Prove $G$ is abelian if $f(f(x)) = x$?

Let $G$ be a finite group and $f$ an automorphism such that

1. $f(f(x)) = x$, and

2. $f(x) = x$ if and only if $x=e$.

Prove that $G$ is abelian and $f(x) = x^{-1}$.

My attempt:

Since $f(f(x)) = x$, we can divide the elements of $G$ (except e) into pairs $(a,b)$ such that $f(a)=b$ and $f(b)=a$. So $G$ has odd order. What to do next.. Thanks in advance for any hints.

Solution 1:

The map $h : G \to G$, $h(x) = f(x) x^{-1}$ is injective, because the only fixed point of $f$ is $e$. So it's injective, and therefore surjective because $G$ is finite. So you can write any $y \in G$ as $y = h(x) = f(x) x^{-1}$. But then $$f(y) = f(f(x)) f(x)^{-1} = x f(x)^{-1} = \left( f(x) x^{-1} \right)^{-1} = y^{-1}$$

And this for all $y \in G$. Now it's a standard argument that if $f(x) = x^{-1}$ is a homomorphism, then the group is abelian: $xy = (y^{-1} x^{-1})^{-1} = (y^{-1})^{-1} (x^{-1})^{-1} = yx$.