TITLE: $\operatorname{Aut}(G)$ contains an involution $\sigma$ with no nontrivial fixed point
QUESTION [8 upvotes]: I am just reading some algebra books on my own, and it seems the following exercise appears in so many of them:

Let $G$ be a finite group with $\sigma\in\operatorname{Aut}(G)$ satisfying 1)$\sigma^2=\text{id}$ and 2)$\sigma(g)\neq g$ for $g\neq e$. Prove $G$ is abelian.

The proof itself is not so difficult, but I am wondering whether this statement is of some importance. Searching for such a $\sigma\in\operatorname{Aut}(G)$ does not seem to me to be a very efficient way to show $G$ is abelian.
So does this statement have some nice applications? Why does it appear in so many algebra books?
Thanks!

REPLY [2 votes]: The only place I've ever seen this problem is on a Harvard preliminary exam for the Ph.D (I remember because I thought about it for a while!)... would you mind telling us where else you found it? (I'm not questioning that you found it in many places!)
That being said, I don't think it's of any practical use. But the proof is fun:
Define a map (a priori not a homomorphism) $g: G \to G$ by $g(a)=\sigma(a)\cdot a^{-1}$. Then $\sigma(g(a))\cdot g(a) = a \cdot \sigma(a^{-1}) \cdot \sigma(a) \cdot a^{-1} = 1$, so, for any element $b$ in the image of $g$, we have $b^{-1}=\sigma(b)$. 
But $g$ is injective: indeed, if $\sigma(a)\cdot a^{-1} = \sigma(c)\cdot c^{-1}$ then $\sigma(c^{-1}a)=c^{-1}a$ and by assumption the only fixed point of $\sigma$ is $1$, so $a=c$.
Since $g$ is injective and $G$ is finite, $g$ is also surjective. Therefore $b^{-1}=\sigma(b)$ for every $b\in G$. But the map $b \mapsto b^{-1}$ is a homomorphism if and only if $G$ is abelian.