# Normalizer of a cyclic group

• Oct 9th 2011, 09:24 AM
dwsmith
Normalizer of a cyclic group
Let G be a finite group and let x be in G.

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

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

Let $|x|=n$

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

Now what?
• Oct 9th 2011, 11:56 AM
zoek
Re: Normalizer of a cyclic group
Quote:

Originally Posted by dwsmith
Let G be a finite group and let x be in G.

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

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

Let $|x|=n$

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

Now what?

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

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

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