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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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Conjugacy  

Slides from a lecture on Conjugacy by Shirleen Stibbe, given at an Open University pure mathematics summerschool.

Conjugacy  

Slides from a lecture on Conjugacy by Shirleen Stibbe, given at an Open University pure mathematics summerschool.

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