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**didact273** 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.