Commutator bijection from a fixed-point-free automorphism
= Commutator bijection from a fixed-point-free automorphism
{title2=$y\longmapsto\alpha(y)^{-1}y$}
Let $N$ be a <finite group> and $\alpha$ an <automorphism> whose only fixed element is the identity. If $\alpha(a)^{-1}a=\alpha(b)^{-1}b$, then $\alpha(ba^{-1})=ba^{-1}$, so $a=b$. The displayed map is therefore a bijection. For conjugation by a nonidentity element of a <Frobenius complement>, it gives the commutator map $y\mapsto h^{-1}y^{-1}hy$.