Thread: Normalizer of a cyclic group

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

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

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

Let $\displaystyle |x|=n$

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

Now what?

Originally Posted by dwsmith

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

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

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

Let $\displaystyle |x|=n$

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

Now what?

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

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

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