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Suppose I have a group $G,$ with normal subgroups $K, L$ such that $G=KL$ but $K\cap L \neq \{1\}.$ If the intersection were trivial, we would say that $G$ is the direct product of $K$ and $L,$ but this operation to direct product is like free product with amalgamation is to free product. Any names for it out there?

I am really more interested in the properties of this more than the name (though the latter might help me find the former) -- for example, given three groups $H_1, H_2, H3,$ is there a way to classify $G$ where $K\simeq H_1, L\simeq H_2, K\cap L \simeq H_3?$ What if $H_1, H_2$ are finitely generated free groups, and $H_3$ is an infinitely generated free group?

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Even when $K\cap L = 1$, you should not call $G=KL$ a "direct product" unless $K,L$ commute. A good general word in this case is "crossed product". – Theo Johnson-Freyd Mar 30 '11 at 2:21
When $K$ and $L$ are normal (and intersect trivially) they DO commute (the commutator of $k, l$ is in both subgroups, thus trivial). – Igor Rivin Mar 30 '11 at 2:25

I do not know whether this answers you question, but such a group is an extension of $K/K \cap L \times K / K \cap L$ by $K \cap L$, i.e.

$$1 \to K \cap L \to KL \to K/K \cap L \times L/K \cap L \to 1.$$

Conversely, any extension of the form $$1 \to G_3 \to G \stackrel{\pi}{\to} G_1 \times G_2 \to 1$$ admits normal subgroups $K=\pi^{-1}(G_1)$ and $L=\pi^{-1}(G_2)$ such that $KL=G$ and $K \cap L =G_3$.

I think that there is not much more to say in general. Classification of group extensions in general is complicated and involves non-abelian cohomology. I do not know if this is any easier in the case where $K$ and $L$ are free groups.

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One phrase is just "normal product", but people who use such a phrase usually don't pay attention K∩L, so I don't think it will help you very much.

These normal products are much weirder than direct products with amalgamation though!

For instance, any (finite) direct product of supersolvable groups is supersolvable, and any quotient of a supersolvable group is supersolvable, so any (finite) direct product with amalgamation of supersolvable groups is supersolvable. However, a normal product of supersolvable groups need not be supersolvable. While the groups may normalize each other, they may not normalize their chief series, and so the result may not be supersolvable.

The key problem is that [K,L] ≤ K ∩ L only implies K and L commute when K ∩ L = 1.

People do say very non-trivial things about normal products, but I've not heard anything that sounds relevant to your situation.

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If $K$ and $L$ are supersoluble, then $KL$ need not be, but $[K,K]$ and $[L,L]$ are nilpotent, and the normal product of two nilpotent groups is still nilpotent. So $KL$ is at worst nilpotent-by-(nilpotent of class 2). No idea what can be said about normal products of free groups, though. – Colin Reid Apr 3 '11 at 14:28

As already mentioned above, the group in question is an extension

$$1\rightarrow K\cap L \rightarrow KL \rightarrow K/K\cap L \times L/K\cap L \rightarrow 1,$$

where we abbreviate the rightmost (a priori) non trivial group by $H$.

In particular $K\cap L$ is normal in $KL=G$, so $G$ is isomorphic to a subgroup of the wreath product $K\cap L \wr H$. I don't know if this point of view is helpful, but I for myself would try to use wreath products here.

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Thanks! I will try to figure out the connection to the wreath product... – Igor Rivin Mar 30 '11 at 17:46

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