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.

  • $\begingroup$ I should also mention that I am influenced by the construction of central products. In central products the bracket map is trivial. $\endgroup$ – Jack Schmidt Aug 23 '13 at 15:51

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.

  • $\begingroup$ I verified all the claims, so in a sense this works. I'm not sure how to use it to answer even the very reduced math.stackexchange.com/questions/474436/… - the compatibility conditions still look very complicated to me. I guess in this setup, you choose $\phi$ first, and then $\varphi$. $\endgroup$ – Jack Schmidt Aug 24 '13 at 14:44
  • $\begingroup$ Yeah, I don't know if this is of any practical use either. It shows that normal products are determined by certain basic ingredients, and that there is some way to construct all possible normal products of two groups. I dunno if there is a different construction that is better, for this one the assumptions are all necessary. $\endgroup$ – Mikko Korhonen Aug 24 '13 at 20:09

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.


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