Exercise 4.3.30 (If $|G|$ is odd, then $x$ and $x^{-1}$ are not conjugate in $G$)

Proof. Let $G$ be a group of odd order and let ${𝒪}_x$ be the conjugation class of $x$. Suppose for the sake of contradiction that $x$ and $x^{-1}$ are conjugate. Then $x^{-1}\in {𝒪}_x$, so we obtain

If $g\in {𝒪}_x$, then $g=\mathit{𝑎𝑥}{a}^{-1}$ for some $a\in G$, thus $g^{-1}=a{x}^{-1}{a}^{-1}\in {𝒪}_{x^{-1}}={𝒪}_x$, so

Consequently we may consider the map

We claim that $\phi$ is an involution without fixed point:

• For all $g$ in ${𝒪}_x$, $\phi (\phi (g))=\phi (g^{-1})={(g^{-1})}^{-1}=g$, so $\phi \circ \phi ={\mathrm{id}}_G$.
• If $g\in {𝒪}_x$ satisfies $\phi (g)=g$, then $g^2=1$. If $g\ne 1$, then the order of $g$ is $2$: since $|G|$ is odd, this is impossible by Lagrange’s Theorem. So $g=1\in {𝒪}_x$. This implies $x=1$, which is false by hypothesis. So the involution $\phi$ has no fixed point.

Since $\phi$ is an involution without fixed point, we can group the elements of ${𝒪}_x$ by pairs $\{g,g^{-1}\}$, hence $|{𝒪}_x|$ is even.

But $|{𝒪}_x|=|G:C_G(x)|$ is a divisor of $|G|$, which is odd, thus $|{𝒪}_x|$ is odd. This contradiction shows that $x$ is not conjugate to $x^{-1}$.

If $G$ is a group of odd order and $x\in G,x\ne 1$, then $x$ and $x^{-1}$ are not conjugate in $G$. □