Automorphism that is an Involution of a finite group 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. 
 A: 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: 


*

*$G$ cannot be even for if it where it would have an element of order $2$, which would map to itself under $\sigma$ for it is its own inverse.

*$G$ is abelian, for $\sigma(gh)=g^{-1}h^{-1}$ since $\sigma$ is homomorpphism, but $\sigma(gh)=(gh)^{-1}=h^{-1}g^{-1}$. thus $g^{-1}h^{-1}=h^{-1}g^{-1}\implies hg=gh$ (just take inverses con both sides).

A: For the second part, note that the elements $x\in G$ with $x\ne e$ can be paired off $\{x, \sigma(x)\}$. If there are $k$ such pairs, then there are $2k+1$ elements in $G$.
In general, given an involution $\sigma$ on a finite group $G$, the parity of $|G|$ is the same as the parity of the number of fixed points of $\sigma$.
A: Note that $ \sigma \circ \sigma =\iota,$ the identity automorphism is given. Thus, for all $g\in G,$ we have 
$$g=\sigma ^2 (g)=\sigma(\sigma(g)). $$Since $\sigma $ is an automorphism , $\sigma ^{-1}$ exists and is a homomorphism. Therefore, $\sigma ^{-1}(g)=\sigma^{-1}(\sigma(\sigma(g)))=\sigma(g).$ Because $\sigma$ is a homomorphism, an elementary exercise shows that $\sigma ^{-1}(g)=\sigma (g^{-1}).$ We then have $\sigma (g^{-1}h^{-1})=\sigma(g^{-1})\sigma(h^{-1})=\sigma (g)\sigma(h)=\sigma((hg)^{-1})=\sigma(hg).$ But this yields the equality $\sigma(hg)=\sigma(gh)$ which in turn yields $hg=gh$ upon applying $\sigma ^{-1}$ both sides. Since the only fixed point of the automorphism is $1,$ for non trivial $g, g$ and $\sigma (g)$ are different so $$G=\cup _{ g \in G\setminus {1}}\{g, \sigma(g)\} \cup \{1\}$$ giving $|G|=2|G\setminus \{1\}|+1$ which is odd .
