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This question probably has a simple and immediate answer which escapes me now. (And, I should admit, it's more my curiosity than anything else.) The only natural way to construct a group structure on the cartesian product $G\times H$ of two groups $G$ and $H$ (in particular, ``natural'' to me means that on each factor the group product should be the original one) is the semi-direct product in the case when one group acts on another one by automorphisms. Are there any natural constructions of a group structure on $G\times H$ where neither factor is a normal subgroup?

Update: I was pointed out that the notion of Zappa-Szep product that appears in the answer given by Steven Gubkin is also mentioned in an earlier MO discussion; I thought I'd link it here for some sort of connectivity.

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up vote 9 down vote accepted

Wikipedia to the rescue!

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Dear Steven, thanks! That (more precisely, the part on external Zappa-Szep products) is exactly what I was looking for. – Vladimir Dotsenko Jun 13 '10 at 19:37
Interstingly enough, this construction -- which I knew but not by name -- is intimately related to the subject of "dressing transformations" in integrable systems. The actions of H on K and of K on H are abstract analogues of the dressing actions in Poisson-Lie groups. – José Figueroa-O'Farrill Jun 13 '10 at 20:30
@Jose: Yes, and also there is an analogous version for Hopf algebras. Indeed, the "Drinfeld Double" of a Hopf algebra is a special example. – Theo Johnson-Freyd Jun 13 '10 at 22:01
As José and Theo remarked, this is certainly an interesting notion, but I don't see any general (non-tautological) $\textit{construction}.$ – Victor Protsak Jun 13 '10 at 23:54
@Victor: I was inclined to say the same after I first read this, but then I looked again into the part on external Zappa-Szep products, and I should admit that in a sense this is a construction in the same way the semi-direct product is a construction. The semi-direct product depends on some data (action of H on K); similarly, this product depends on the mappings $\alpha$ and $\beta$ satisfying compatibility condition... – Vladimir Dotsenko Jun 14 '10 at 1:34

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