Commutator map with fixed-point-free automorphism is injective

Statement

Suppose $G$ is a group and $\sigma$ is a fixed-point-free automorphism of $G$, i.e., the only fixed point of $\sigma$ is the identity element. Then, the mapping:

$x\mapsto x\sigma (x^{-1})$

is an injective self-map from $G$ to itself.

Related facts

Applications

• Fixed-point-free involution on finite group is inverse map
• Semidirect product of finite group by fixed-point-free automorphism implies all elements in its coset have same order

Proof

Given: Group $G$, fixed-point-free automorphism $\sigma$, elements $x,y\in G$ such that $x\sigma (x^{-1})=y\sigma (y^{-1})$

To prove: $x=y$

Proof: With some manipulation, we can rewrite the condition as:

$y^{-1}x=\sigma (y^{-1})(\sigma (x^{-1})^{-1}$

Simplify the right side using that $\sigma$ is an automorphism, and obtain:

$y^{-1}x=\sigma (y^{-1}x)$

Since $\sigma$ is fixed-point-free, and $y^{-1}x$ is a fixed point of $\sigma$, this forces that $y^{-1}x$ is the identity element of $G$, yielding $x=y$.