**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 1.6 Exercise 1.6.23**

Let $G$ be a finite group which possesses an automorphism $\sigma$ such that $\sigma(g) = g$ if and only if $g = 1$. If $\sigma^2 = \sigma \circ \sigma$ is the identity map on $G$, prove that $G$ is abelian. Such an automorphism $\sigma$ is called *fixed point free* of order $2$. [Hint: Show that every element of $G$ can be written in the form $x^{-1} \sigma(x)$ and apply $\sigma$ to such an expression.]

Solution: We define a mapping $f : G \rightarrow G$ by $f(x) = x^{-1} \sigma(x)$.

Proof of claim: Suppose $f(x) = f(y)$. Then $y^{-1} \sigma(y) = x^{-1} \sigma(x)$, so that $xy^{-1} = \sigma(x)\sigma(y^{-1})$, and $xy^{-1} = \sigma(xy^{-1})$. Then we have $xy^{-1} = 1$, hence $x = y$. So $f$ is injective.

Since $G$ is finite and $f$ is injective, $f$ is also surjective. Then every $z \in G$ is of the form $x^{-1} \sigma(x)$ for some $x$. Now let $z \in G$ with $z = x^{-1} \sigma(x)$. We have $$\sigma(z) = \sigma(x^{-1} \sigma(x)) = \sigma(x)^{-1} x = (x^{-1} \sigma(x))^{-1} = z^{-1}.$$ Thus $\sigma$ is in fact the inversion mapping, and we assumed that $\sigma$ is a homomorphism. By a previous example, then, $G$ is abelian.