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Let $H$ be a nilpotent group of class $a$ and $K$ a nilpotent group of class $b$. If $H$ and $K$ are normal subgroups of a group $G$, then we know that $HK$ is a normal nilpotent subgroup of $G$ and $HK$ has class $\leq a+b$.

Can we always find examples where this upper bound is attained? That is: given $a$ and $b$, can we always find groups $H$, $K$ and $G$ such that:

  • $H$ is nilpotent of class $a$
  • $K$ is nilpotent of class $b$
  • $H$ and $K$ are normal subgroups of $G$, and $HK$ is nilpotent of class $a+b$

I think this is possible when $b = 1$. Let $G = D_{2^{a+2}} = \langle x, y: x^2 = y^{2^{a+1}} = 1, xyx^{-1} = y^{-1} \rangle$ be the dihedral group of order $2^{a+2}$. Let $H = \langle x, y^2 \rangle \cong D_{2^{a+1}}$ and $K = \langle y \rangle$. Then $H$ and $K$ are normal subgroups of $G$, $H$ has class $a$, $K$ has class $1$ and $HK = G$ has class $a+1$.

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    $\begingroup$ If $a=b$ or $a=b+2$, I think a wreath product $D_{2^{(a+b)/2+1}} \wr C_2$ works. $\endgroup$ – Jack Schmidt Aug 19 '13 at 15:29
  • $\begingroup$ @JackSchmidt: Thanks for the bounty. Constructing examples seems a bit tricky.. is there anything like an "external normal product" of groups? Ie. given groups $H$ and $K$, is there a general way to construct all groups $G = HK$, where $H$ and $K$ are normal in $G$? For example, semidirect products are determined by $H$, $K$ and a homomorphism $H \rightarrow \operatorname{Aut}(K)$. Is there something like this for normal products? $\endgroup$ – spin Aug 23 '13 at 11:12
  • $\begingroup$ I don't know if something like that would even be useful. Just trying to figure out what is the best way to understand the structure of a normal product of two groups. $\endgroup$ – spin Aug 23 '13 at 11:22
  • $\begingroup$ Lately I have found normal products fairly surprising. The basic ingredients for the normal product must include an isomorphism between a subgroup of $H$ and a subgroup of $K$, along with homomorphisms from $H$ and $K$ to the automorphism group of that subgroup. When the subgroup is trivial, one gets the direct product. $\endgroup$ – Jack Schmidt Aug 23 '13 at 13:35
  • $\begingroup$ @JackSchmidt: Do you also need a homomorphism $H \rightarrow \operatorname{Aut}(K)$ or $K \rightarrow \operatorname{Aut}(H)$ to completely determine the normal product? $\endgroup$ – Mikko Korhonen Aug 23 '13 at 14:45
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I made some progress with this. I am virtually certain that the answer is yes, and that I could produce an example for any reasonably sized $a$ and $b$ but I am not confident of being able to prove it.

Fix some $n$ and let $G$ be a Sylow 2-subgroup of $S_{2^n}$. So $G$ is an iterated wreath product of cyclic groups of order 2. The (nilpotency) class of $G$ is $2^{n-1}$. It has $n$ generators, which we can take to be $g_1=(1,2)$, $g_2=(1,3)(2,4)$, $g_3=(1,5)(2,6)(3,7)(4,8)$, $\ldots$, $g_n=(1,2^{n-1}+1)(2,2^{n-1}+2)\cdots(2^{n-1},2^n)$.

For $1 \le i \le n$, let $H_i$ be the normal closure of $g_i$ in $G$. Then $H_i$ has class $2^{i-2}$ (except that class of $H_1$ is 1). More generally, if $S$ is any subset of $\{1,2,\ldots,n\}$, then the the class of the normal closure in $G$ of $\{g_i : i \in S\}$ is the sum of the classes of $H_i$ for $i \in S$. I am moderately confident that I could prove the above claims if forced!

So, if the binary expansions of $a$ and $b$ have no 1s in the same positions, then we can solve the problem using these normal closures. For example, if $a=11$, $b=20$, then we take $n=6$, and the subgroups $\langle H_2,H_3,H_5 \rangle$, $\langle H_4,H_6 \rangle$, which have classes 11 and 20, and their product has class 31.

In the general case, I am convinced that we can always solve the problem in a suitable quotient of some such $G$ by a term in its lower central series, and I have had no difficulty doing this for lots of different values of $a,b$, but I am not confident of being able to prove that this is always possible.

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  • $\begingroup$ Thanks! I still haven't had time to verify the ideas of the $a \cup b$ construction, but your answer for that one is just fine (my problem; start of teaching semester here). $\endgroup$ – Jack Schmidt Aug 28 '13 at 17:57
  • $\begingroup$ I more or less understand what is going on now, without having proven anything formally. If you want to talk more about the details some time then that would fine. $\endgroup$ – Derek Holt Aug 28 '13 at 18:52
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Not always. Take $H,K$ abelian, ${\rm gcd}(|H|,|K|)=1$. Then $H\cap K=1$, so $HK$ is abelian also.

(This is true also for nilpotent groups of coprime orders.)

Addendum: This is not complete answer (thank to Jack Schmidt).

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  • $\begingroup$ I think he allows $H$ and $K$ to vary (so I would guess the answer is yes or at least close to yes). $\endgroup$ – Jack Schmidt Aug 19 '13 at 14:57
  • $\begingroup$ @Jack Schmidt: You are right, thank you. $\endgroup$ – Boris Novikov Aug 19 '13 at 15:15
  • $\begingroup$ @JackSchmidt: Right, that's what I meant. I'll edit to make it more clear $\endgroup$ – spin Aug 19 '13 at 15:21

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