Showing that every element in G can be written a certain way

[Mr Davis 97]

Let G be a finite group which possesses an automorphism $\sigma$ that has no nontrivial fixed points and $\sigma ^2$ is the identity map. Prove that $G$ is abelian.

So there's a hint, and it tells me first to established that every element in $G$ can be written as $x^{-1} \sigma (x)$ for some $x \in G$. I have two questions. Why does showing that $f(x) x^{-1} \sigma (x)$ is injective prove the hint, and how could I ever approach this problem without knowing the hint?

 

[fresh_42]

If $f\, : \,x \mapsto x^{-1}\sigma(x)$ is injective, then it is also surjective, because $G$ is finite, hence every element can uniquely be written in this form. For injectivity use the fact, that $\sigma$ has only one fixed point.

How to find the trick is a difficult question. It's usually a mixture of experience, practice, fantasy, and sometimes methodology. I like to say: first list all what is given. Also a backwards approach can help: Write down the statement which has to be proven and see if there are sufficient conditions, from which it will follow. Then try to prove such a condition. This method is often the first step in very complex proofs.

For the above we had used that $\sigma$ is a homomorphism, has only $e$ as fixed point, and that $G$ is finite. So somehow we need to use the fact, that $\sigma^2= 1$. This means $\sigma = \sigma^{-1}$ so inverses might play a role. From your own experience you already know, that inversion together with automorphism implies commutativity, so to consider elements $x^{-1}\sigma (x)$ isn't far fetched. This is only one way, but to gather all given facts in the first place is always a good start. Unfortunately a rarely used habit, if I think about all the empty places under section 2 of our homework template. In real life one probably tries a lot of ideas and among the scribbling will turn up the solution. The expressions which can help aren't too many in this example.