Really struggling on where to start with this question:
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 }=gx^{n}g^{-1 }for all n ϵ Z, and hence (or otherwise) that o(gxg-1) = o(x)
Any help would be really appreciated thanks
for part b): if n > 0
(n times)
if n = 0:
.
if n < 0, so n = -k, for k > 0:
now if |x| = n, then x^{n} = e, so (gxg^{-1})^{n} = gx^{n}g^{-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 gx^{k} = eg = g, whence x^{k} = g^{-1}g = 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}.