Normalizer of a cyclic group

**dwsmith** · Oct 9th 2011, 09:24 AM

Let G be a finite group and let x be in G.

Prove that if $g\in N_G(<x>)$ , then $gxg^{-1}=x^a$ for some $a\in\mathbb{Z}$ .

$N_G(<x>)=\{g\in G:g<x>g^{-1}=x\}$

Let $|x|=n$

$<x>=\{e,x,x^2,\cdots x^{n-1}\}$

Now what?

**zoek** · Oct 9th 2011, 11:56 AM

Originally Posted by dwsmith

Let G be a finite group and let x be in G.

Prove that if $g\in N_G(<x>)$ , then $gxg^{-1}=x^a$ for some $a\in\mathbb{Z}$ .

$N_G(<x>)=\{g\in G:g<x>g^{-1}=x\}$

Let $|x|=n$

$<x>=\{e,x,x^2,\cdots x^{n-1}\}$

Now what?

Note that $N_G(<x>)=\{g\in G:g<x>g^{-1}=<x>\}$ .

Now, $x\in<x> \implies gxg^{-1}\in g<x>g^{-1} \overset{g\in N_G(<x>)}{==}<x>$ .

So, $gxg^{-1}=x^a$ for some $a\in\mathbb{Z}$ .

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