$\newcommand{\Aut}{\operatorname{Aut}}$A classmate of mine recently posted an interesting question on Facebook. It didn't get an answer, and I couldn't get anywhere myself, so I'm hoping that someone here will be able to help out.
Here's a paraphrase of the question:
The Zappa–Szép product generalizes the semi-direct product, which itself is a generalization of the direct product. They've all got roughly the same conditions for the "internal" product:
- $H$ and $K$ are subgroups of some parent group $G$
- $H \cap K = \{ e \}$
- $HK = G$
- Normality:
- Direct: $H$ and $K$ are normal in $G$
- Semidirect: $K$ is normal in $G$
- Zappa-Szép: no further assumptions
But usually, people use this products in their "external" forms. For these three, we need different amounts of additional data describing how the groups commute.
For the direct product $H \times K$, we need no further information. For a semidirect product $H \rtimes_\varphi K$, we also need a homomorphism $\varphi : H \to \Aut(K)$.
So for the Zappa-Szép product, one might expect to have two homomorphisms: $\alpha : K \to \Aut(H)$ and $\beta : H \to \Aut(K)$. But as the Wikipedia article explains, one of the conditions we need is that $$ \alpha(k, h_1 h_2) = \alpha(k, h_1) \cdot \alpha(\beta(h_1, k), h_2) $$
But if $\alpha(k)$ were an automorphism of $H$, then the condition would just be $$ \alpha(k, h_1 h_2) = \alpha(k, h_1) \cdot \alpha(k, h_2) $$
This makes sense from a multiplication perspective, because as we pull the $k$ to the right of $h_1$, it will become a different element of $K$. But it's rather unappetizing from a morphism perspective.
Is there a way to phrase this additional data more elegantly?
Here's my thoughts so far: $\alpha$ might not be a homomorphism into $\Aut(H)$, but it's still a left group action of $K$ on $H$. Same for $\beta$, except it's a right group action. So that takes care of the first five conditions from the wiki page. So now we just need to rephrase the last two conditions in a group-action-friendly way. But I can't seem to do so.
Alternatively, maybe we should think of this as a single map $\psi : K \times H \to H \times K$. But I can't think of any natural, morphism-ish qualities that this map must have.