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. Commutator map with fixed-point-free automorphism is injective
2. Inverse map is automorphism iff abelian
3. Cauchy's theorem

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.

References

• Topics in Algebra by I. N. Herstein, ^{More info}, Page 70, Problems 10-11
• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 41, Exercise 23, Section 1.6 (Homomorphisms and isomorphisms)
• Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 336, Theorem 1.4, Section 10.1 (Elementary properties)