## In a finite group of odd order, no nontrivial element is conjugate to its inverse

If $G$ is a group of odd order, prove that for any nonidentity element $x \in G$, $x$ and $x^{-1}$ are not conjugate.

Suppose to the contrary that $axa^{-1} = x^{-1}$ for some $a \in G$. Note then that $ax = x^{-1}a$ and that $ax^{-1} = xa$.

We now prove a few facts.

Fact 1: If $m$ is odd, then $a^mx = x^{-1}a^m$. Proof: If $m = 1$, then clearly $ax = x^{-1}a$. Suppose that the conclusion holds for some odd $m \geq 1$. Then $a^{m+2}x = a^2a^mx = aa x^{-1} a^m = axaa^m = x^{-1}a^{m+2}$. By induction, the conclusion holds for all odd positive $m$. $\square$

Fact 2: If $m$ is even, then $a^mx = xa^m$. Proof: If $m = 0$, then clearly $x = x$. Suppose that the conclusion holds for some even $m \geq 0$. Then $a^{m+2}x = a^2a^mx = aaxa^m = ax^{-1}aa^m = xa^{m+2}$. By induction, the conclusion holds for all even nonnegative $m$. $\square$.

Fact 3: If $m$ is odd, then $(ax)^m = a^mx$. Proof: If $m = 1$, we have $ax = ax$. Now suppose the conclusion holds for some odd $m \geq 1$; we have $(ax)^{m+2} = axax(ax)^m = axx^{-1}aa^mx = a^{m+2}x$. By induction, the conclusion holds for all odd positive $m$. $\square$

Fact 4: If $m$ is even, then $(ax)^m = a^m$. Proof: If $m = 0$, we have $1 = 1$. Now suppose the conclusion holds for some even $m \geq 0$. Then $(ax)^{m+2} = axax(ax)^m = axx^{-1}aa^m = a^{m+2}$. By induction, the conclusion holds for all even nonnegative $m$. $\square$

Suppose $|a| = m$; now $m$ is finite and moreover odd since $|G|$ is odd.

Note that $(ax)^{2m} = ((ax)^2)^m = (a^2)^m = (a^m)^2 = 1^2 = 1$. On the other hand, $(ax)^{2m} = ((ax)^m)^2 = (a^mx)^2 = x^2$. Thus $|x| = 2$, a contradiction.

Hence $x$ and $x^{-1}$ are never conjugate in a finite group $G$ of odd order.