An automorphism that has no fixed points except for the identity and is its own inverse implies commutativity

Question

Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.

While trying to do this exercise I noticed two facts.

First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.

Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.

Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.

Is this the right way, or is there an easier manner to solve this?

Answer 1

You have already noticed that $g$ and $f(g)$ commute. Excellent. Now, where does $f$ map $g f(g)$?

Answer 2

This is a classic problem. If we can show that $\sigma(g) = g^{-1}$ for all $g$ then it easily follows that $G$ is abelian. Remark that if $g = h^{-1} \sigma(h)$ for some $h$, then, since $\sigma$ has order $2$,

$$\sigma(g) = \sigma(h)^{-1} h = g^{-1}.$$

Now I claim that every element of $G$ is of the form $h^{-1} \sigma(h)$ for some $h$. Indeed consider the map (of sets) $\varphi : G \to G$ given by $h \mapsto h^{-1} \sigma(h)$. I claim that $\varphi$ is injective. Indeed if $h^{-1} \sigma(h) = k^{-1} \sigma(k)$, then $\sigma(hk^{-1}) = hk^{-1}$; but by assumption $\sigma$ fixes only the identity so $h = k$ and $\varphi$ is injective. Since $G$ is finite $\varphi$ is also surjective and therefore every element of $G$ is of the form $h^{-1} \sigma(h)$, and therefore $\sigma(g) = g^{-1}$ for every $g \in G$ and we are done.

Remark that it's not true without the finiteness assumption. For instance if $G= F_2$ is the free group on $x, y$, then the automorphism which switches $x$ and $y$ has order $2$ and fixes only the identity, but $F_2$ is far from abelian.