Let $f: G \rightarrow G$ be an isomorphism, with $|G| < \infty$, $f$ has no non-trivial fixed points and $f \circ f = id_G$. Show $f(x)=x^{-1}$ and $G$ is abelian. (Hint. Prove that every element of G has the form $xf(x)^{-1}$.
This is exercise 1.50 ii from Rotman - Introduction to the theory of groups. I have tried to do this exercise, but I couldn't get it.
What I have done already
If $G$ is abelian, then $f(xf(x)) = f(x)x=xf(x)$. This showns $xf(x) = e$ because $f$ has no non-trivial fixed points and thus $f(x) = x^{-1}$.
Exercise 15.50 i states $G$ is abelian iff $f(x) = x^{-1}$ is a homomorphism. I didn't succeed in proving $f(x) = x^{-1}$ so I thought I tried showing $G$ is abelian.
In conclusion, I showed:
$$G \text{ abelian} \Leftrightarrow f(x)=x^{-1}$$
However, I cannot prove either statement.
About the hint
I thought about $g=xf(x)^{-1}$. Suppose we already know $f(x)=x^{-1}$, this means that every $g\in G$ can be written as $x^2$. Thus every element has a root, which seems kinda weird. I worked out $\mathbb{Z}_7$ and there you have two "cycles": (1, 2, 4) and (3,5,6). I thought you can do some kind of argument where you take $x, x^2, x^4, x^8, ...$ where you know that this is bound to have some repetition ($G$ is finite). But I don't really know the point of doing this, gets me nowhere.
The only way of proving the hint (I think) is just showing that the set $\{x f(x)^{-1} | x \in G\}$ has the same cardinality as $G$ (we can do that since $G$ is finite). Thus I need to show $xf(x)^{-1}=yf(y)^{-1} \Rightarrow x = y$. Could not do it. I think I need to use that $f$ has no non-trivial fixed points, but that just comes down to what I stated above.
Hint from you guys, no answers!
I don't want the answer, I want to do it myself. I studied math long time ago and group theory was my favorite subject. I picked up the book to just work through all the exercises (beginning I know already, but there is some new stuff at the end). I try to do all exercises from all the stuff I already know to get into the subject. I don't want the answer itself. If you can just give a hint in the right direction, that would be awesome :)
