Question:Semidirect product resemblance of complements

Q: '''Suppose $$N$$ is a normal subgroup of a group $$G$$ and $$H, K $$ are both permutable complements to $$N$$ in $$G$$ (i.e., they both have a common normal complement). How much does $$H$$ resemble $$K$$?'''

A: $$H$$ and $$K$$ are isomorphic groups, because answer references::complement to normal subgroup is isomorphic to quotient (this is a corollary of the second isomorphism theorem).

When $$N$$ is non-abelian, it is not necessary in general that there be an automorphism of $$G$$ sending $$H$$ to $$K$$. See answer references::complements to normal subgroup need not be automorphic. Further, it is not even necessary that the induced action by conjugation of $$K_1$$ on $$H$$ be the same as that of $$K$$ on $$N$$, although the induced map on the outer automorphism group is the same.

We can go even further and note that when $$H$$ is non-abelian, it is possible that $$H$$ be a normal subgroup and $$K$$ be a non-normal subgroup so that the action of $$H$$ by conjugation on $$H$$ is trivial whereas the action of $$K$$ by conjugation is nontrivial. The simplest example is a square of a group $$A \times A$$, with $$N = A \times \{ e \}$$, $$H = \{ e \} \times A$$, and $$K$$ the diagonal subgroup $$\{ (a,a) : a \in A \}$$.

On the other hand, in the special case that $$N$$ is an abelian normal subgroup, there is an automorphism of $$G$$ sending $$H$$ to $$K$$. See answer references::complements to abelian normal subgroup are automorphic. In particular, there is a well-defined action of the quotient group on the normal subgroup, independent of the choice of complement (and even if no complement exists): answer references::quotient group acts on abelian normal subgroup.