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.

Related facts

 * Semidirect product of finite group by fixed-point-free automorphism implies all elements in its coset have same order
 * Frobenius conjecture: This is a generalization of sorts of the above result. It states that if a finite group possesses a fixed point-free automorphism of prime order, the finite group must be nilpotent. This was proved by Thompson in 1959.

Facts used

 * 1) uses::Commutator map with fixed-point-free automorphism is injective
 * 2) uses::Inverse map is automorphism iff abelian
 * 3) uses::Cauchy's theorem

Proof
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: