# Math Help - Question with proof involving conjugacy classes

1. ## 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.

2. Originally Posted by didact273
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.