TITLE: Show that a finite group with certain automorphism is abelian
QUESTION [11 upvotes]: Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)

With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.
Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50

REPLY [11 votes]: For proving the second part, that is $f \circ f = I$ implies $G$ is abelian you can proceed as follows:
Consider $g \in G$. By the above $\mathsf{Hint}$ every $g \in G$ can be written as $g = x^{-1} f(x)$ for some $x \in G$. 
We then have 
\begin{align*} g = x^{-1} f(x) \Longrightarrow f(g) &= f(x^{-1}f(x))= f(x^{-1}) \cdot f(f(x)) \\ &=(f(x))^{-1} \cdot x = \bigl[x^{-1} f(x)\bigr]^{-1}=g^{-1}
\end{align*}
Now we prove that if $f: G \to G$ defined by $f(g) = g^{-1}$ is a Automorphism if $G$ is abelian. 
To see this note that $f(ab)=(ab)^{-1} = b^{-1} \cdot a^{-1} $. But $$f(ab) = f(a) \cdot f(b) \Rightarrow f(ab) = a^{-1}b^{-1}$$ From above we have $$b^{-1}a^{-1} = a^{-1}b^{-1} \Rightarrow (ab)^{-1} = (ba)^{-1} \Rightarrow ab=ba$$ Hence $G$ is abelian.