Fixed-point-free involution on finite group is inverse map

Statement

Let $G$ be a finite group and $\sigma:G \to G$ be an automorphism that is involutive i.e. $\sigma^2$ is the identity map. Suppose, further, that $\sigma$ is fixed-point-free. Then, $\sigma$ is the inverse map from $G$ to itself and $G$ is an odd-order abelian group.

Proof

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Given: A finite group $G$, a fixed-point-free involution $\sigma$ of $G$

To prove: $G$ is an abelian group and $\sigma$ sends every element to it inverse.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The map $x \mapsto x \sigma(x^{-1})$ is injective from $G$ to itself. Fact (1) $\sigma$ is fixed-point-free. Fact-given direct.
2 The map $x \mapsto x\sigma(x^{-1})$ is surjective from $G$ to itself. $G$ is finite Step (1) Step-given direct.
3 If $a \in G$ is of the form $a = x \sigma(x^{-1})$ for some $x \in G$, then $\sigma(a) = a^{-1}$. $\sigma$ has order two and is an automorphism We have $\sigma(a) = \sigma(x\sigma(x^{-1}) = \sigma(x) \sigma^2(x^{-1} = \sigma(x)x^{-1} = (x \sigma(x^{-1}))^{-1} = a^{-1}$.
4 $\sigma$ sends every element to its inverse. Steps (2), (3) Step-combination direct.
5 $G$ is abelian and $\sigma$ is its inverse map. Fact (2) Step (4) Step-fact combination direct.
6 $G$ is abelian of odd order and $\sigma$ is its inverse map. Fact (3) Step (5) By Fact (3), if $G$ had even order, it would have an element of order two. This would be a fixed point under $\sigma$, contradicting the fixed-point-free nature of $\sigma$. Thus, $G$ has odd order.