Expression for normal product similar to semidirect product

If $H$ and $K$ are groups then a normal product is a collection $(\iota_H,\iota_K,G)$ where $\iota_X:X \to G$ is a monomorphism with $\iota_X(X) \lhd G$ and $G=\iota_H(H) \iota_K(K)$.

In such a case, $L_X = \iota_X^{-1}(\iota_H(H) \cap \iota_K(K))$ is a normal subgroup of $X$ and $\phi:L_H \to L_K : h \mapsto \iota_K^{-1}( \iota_H(h))$ is an isomorphism.

So given groups $H,K$; normal subgroups $L_H \lhd H$, $L_K \lhd K$; and an isomorphism $\phi:L_H \to L_K$ what more is needed in order to describe a normal product $(\iota_H,\iota_K,G)$ up to some obvious sense of isomorphism?

If $L_H=L_K=1$, then there is only one possible $\phi$ and all $(\iota_H,\iota_K,G)$ are isomorphic to $(\iota_1,\iota_2,H\times K)$. Hence the issue really is what do we need to say about that intersection, $L$.

A homomorphism $H \to \operatorname{Aut}(K)$ would be sufficient, but such a homomorphism would have to satisfy a fair number of compatibility conditions to ensure such a normal product existed. Thinking inside $G=HK$, we need to rewrite $kh = h k^h$ and this homomorphism serves to define $k^h$, but if $k \in H \cap K$, then the action of $h$ on $K$ has to agree with the action of $H$ on $H$.

Another idea is that we can write $kh = hk[k,h]$ so we just need to describe the somewhat bilinear map $[] : H \times K \to L$, where $L=H \cap K$, and such a map is probably related to derivations in some reasonable way.

Can anyone supply clean details for either approach?

The goal is to provide a better framework to answer the issues raised in Normal products of groups with maximal nilpotency class and Normal products and radicals in finite groups.

Solution 1:

Here's an attempt for the first approach.. I hope this is at least somewhat correct, I'll check through all the details later.

Let $H$ and $K$ be groups, let $T$ be normal in $H$ and $S$ normal in $K$. Let $\varphi: T \rightarrow S$ be an isomorphism, and let $\phi: H \rightarrow \operatorname{Aut}(K)$ be group action (denote $h^k = \phi(h)(k)$). Assume that

\begin{align*} &\varphi(t^h) = \varphi(t)^h \\ &k^t = k^{\varphi(t)} \\ &k^h S = k S \\ \end{align*}

for all $t \in T$, $h \in H$ and $k \in K$ (here $t^h$ is conjugation).

Now define $M = H \ltimes_{\phi} K$ and $N = \{d\varphi(d)^{-1}: d \in T \}$. Then $N$ should be a normal subgroup of $M$, and $M/N$ is the normal product of $H$ and $K$ that we seek. Let $\bar{H}$ and $\bar{K}$ be the images of $H$ and $K$ in $M/N$, respectively. Then $\bar{H}$ and $\bar{K}$ are normal in $M/N$, and $M / N = \bar{H} \bar{K}$. Furthermore, $\bar{H} \cap \bar{K}$ is the image of $T$ and thus $\bar{H} \cap \bar{K} \cong T$.

When $T$ and $S$ are trivial, this is the direct product of $H$ and $K$. When $\phi$ is trivial, we get the central product of $H$ and $K$.

If $G = HK$ and $H$ and $K$ are normal in $G$, then let $\phi: H \rightarrow \operatorname{Aut}(K)$ be the conjugation action, let $S = H \cap K$ and let $\varphi: S \rightarrow S$ be the identity map. Using the construction above (all the assumptions for the action and the isomorphism hold, the third one since $[H, K] \leq S$ by normality) we get a normal product and it will be isomorphic to $G$. Thus all normal products arise in this way.