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.
 A: 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.
A: This is what I was working on. It's a start, but the end looks pretty messy to me.
Assume we do have a normal product $G=HK$ with $L= H \cap K$.
Now $kh = h k[k,h]$ so to write down the multiplication in $HK$ we need to know $[k,h] \in L$. Now $[k,xy] = [k,y][k,x]^y$ and $[xy,h] = [x,h]^y [y,h]$ offer a slightly more twisted version of bi-linearity.
So given abstract groups $H,K$ and injections $\ell_H:L \hookrightarrow H$ and $\ell_K:L \hookrightarrow K$ with $\ell_H(L) \lhd H$ and $\ell_K(L) \lhd K$ we automatically get an $H$ action on $L$ with $l^h$ defined as the unique solution to $\ell_H(l^h) = \ell_H(l)^h$. Similarly we get a $K$ action on $L$.
Given any function $b:K \times H \to L$ such that $b(k,h_1 h_2) = b(k,h_2) b(k,h_1)^{h_2}$ and $b(k_1 k_2,h) = b(k_1,h)^{k_2} b(k_2,h)$ we define a group $G$ on the set $H/\ell_H(L) \times L \times K/\ell_K(L)$ using the following rule: $$(h_1,l_1,k_1) \cdot (h_2,l_2,k_2) = \left(h_1 \cdot h_2, l_1^{h_2} \cdot b(k_1,h_2)^{k_1^{-1}} \cdot l_2^{k_1^{-1}}, k_1 \cdot k_2\right)
$$
Now handle the section problem on the two quotient sides and check associativity.
