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## Homework Statement

G is a group, and H is a subgroup of G.

(1) Show H is a subgroup of its Normalizer. Give an example to show that this is NOT necessarily true if H is NOT a subgroup.

(2) Show H is a subgroup of its Centralizer iff H is Abelian

## Homework Equations

normalizer NG(H) = {g in G|gHg(^-1)=H}

centralizer CG(H) = {g in G|gh = hg for all h in H}

## The Attempt at a Solution

Okay, for part (1) I showed the left coset of H equals to the right coset of H, that is g*H = H*g = H <=> h in H

Can anyone show me an example that it is NOT necessarily true if H is NOT a subgroup? (If H is NOT a subgroup of G, isn't that obvious that it's not necessary a subgroup of its Normalizer?)

For part (2) by definition of the centralizer CG(H)={g in G|gh = hg for all h in H}.

(=>) Suppose H < CG(H), WTS H is Abelian, that is for any elments h1, h2 in H, h1*h2=h2*h1.

so by definition

g*h1 = h1*g => g = h1*g*h1^(-1)

g*h2 = h2*g => g = h2*g*h2^(-1)

So h1*g*h1^(-1) = h2*g*h2^(-1)

Also h1*h2^(-1) is in H by definition of being subgroup of it's centralizer, so g*h1*h2^(-1)=h1*h2^(-1)*g

Okay, I've been trying to drive to h1*h2=h2*h1 but somehow stuck. Can anyone help? I know it's not a complicated problem...

(<=) Suppose H is Abelian, then for any elments h1, h2 in H, h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1. WTS H < CG(H)

WTS two things:

1) H is nonempty

2) for any h1, h2 in H => h1*h2^(-1) is in H

Obviously H is nonempty since the identity is there.

Let h1, h2 in H

Then since h1*h2=h2*h1 => h1*h2^(-1) = h2^(-1)*h1

How do I show from here that it implies that h1*h2^(-1) is in H?

Thanks!