Show that elements are not conjugate to their inverse in groups of odd order.

Question

Show that if G is a group of odd order, then no $x\in G$ other than the identity is conjugate to its inverse.

We can't have elements of order 2, since by Lagrange theorem this would mean we would have the subgroup generated by that element would have to divide the order of the group, but G is odd so this can't happen. Hence we can't have $x = x^{-1}$.

Now suppose there exists element $x\in G$ such that $x^{-1}=gxg^{-1}$ for some $g\in G$. Suppose that $y\in\operatorname{Orb}_G(x)$, so we have $y=qxq^{-1}$. Then $y^{-1}=qx^{-1}q^{-1}=qgxg^{-1}q^{-1}$, which implies that $y^{-1}\in\operatorname{Orb}_G(x)$.

So for each $y\in\operatorname{Orb}_G(x)$ we will also have $y^{-1}\in \operatorname{Orb}_G(x)$, so $\operatorname{Orb}_G(x)$ must be even, but by orbit-stabilizer theorem we have $|\operatorname{Orb}_G(x)|=[G:C_G(x)]$, where $C_G(i)$ is the centralizer of group $G$ for element $x$, but by Lagrange this must divide order of group $G$, which is odd, so this can't happen. Therefore our original assumption is wrong and we get our result.

Answer 1

Your line of reasoning is correct, but the way the argument is written out is not entirely correct:

1. You first conclude that we cannot have $x=x^{-1}$, but we have $e=e^{-1}$. Maybe you mean for any $x\in G$ other than $x=e$?
2. From $y,y^{-1}\in\operatorname{Orb}_G(x)$ you conclude that the order of $\operatorname{Orb}_G(x)$ must be even. But what if $y=y^{-1}$ for some $y\in\operatorname{Orb}_G(x)$? Again you must treat the case $x=e$ separately.

Apart from this your proof is sound. As a side note, as user Jack Yoon comments, a more common notation for $\operatorname{Orb}_G(x)$ would be $\operatorname{Cl}_G(x)$, which is the conjugacy class of $x$ in $G$.