An element of a group $G$ is not conjugate to its inverse if $\lvert G\rvert$ is odd
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.
Suppose $x \neq 1$ is conjugate to $x^{-1}$. Let $C(x)$ be the conjugate class containing $x$. If $x = x^{-1}$, then $x^2 = 1$. Hence $|G|$ is divisible by $2$. This is a contradiction. Hence $x \neq x^{-1}$. Since $|C(x)|$ is odd, $C(x)$ contains $y$ which is neither $x$ nor $x^{-1}$. Since $x$ is conjugate to $y$, $x^{-1}$ is conjugate to $y^{-1}$. Hence $y^{-1} \in C(x)$. Since $y \neq y^{-1}$, we must conclude that $|C(x)|$ is even. This is a contradiction.
Your point 1. is very useful. I don't think I need the other points; here's how I would proceed:
Try listing the elements of $x^G$. Well, we know $x$ is in there, and $x^{-1}$. If these two are distinct, then we know they're not the only two, because you told us that $|x^G|$ was odd. So some $y$ is in there too. But now what about $y^{-1}$?
Can you come up with a good reason why $x$ and $x^{-1}$ are distinct?