Chapter 4 CONJUGACY

Question 1

Conjugate of S

Answer 1

S^g={g^-1 x g | x ∈ S}

Question 2

Conjugacy Class of S

Answer 2

x^g={g^-1 x g | g ∈ G}

Question 3

|S| = …

Answer 3

|S| = |S^g|

Question 4

If S ≤ G, then …

Answer 4

S^g ≤ G

Question 5

if x and y are conjugate in G then x and y…

Answer 5

have the same order

Question 6

1^G = …

Answer 6

{1}

Question 7

Suppose G a group, then G is …

Answer 7

G is the disjoint union of its conjugacy classes.

Proof: x~y <=> x^g=y

prove that ~ is an equivalence relation. REFLEXIVE, SYMMETRIC, TRANSITIVE

Question 8

let G group, N = NG(S)
let {gi | i ∈ I} be a complete set of right coset representatives for N in G.
conjugates of S are {S^gi | i ∈ I} and S^gi = S^gj <=> …

Answer 8

gi = gj.

If further, G is finite, number of conjugates of S is [G:N] = |G|/|N|

start of proof: S^g is a typical conjugate. g=ngi for some gj.

then S^g = S^ngj = (S^n)gj = S^gj

Question 9

NG({x}) = …

Answer 9

CG({x}).

[x^g= x becomes g-1 x g =x which is the centralizer]

Question 10

let gi be a complete set of right coset representatives for C in G. then x^gi =x^gj <=> …

Answer 10

gi = gj

Furthermore if G is finite, then |x^G| = [G: CG(x)] and |x^G| | |G|

Question 11

S≤ NG(S)≤ G and so by lagrange’s theorem what can we say NG(S) is equal to

Answer 11

either S=NG(S) or G=NG(S)

Question 12

Class equation for finite groups

Answer 12

let x1…xk be representatives from each of the k conjugacy classes of G. set ni = |xi ^ G|

assume n1=n2=….=nl =1 , nj>1 for j>l

|G| = (1 to n)Σni = (1 to n)Σ[G:CG(xi)]
|G| = |Z(G)| + (l+1 to k)Σni
|G| = |Z(G)| + (l+1 to k)Σ[G:CG(xi)]

Question 13

if p is prime, what is a p group?

Answer 13

|G| = p^r with r∈NU{0}

example: |G| = 125= 5^3. G is a 5-group

Question 14

If G is a p group, what can be said about Z(G)

Answer 14

Z(G) =/= 1.

proof using class equation of finite groups

Question 15

if f:a->b then g-1 f g = …

(a,b belong to omega = {1,….,n}

Answer 15

ag->bg

Question 16

what is cycle type of a permutation (sigma)

Answer 16

a permutation has cycle type 1^c1 2^c2 …. n^cn where the permutation is written as a product of pairwise disjoint cycles. it has c1 cycles of length 1, c2 cycles of length 2 etc…

Question 17

f,f’ are conjugate in Sn <=>

Answer 17

f and f’ have the same cycle type