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Thread: Question with proof involving conjugacy classes

  1. #1
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    Question with proof involving conjugacy classes

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

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

    I'm thinking that since $\displaystyle |x^G|$ divides $\displaystyle |G|$, both $\displaystyle |x^G|$ and $\displaystyle |C_G(x)|$ are odd since $\displaystyle |x^G|*|C_G(x)|=|G|=2k+1$. If I assume that there is $\displaystyle x \in G$ conjugate to $\displaystyle x^{-1}$ other than 1, does it then follow that either $\displaystyle |x^G|$ or $\displaystyle |C_G(x)|$ is even? How can I find $\displaystyle x^G$ or $\displaystyle |C_G(x)|$ 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 $\displaystyle gxg^{-1}=x^{-1}.$ then for any natural number $\displaystyle n$ we have $\displaystyle g^nxg^{-n}=\begin{cases} x & \text{if} \ 2 \mid n \\ x^{-1} & \text{otherwise} \end{cases}.$ let $\displaystyle k=o(g).$ then $\displaystyle k$ is odd because |G| is odd. hence $\displaystyle x=g^kxg^{-k}=x^{-1},$ which gives us $\displaystyle x^2=1.$

    thus $\displaystyle x=1$ because |G| is odd.
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