 Solution

Given $N⊴G$ and $H\le G$, the conditions are $\boxed{G=NH,N\cap H=\{1\}}$. Every element then has a unique expression $nh$. The extra datum is the action $\alpha :H\to \mathrm{Aut}(N)$ given by conjugation, ${\alpha }_h(n)=hn{h}^{-1}$. On pairs $(n,h)$ the reconstructed multiplication isFor example $S_3=⟨(123)⟩⋊⟨(12)⟩$. The order-three subgroup is normal, the intersection is trivial, and their sizes multiply to six. Conjugation by $(12)$ inverts $(123)$, so the action is nontrivial. Consequently this is not the direct product of these subgroups: $C_3\times C_2$ is Abelian, whereas $S_3$ is not.