Group Theory

If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse

Problem 106

Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element.

Show that $x$ is not conjugate to $x^{-1}$.

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Proof.

Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$.
Then we have
\[xg=gx^{-1}. \tag{*}\] Then we compute
\begin{align*}
(xg)^2&=(gx^{-1})(xg)=g^2\\
(xg)^3&=(xg)(xg)^2=(xg)(g^2)=xg^3\\
(xg)^4&=(xg)^2(xg)^2=g^2g^2=g^4.
\end{align*}

In general, we have by induction
\begin{align*}
(xg)^k=
\begin{cases}
xg^k & \text{ if } k \text{ is odd}\\
g^k & \text{ if } k \text{ is even}. \tag{**}
\end{cases}
\end{align*}

Let $m$ be the order of $g$. Since $G$ is a finite group of odd order, $m$ is odd.
We have by (**)
\[(xg)^{m-1}g=g^{m-1}g=g^m=1.\] Thus we have $g^{-1}=(xg)^{m-1}$. Multiplying by $xg$ from the right, we have
\begin{align*}
g^{-1}xg &=(xg)^m\\
&=xg^m \text{ by (**)}\\
&=x.
\end{align*}

Thus we have $gx=xg$, and combining with (*) we obtain $gx=gx^{-1}$.
This implies that $x^2=1$, and since the order of $G$ is odd, this implies $x=1$.

This contradicts our choice of $x$.
Therefore $x$ cannot be conjugate to $x^{-1}$.

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