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Thread: Algebra, Problems For Fun (27)

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    Algebra, Problems For Fun (27)

    Let $\displaystyle G$ be a finite group of odd order and $\displaystyle 1 \neq g \in G.$ Prove that $\displaystyle g$ is not conjugate to $\displaystyle g^{-1}.$
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    MHF Contributor Bruno J.'s Avatar
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    Suppose $\displaystyle g=ag^{-1}a^{-1}$ .......... (1)

    Then

    $\displaystyle g^{-1}=aga^{-1}$

    Combining with (1) : $\displaystyle g=ag^{-1}a^{-1} = a(aga^{-1})a^{-1}=a^2ga^{-2}=a^2(a^2ga^{-2})a^{-2}=a^4ga^{-4}=...$

    So $\displaystyle g=a^2ga^{-2}=a^4ga^{-4}=a^6ga^{-6}...$

    But since $\displaystyle |G|$ is odd, $\displaystyle <a>=<a^2>$. In particular we find $\displaystyle k$ such that $\displaystyle a^{2k}=a$. This yields $\displaystyle g=a^{2k}ga^{-2k}=aga^{-1}$, so $\displaystyle g,a$ commute and $\displaystyle g^{-1}=g\Rightarrow g^2=1 \Rightarrow g=1$.
    Last edited by Bruno J.; Jul 2nd 2009 at 02:38 PM.
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    Quote Originally Posted by Bruno J. View Post
    Suppose $\displaystyle g=ag^{-1}a^{-1}$ .......... (1)

    Then

    $\displaystyle g^{-1}=aga^{-1}$

    Combining with (1) : $\displaystyle g=ag^{-1}a^{-1} = a(aga^{-1})a^{-1}=a^2ga^{-2}=a^2(a^2ga^{-2})a^{-2}=a^4ga^{-4}=...$

    So $\displaystyle g=a^2ga^{-2}=a^4ga^{-4}=a^6ga^{-6}...$

    But since $\displaystyle |G|$ is odd, $\displaystyle <a>=<a^2>$. In particular we find $\displaystyle k$ such that $\displaystyle a^{2k}=a$. This yields $\displaystyle g=a^{2k}ga^{-2k}=aga^{-1}$, so $\displaystyle g,a$ commute and $\displaystyle g^{-1}=g\Rightarrow g^2=1 \Rightarrow g=1$.
    you can also use induction to show that $\displaystyle g=a^ng^{-1}a^{-n}$ for all odd numbers $\displaystyle n$ and then choose $\displaystyle n=o(a)$ to get $\displaystyle g^2=1$ and thus $\displaystyle g=1.$
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