I was studying some abstract algebra and I ran into this exercise and I'm like stuck on it for almost three days now, I would appreciate if I could get some help.
Given $\sigma \in \operatorname{Aut}(G)$ for $G$ a finite group and $\operatorname{Aut}(G)$ the automorphisms group, if $\sigma^2 = 1$ and $x^\sigma \neq x$ for all $x \neq 1$, prove that the group G is abelian.
The notation $x^\sigma$ is just $\sigma(x)$. Thanks in advance!
Also, the book gives a hint to help solving this, but I cannot understand how this hint gives the result in any way:
To solve this part, try to prove that the set $\{x^{-1}x^\sigma : x \in $ G$\}$ is the entire group G by using the map $x \mapsto x^{-1}x^\sigma$.
Following the hint:
Observe that the map $\phi:x\mapsto x^{-1}x^\sigma$ is injective. In fact from $x^{-1}x^\sigma=y^{-1}y^\sigma$ we get $$ yx^{-1}=y^\sigma(x^{-1})^\sigma=(yx^{-1})^\sigma $$ and thus $x=y$ since $1_G$ is the only element fixed by $\sigma$.
Therefore $\phi$ is bijective because $G$ is finite.
Given $g=x^{-1}x^\sigma$ observe that $g^\sigma=(x^\sigma)^{-1}x=g^{-1}$ since $\sigma^2=1$. By the surjectivity of $\phi$ this means that $\sigma$ is the involution $g\mapsto g^{-1}$ which is an antiautomorphism.
We conclude recalling that a non trivial antiautomorphism is an automorphism only when the group is abelian.
Moreover we can also say that $|G|$ is odd (involutions on an set of even order have an even number of fixed points).