1. ## Subsets and subgroups

a. Let G be a group, let H be a subgroup of G and let g be a fixed element of G. Show that the subset gHg-1 = {ghg-1: h ϵ H} is also a subgroup of G.

b. Suppose g, x ϵ G. Show that (gxg-1 )n =gxng-1 for all n ϵ Z, and hence (or otherwise) that o(gxg-1) = o(x)

Any help would be really appreciated thanks

2. ## Re: Subsets and subgroups

Originally Posted by NathanBUK
a. Let G be a group, let H be a subgroup of G and let g be a fixed element of G. Show that the subset gHg-1 = {ghg-1: h ϵ H} is also a subgroup of G.
Here is a useful theorem on subgroups.
If $G$ is a group and $H\subseteq H$ then $H$ is a subgroup if annd only if $\{x,y\}\subseteq H$ impies $xy^{-1}\in H$.

Take $x=gh_1g^{-1}~\&~h=gh_2g^{-1}$.

3. ## Re: Subsets and subgroups

for part b): if n > 0

$(gxg^{-1})^n = (gxg^{-1})(gxg^{-1})\cdots(gxg^{-1})$ (n times)

$= gx(g^{-1}g)x(g^{-1}g)\cdots xg^{-1} = gxexe\cdots xg^{-1} = gx^ng^{-1}$

if n = 0:

$(gxg^{-1})^0 = e = gg^{-1} = geg^{-1} = gx^0g^{-1}$.

if n < 0, so n = -k, for k > 0:

$(gxg^{-1})^n = (gxg^{-1})^{-k} = ((gxg^{-1})^{-1})^k = ((g^{-1})^{-1}x^{-1}g^{-1})^k$

$= (gx^{-1}g^{-1})^k = g(x^{-1})^kg^{-1} = gx^{-k}g^{-1} = gx^ng^{-1}$

now if |x| = n, then xn = e, so (gxg-1)n = gxng-1 = geg-1 = gg-1 = e.

this means that |gxg-1| is at most n.

suppose that (gxg-1)k = e, for some 0 < k < n.

then gxk = eg = g, whence xk = g-1g = e, contradicting the fact that the order of x is n.

thus n is the smallest positive integer for which (gxg-1)n = e, so is the order of gxg-1.