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Math Help - Question with proof involving conjugacy classes

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    Question with proof involving conjugacy classes

    Prove that if a group G has an odd order than no x \in G other than x=1 is conjugate to its inverse.

    I'm thinking that since |x^G| divides |G|, both |x^G| and |C_G(x)| are odd since |x^G|*|C_G(x)|=|G|=2k+1. If I assume that there is x \in G conjugate to x^{-1} other than 1, does it then follow that either |x^G| or |C_G(x)| is even? How can I find x^G or |C_G(x)| so I can get a contradiction?

    I think i'm going in the right direction but i'm not sure. Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by didact273 View Post
    Prove that if a group G has an odd order than no x \in G other than x=1 is conjugate to its inverse.

    I'm thinking that since |x^G| divides |G|, both |x^G| and |C_G(x)| are odd since |x^G|*|C_G(x)|=|G|=2k+1. If I assume that there is x \in G conjugate to x^{-1} other than 1, does it then follow that either |x^G| or |C_G(x)| is even? How can I find x^G or |C_G(x)| so I can get a contradiction?

    I think i'm going in the right direction but i'm not sure. Any help would be appreciated.
    suppose gxg^{-1}=x^{-1}. then for any natural number n we have g^nxg^{-n}=\begin{cases} x & \text{if} \ 2 \mid n \\ x^{-1} & \text{otherwise} \end{cases}. let k=o(g). then k is odd because |G| is odd. hence x=g^kxg^{-k}=x^{-1}, which gives us x^2=1.

    thus x=1 because |G| is odd.
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