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

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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

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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.
 

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