Complements to normal subgroup need not be automorphic

Statement

Suppose $G$ is a group, $N$ is a normal subgroup, and $H$ and $K$ are permutable complements to $N$ in $G$. Then, it is not necessary that there exists an automorphism of $G$ sending $H$ to $K$.

Related facts

• Schur-Zassenhaus theorem
• Complements to Abelian normal subgroup are automorphic
• Complement to normal subgroup is isomorphic to quotient

Proof

A generic example

Let $A$ be any non-Abelian group. Consider $G=A\times A$ and the subgroup $N=A\times \{e\}$. Let $H$ be the subgroup $\{e\}\times A$ and $K$ be the subgroup $\{(a,a)\mid a\in A\}$.

Note that:

• $N$ is normal in $G$: In fact, it is a direct factor of $G$.
• $H$ is a permutable complement to $N$ in $G$.
• $K$ is a permutable complement to $N$ in $G$.
• $H$ is normal in $G$: In fact, it is a direct factor of $G$.
• $K$ is not normal in $G$: Pick $a,b\in A$ such that $a,b$ do not commute. Then, we have $(b,e)(a,a)(b,e{)}^{-1})=(ba{b}^{-1},a)$. Thus, a conjugate of an element in $K$ lies outside $K$.