A group of odd order has no non-identity elements which are conjugate to their inverse. I want some verification for my proof to a homework problem. (Is it correct? Is there a simpler way to do this?)
Let $G$ be a finite group of odd order and suppose there is an element $g$ that is conjugate to its own inverse. In other words, there is $h \neq e$ such that $h^{-1}gh = g^{-1}$. We will show $g=e$ by supposing it's not and finding a contradiction.
We see that $gh = hg^{-1}$ and $hg = g^{-1}h$. This means given any word composed of $g$, $h$, $g^{-1}$, and $h^{-1}$, we can always "push" the $g^r$'s to one side and the $h^s$'s to the other, giving us a canonical spelling $g^rh^s$. That is to say, $\langle g, h\rangle \cong \langle g \rangle \oplus \langle h \rangle$.  
By Lagrange, finding any non-trivial even subgroup will prove that $|G|$ was even after all, giving us the desired contradiction. Since both $\langle g \rangle$ and $\langle h \rangle$ are nontrivial, neither may be even. But if that is the case, then they are both odd, and $\langle g \rangle \oplus \langle h \rangle \cong \langle g, h \rangle$ is now non-trivial even.
EDIT Critical error at the end. The order of the direct sum is the product, not the sum.
 A: HINT: Note that $gh=hg^{-1}$. What is $(gh)^2$?
A: Assume there exists $x \neq 1$ in $G$ with $g^{-1}xg=x^{-1}$ for some $g \in G$. Then $g^{-1}x^{m}g = x^{-m}$ for any $m$, so we can let $G$ act on $\langle x \rangle$ by conjugation.
Consider the conjugacy class $x^G$, we know if $y \in x^G$ then also $y^{-1} \in x^G$. In addition, $y \neq y^{-1}$, otherwise $y^2=1$, contradicting $|G|$ is odd.
Thus any element in $x^G$ can be paired with its inverse, hence $x^G$ must contain even elements. Therefore, $|G|$ is also even since it is a multiple of $|x^G|$, a contradiction.
Remark: $\,$ Actually we can let $G$ act on itself by conjugation, and deduce that $x^G$ has even elements, because if $x$ is conjugate to $x^{-1}$ then any other element $g^{-1}xg$ conjugating to $x$ must also be conjugate to $x^{-1}$, and henceforth to its inverse $g^{-1}x^{-1}g$. But I did not realize that at first, and I think in order to force all elements in $x^G$ to conjugate to its inverse we must put some constraint on $x^G$ so that any other element will be in some relation with $x$.
