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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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Wikipedia to the rescue!

http://en.wikipedia.org/wiki/Zappa-Szep_product

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  • $\begingroup$ Dear Steven, thanks! That (more precisely, the part on external Zappa-Szep products) is exactly what I was looking for. $\endgroup$
    – Vladimir Dotsenko
    Commented Jun 13, 2010 at 19:37
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    $\begingroup$ 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. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Jun 13, 2010 at 20:30
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    $\begingroup$ @Jose: Yes, and also there is an analogous version for Hopf algebras. Indeed, the "Drinfeld Double" of a Hopf algebra is a special example. $\endgroup$
    – Theo Johnson-Freyd
    Commented Jun 13, 2010 at 22:01
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    $\begingroup$ As José and Theo remarked, this is certainly an interesting notion, but I don't see any general (non-tautological) $\textit{construction}.$ $\endgroup$
    – Victor Protsak
    Commented Jun 13, 2010 at 23:54
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    $\begingroup$ @VladimirDotsenko: The relationship with semi-direct product is closer than your last remark hints at: $\alpha$ must be a left action of $K$ on $H$, and $\beta$ must be a right action of $H$ on $K$. (And then those two actions must satisfy some compatibility assumption.) And indeed, if either of these actions is trivial, then one recovers the semi-direct product. $\endgroup$
    – Joshua Grochow
    Commented Mar 20, 2017 at 15:19

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