Normal products of groups with maximal nilpotency class

Question

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 $a+1$.

Answer 1

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).

Answer 2

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.