2 colours:

1

/8[24 + 2 × 2 + 3 × 22 + 2 × 23] = 6

replace xa in the cycle index by m

3 Polya's Theorem:

Pattern inventory:

1

/8[(r + y)4 + 2(r4 + y4) + 3(r2 + y2)2 + 2(r + y)2(r2 + y2)]

=

1

/8[8r4 + 8r3y + 16r2y2 + 8ry3 + 8y4]

=

r4 + r3y + 2r2y2 + ry3 + y4

replace xab in the cycle index by (ra + y a)b

Note: The coefficient of the term rayb is the number of orbits with a red triangles and b yellow triangles.

This shows that there is:

one orbit with 4 red triangles one orbit with 3 red triangles two orbits with 2 red triangles one orbit with 1 red triangle one orbit with no red triangles

2

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