# Nilpotency class of normal product of dihedral 2-groups

Pessimistically paraphrasing Polya: “if Jack cannot answer a question, there is an easier question Jack also cannot answer.” Hence I ask:

Given positive integers $a,b$ describe the set $$N_{a,b} = \left\{ n(G) : G=HK; H,K \lhd G; |H| = 2^{a+1}; |K|=2^{b+1}; H,K \text{ are dihedral }\right\}$$ where $n(G)$ is nilpotency class of $G$.

In other words, exactly what are the possible nilpotency classes of normal products of dihedral groups?

I believe this is a special case of the issues raised in Expression for normal product similar to semidirect product and Normal products of groups with maximal nilpotency class.

I know that $\max(a,b) \in N_{a,b}$ and $N_{a,b} \subseteq \{ i : \max(a,b) \leq i \leq a+b \}$, but not much else.

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$N_{a,b} = \{\max(a,b)\}$ if $\{a,b\} \in \{ \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\} \}$ and $N_{a,a} = \{a,a+1\}$ if $a \leq 3$. I conjecture the first holds for all $a \neq b$, and the second for all $a=b$. –  Jack Schmidt Aug 24 '13 at 3:19

What Jack conjectured in the comments seems to be correct.

Let $H$ be dihedral of order $2^{a+1}$ and let $K$ be dihedral of order $2^{b+1}$. Suppose that $H$ and $K$ are normal in some larger group $G = HK$.

Claim: If $a \neq b$, then $G$ is nilpotent of class $\operatorname{max}(a,b)$. If $a = b$, then $G$ is nilpotent of class $a$ or $a + 1$.

(Note that for example, $D_{2^{a+2}}$ contains two distinct normal subgroups isomorphic to $D_{2^{a+1}}$, so in that case $a+1$ is attained.)

Proof: If $H \cap K = H$ or $H \cap K = K$, then $G = H$ or $G = K$ and thus $G$ is of class $\operatorname{max}(a,b)$. So let's assume that $H \cap K$ is a proper normal subgroup of both $H$ and $K$. If $H \cap K$ is dihedral, then it is of index $2$ in both $H$ and $K$. In this case $a = b$, so $HK$ has order $2^{a+2}$ and thus class $\leq a+1$.

If $H \cap K$ is not dihedral, then it must be cyclic. If $H \cap K$ is of index $2$ in both $H$ and $K$, again $a = b$ and the previous argument applies. So we assume that this is not that case, and next prove that $\gamma_i(HK) = \gamma_i(H)\gamma_i(K)$ for all $i$, which of course implies that $HK$ has class $\operatorname{max}(a,b)$ (here $\gamma_i$ denotes the lower central series). This is true when $i = 1$. Recall the identity

$$[AB, C] = [A,C][B,C]$$

when $A, B$ and $C$ are normal subgroups of $G$. Now if $\gamma_i(HK) = \gamma_i(H)\gamma_i(K)$, then applying the identity repeatedly shows that

\begin{align*} \gamma_{i+1}(HK) &= [\gamma_i(H)\gamma_i(K), HK] \\ &= [\gamma_{i}(H), H][\gamma_i(H), K][\gamma_i(K), H][\gamma_i(K), K] \\ &= \gamma_{i+1}(H)[\gamma_i(H), K][\gamma_i(K), H] \gamma_{i+1}(K) \\ \end{align*}

So it suffices to prove that for all $i$, the subgroups $[\gamma_i(H), K]$ and $[\gamma_i(K), H]$ are contained in $\gamma_{i+1}(H)$ or $\gamma_{i+1}(K)$.

Note that $\gamma_1(H) = H$, and when $2 \leq i \leq a+1$, the term $\gamma_i(H)$ is the unique normal cyclic subgroup of order $2^{a-i+1}$ in $H$. So any subgroup of $\gamma_2(H)$ is a term of the lower central series of $H$. The only cyclic normal subgroup which is not a term of the lower central series of $H$ is the cyclic subgroup of index $2$. The same statements hold for $K$ and its lower central series.

Since we assumed that $H \cap K$ is a cyclic normal subgroup, not of index $2$ in both $H$ and $K$, it follows that $[\gamma_1(H), K] = [\gamma_1(K), H] = [H, K]$ is contained in $\gamma_2(H)$ or $\gamma_2(K)$. Next we show that $[\gamma_i(H), K] \leq \gamma_{i+1}(H)$ and $[\gamma_i(K), H] \leq \gamma_{i+1}(K)$ for all $i \geq 2$.

First of all, $[\gamma_i(H), K] \leq \gamma_i(H)$ by normality. If $[\gamma_i(H), K]$ is a proper subgroup of $\gamma_i(H)$, then $[\gamma_i(H), K]$ is contained in $\gamma_{i+1}(H)$ because $i \geq 2$. Otherwise $[\gamma_i(H), K] = \gamma_i(H)$. In this case $\gamma_i(H) \leq K$, so

$$\gamma_i(H) = [\gamma_i(H), K] \leq [K, K] = \gamma_2(K).$$

This implies that $\gamma_i(H) = \gamma_t(K)$ for some $t \geq 2$, since every subgroup of $\gamma_2(K)$ is a term of the lower central series of $K$. Now

$$\gamma_t(K) = [\gamma_i(H), K] = [\gamma_t(K), K] = \gamma_{t+1}(K)$$

so $\gamma_t(K) = 1$, and thus $[\gamma_i(H), K] = 1$.

With the same method we can show that $[\gamma_i(K), H] \leq \gamma_{i+1}(K)$ when $i \geq 2$.

This is all pretty specific to the dihedral groups, but for different normal products too it might be useful to just go with the straightforward approach and to look at the lower central series of $HK$. The idea is that for all $i$, the term $\gamma_i(HK)$ is a product of $2^i$ commutator subgroups of the form $[L_1, L_2, \ldots, L_i]$, where for all $j$, each $L_j$ is either $H$ or $K$. This is used in the usual (?) proof of the fact that $n(HK) \leq a+b$ when $n(H) = a$ and $n(K) = b$. If $H$ appears $n$ times in a commutator subgroup $[L_1, L_2, \ldots, L_i]$, then it must be contained in $\gamma_n(H)$ (similarly for $K$). Thus if $i \geq a+b+1$, all these iterated commutators must vanish, since either $H$ will appear $\geq a+1$ times or $K$ will appear $\geq b+1$ times in each iterated commutator.

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Cool, I think this technique has a lot of potential. If $H \cap K$ is small, then those iterated commutators are all small except for the ones lying solely in $H$ or $K$ (and then the nilpotency class of $G$ is the same as that of $H$ or $K$). The only time one gets a big bump is when $H \cap K$ is large, but then the bump itself is controlled by the composition length of $H/(H \cap K)$. –  Jack Schmidt Aug 24 '13 at 20:20
Okay, with some modifications the same idea seems to work in the general case. –  Mikko Korhonen Aug 25 '13 at 13:20