I am studying for a final and am trying to solve this problem:
Let $G$ be a finite group with an automorphism $\sigma:G\rightarrow G$ such that $\sigma \circ \sigma=1$ and whose only fixed point is the identity element. I want to prove that $G$ is abelian and has odd order.
To show $G$ is abelian, since $\sigma \circ \sigma(xy)=xy$, showing that $xy=yx$ is the same as showing $\sigma \circ \sigma(xy)=\sigma \circ \sigma(yx)$. Since $\sigma \circ \sigma(xy)=\sigma(\sigma(x)\sigma(y))$, we want to show $\sigma(\sigma(x)\sigma(y))=\sigma(\sigma(y)\sigma(x))$, so it is enough to show that $\sigma(x)\sigma(y)=\sigma(y)\sigma(x)$, but I am having trouble working out the details.
To prove that $G$ has odd order, I am trying to show this by contradiction and assuming that $G$ is even. If $G$ is even, then it has at least one element $y$ of order 2, but using this, I haven't been able to get a contradiction.
Best Answer
The map $x^{-1}\sigma(x)$ is injective, proof: assume $x^{-1}\sigma(x)=y^{-1}\sigma(y)$ then $yx^{-1}=\sigma(y)\sigma(x)^{-1}=\sigma(yx^{-1})\implies yx^{-1}=e\implies y=x$ since there are no fixed points besides $e$. since $G$ is finite and $x^{-1}\sigma(x)$ is injective it is also surjective.
hence we can write every element of $G$ in the form $x^{-1}\sigma(x)$. Take $g$ in $G$, it can be written as $x^{-1}\sigma(x)$. thus $\sigma(g)=\sigma(x)^{-1}x$. Notice multiplying $g$ with $\sigma(g)$ gives us $e$. we conclude $\sigma(g)=g^{-1}$
Two things are now clear: