Group Theory: Conjugacy
Shirleen Stibbe
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M203 Pure Mathematics Summerschool
This poster hangs on my neighbour's wall. How do you know he's a Group Theorist?
I won't tell you his reaction when his wife bought him a tartan jacket on a Tuesday – too painful!
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Group Theorists are compulsive partitioners
Orders of elements
Orbits
Cosets
Direct vs Indirect
Odd vs Even
Conjugacy Classes 3
Conjugacy is a way of partitioning the group by type of element
Symmetry groups rotations through the same angle reflections in axes through vertices reflections in axes through sides Permutation groups same cycle types Compartments are called Conjugacy Classes 4
Definition: Let x, y ∈ G. Then y is a conjugate of x if there is a g ∈ G such that y = gxg-1. Properties: Conjugacy is an equivalence relation: reflexive, symmetric, transitive so conjugacy classes partition the set The conjugacy class of x is the set {gxg-1 : g ∈ G}
The identity is always in a class on its own
Abelian ⇔ every class contains a single element
(gxg-1)n = gxng-1
so all the elements in a particular class have the same order [NB: converse not true]
Conjugacy is a Group Action G is the group doing the action X is the set of elements of the group g ∧ x = gxg-1 conjugacy classes are orbits of the action 5
Conjugacy in Permutation Groups Classes = cycle types Calculations - the easy way: Let x = (xa…xn)…(xr…xz) Then gxg-1 = (g(xa)…g(xn))…(g(xr)…g(xz) Examples from S8 1 x = (1234)(758) g = (14)(235) gxg-1 = (g(1)g(2)g(3)g(4))
(g(7)g(5)g(8))
= (4351)(78) 2 x = (253)(16) y = (634)(25) y = gxg-1 Which g? Want: g(2)=6, g(5)=3, g(3)=4, g(1)=2, g(6)=5 So g = (126534) [or g = (126534)(78)]
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Let H be a subgroup of G Then, for some g ∈ G, H' = gHg-1 = {ghg-1 : h ∈ H} is a subgroup of G Two subgroups H and H' are conjugate subgroups in G if there is an element g ∈ G, such that H' = gHg-1 | gHg-1| = |H| - i.e. same order H is self conjugate if gHg-1 = H for all g ∈ G
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Cosets vs Conjugate Subgroups (Amusing properties in red) H = { e, (234), (243), (23), (24), (34) } ⊆ S4 Cosets of H: eH = H
fixes 1 (12)H = { (12), (1234), (1243), (123), (124), (12)(34) }
sends 1 to 2 (13)H = { (13), (1324), (1342), (132), (134), (13)(24) }
sends 1 to 3
(14)H = { (14), (1423), (1432), (142), (143), (14)(23) }
sends 1 to 4 Conjugate subgroups of H: eHe-1
= {e, (234), (243), (23), (24), (34)}
fixes 1
(12)H(12)-1 = {e, (134), (143), (13), (14), (34)}
fixes 2
(13)H(13)-1 = {e, (124), (142), (12), (14), (24)}
fixes 3
(14)H(14)-1 = {e, (123), (132), (12), (13), (23)}
fixes 4
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Conjugacy and Normality Normal sugroups N is normal iff gN = Ng for all g ∈ G [left cosets = right cosets] gn1= n2g, n1, n2 ∈ N, g ∈ G ⇒ gn1g-1 = n2 ∈ N This says: gng-1 ∈ N for all n ∈ N, g ∈ G 1 A normal subgroup contains every conjugate of each of its elements 2 A normal subgroup is a union of conjugacy classes 3 Normal subgroups are self-conjugate gNg-1 = N for all g ∈ G 9
The S4 Game
Find all the normal subgroups of S4 Cycle type:
Number
(*) (*) (*) (*)
1
(*) (*) (* *)
6
(*) (* * *)
8
(* *) (* *)
3
(* * * *)
6 24
Rules of the game Each normal subgroup must be a union of conjugacy classes order (# of elements) divides 24 (Lagrange) must be a group must contain e
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Part 1 Exam question G is the group of symmetries of the hexagon, h 2
1
3
6 4
5
h is the reflection in the axis shown, g = (153)(264).
(a) Write down h in cyclic form, and give a geometric description of g. (b) Find the conjugate ghg-1 and identify it geometrically. (c) State whether the elements (14)(25)(36) and (16)(25)34) are conjugate in G and justify you answer briefly.
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Same question – the polite version
h
G is the group of symmetries of the hexagon.
h is reflection in the axis shown
2
↔
1
3
6
and
g = (153)(264).
(a)
If you would be so kind, please would you
4
5
h in cycle form, and perhaps you wouldn t mind giving a geometric description of g? write
(b)
I'd be terribly grateful if you would find the conjugate ghg–1 and identify it geometrically.
(c)
I m dreadfully sorry to trouble you again, but it would be helpful if you could state whether (14)(25)(36) and (16)(25)(34) are conjugate in G, and justify your answer. Thanks awfully.
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Solution (a) h = (12)(36)(45)
Dear examiner, Only too happy to help. x
g is clockwise rotation about the centre of the hexagon through angle 2π/3.
NB: don't forget direction - anticlockwise is (135)(246). (b)
ghg-1 = (g[1]g[2])(g[3]g[6])(g[4]g[5]) = (56)(14)(23) reflection in line bisecting edges 2-3 and 5-6
Quick method HB p48 S4 §2 (c) (14)(25)(36) is a rotation (through π) and (16)(25)(34) is a reflection. The symmetries are different geometric types, and therefore cannot be conjugate in G.
Note: they are conjugate in S6 (same cycle structure) eg k(14)(25)(36)k-1 = (16)(25)(34) , where k = (46), but k is not in G.
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