Results 1 to 4 of 4

- Oct 8th 2011, 10:43 AM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 8

## |x|=|gxg^{-1}|

Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by . For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and have the same order for all x in G and that for any subset A of G, .

I have shown this an isomorphism. I am not sure how to show x and have the same order.

- Oct 8th 2011, 11:05 AM #2

- Oct 8th 2011, 11:09 AM #3

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,546
- Thanks
- 842

## Re: |x|=|gxg^{-1}|

there are different ways to do this:

1). suppose that x has order n. then x^n = e, so (gxg^-1)^n = (gxg^-1)(gxg^-1).....(gxg^-1) (n times)

= gx(g^-1g)x(g^-1g).....xg^-1 = (gx)(x)(x)....(x)g^-1 = g(x^n)g^-1 = geg^-1 = gg^-1 = e.

(ok, technically using induction on n would be better, but you get the idea). thus |gxg^-1| divides n.

now suppose that |gxg^-1| = k < n. then (gxg^-1)^k = g(x^k)g^-1 = e, so x^k = g^-1g = e, a contradiction.

hence |gxg^-1| = |x|.

2). since x-->gxg^-1 is an isomorphism, the order of gxg^-1 must be the order of x. why? suppose not.

case 2a) |x| < |gxg^-1|. in this case we can find a k with x^k = e, but (gxg^-1)^k ≠ e.

using our isomorphism e = geg^-1 = g(x^k)g^-1 = (gxg^-1)^k, contradicting our choice of k.

case 2b) |gxg^-1| < |x|. then we have (gxg^-1)^k = e, but x^k ≠ e, for some k.

since e = (gxg^-1)^k = g(x^k)g^-1, we have that x^k is in the kernel of our isomorphism.

but since an isomorphism is injective, the kernel is {e}, contradiction.

thus |gxg^-1| = |x|.

now, since x-->gxg^-1 is an isomorphism, it is bijective, so |A| and |gAg^-1| have to be equal.

- Oct 8th 2011, 11:12 AM #4

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 8