Prove that if $G$ is a finite group of odd order, then no $x\in$$G$ , other than $x=1$, is conjugate to its inverse.
This question is from Advanced Modern Algebra (exer 2.79) by Joseph J. Rotman.
The hint states that if $x$ and $x^{-1}$ are conjugate, how many elements are in $x^{G}$?
What I know so far:
- $\left\lvert x^{G}\right\rvert$ is odd (greater than 1, otherwise it's in the center) and is a divisor of |$G$|
- |Z($G$)| has a common factor (other than 1) with size of the orbit $\left\lvert x^{G}\right\rvert$ so that the center is not just the identity. This is from the class equation.
- The centralizer of $x$ has odd size and $\lvert C_{G}(x) \rvert \cdot \lvert x^{G}\rvert=\lvert G\rvert$
I don't see the implication of $x$ and its inverse being conjugate has other than their orbits having the same size and laying in the same conjugacy class.
Is the info I have useful? Any help would be appreciated.