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Thread: Normalizer of a cyclic group

  1. #1
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    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?
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  2. #2
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    Re: Normalizer of a cyclic group

    Quote Originally Posted by dwsmith View Post
    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 $\displaystyle a\in\mathbb{Z}$.
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