Group theory problem automorfism

Question

Let $G$ be a finite group. If there exist an automorphism $f$ such that if $f(x)=x \iff x=e$ and $f(f(x))=x$ for all $x$ in $G$, then prove $G$ is Abelian.

Answer 1

Let $\sigma: G \to G$ be the set-theoretic map (not a homomorphism, a priori)

$$\sigma(x) = f(x)x^{-1}.$$

Remark that if $\alpha \in \text{Im}(\sigma)$, say $\alpha = f(y)y^{-1}$, then

$$\sigma(\alpha) = f(\alpha)\alpha^{-1} = f(f(y)y^{-1})yf(y)^{-1}$$

$$=yf(y)^{-1}yf(y)^{-1} = \alpha^{-2}$$

hence $f(\alpha)\alpha^{-1} = \alpha^{-2},$ in other words $f(\alpha)=\alpha^{-1}$. Now if $G$ is any group, the map $x \mapsto x^{-1}$ is an automorphism of $G$ if and only if $G$ is abelian (exercise). Hence to show that $G$ is abelian, it suffices to show that the map $\sigma$ is surjective. Since $G$ is finite, it suffices to show that $\sigma$ is injective. Suppose therefore that

$$f(x)x^{-1} = f(y)y^{-1}.$$

Then $f(xy^{-1}) = xy^{-1}$, but by assumption $f$ fixes only the identity $e$. Hence $xy^{-1} = e$, and $x=y$.