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Math Help - Algebra, Problems For Fun (27)

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    Algebra, Problems For Fun (27)

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

    Then

    g^{-1}=aga^{-1}

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

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

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

    Then

    g^{-1}=aga^{-1}

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

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

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