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I've been struggling with the following problem. I don't really know how to tackle this, but I suspect that it is not hard at all.

Let $G$ be a group and $f\colon G\to G$ an isomorphism such that $f$ is an involution ($f^2=1_G$) and $e$ is the only fixed point of $f$ (that is, $f(x)=x\implies x=e$)

I must prove that $G$ is abelian.

How should I proceed here ?

The problem gives a hint: show that every element of $G$ can be written as $x f(x)^{-1}$. I suspect that then one can use some result like "$G$ is abelian $\iff x\mapsto x^{-1}$ is a group morphism ", but I haven't tried yet because I don't see why the hint holds.

Any help is appreciated, thanks in advance!

EternalBlood
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    let $G=\Bbb Z$ and $f:n \mapsto -n$. Then, $n-f(n) = 2n$, which clearly does not cover every element of $G$, so your hint is wrong. – Kenny Lau Apr 09 '18 at 02:29
  • Is your group assumed to be finite? – Kenny Lau Apr 09 '18 at 02:29
  • No, the problem doesn't say that, but I suspect that it is a missing hypothesis. – EternalBlood Apr 09 '18 at 02:31
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    It seems the map which exchanges the generators of $F_2$ (the free group on two letters) is a counterexample. (Any nontrivial word either starts with $a$ or $b$, but this involution exchanges the two so any non-trivial word is not a fixed point) – PVAL-inactive Apr 09 '18 at 02:32
  • @EternalBlood where does this problem come from? in the link above it comes from "Rotman J.J. Introduction to the theory of groups, exercise 1.50" where finiteness is assumed – Kenny Lau Apr 09 '18 at 02:32
  • It was a problem given during my group theory class, but I suspect that it comes down from Rotman. – EternalBlood Apr 09 '18 at 02:34
  • Of course, $\mathbb Z$ is abelian. The hint might be that this map would cover $G$ if $G$ is not abelian. @KennyLau – Thomas Andrews Apr 09 '18 at 02:36
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    @ThomasAndrews see the duplicate. the original problem, I presume, has $G$ finite, which the OP does not assume (by error or whatever) – Kenny Lau Apr 09 '18 at 02:36

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