Inductive proof showing that nilpotency is preserved by product, question on exact hypothesis

Question

If $H$ and $K$ are normal nilpotent subgroups of $G$, then so is $HK$.

Proof: We know that $HK \unlhd G$ and we proceed by induction on $|G|$ to show that it is nilpotent. If $HK < G$, the assertion follows by induction, so we may assume $G = HK$.

[some computations to show that $Z(G) \ne 1$]

But then $G / Z(G)$ is nilpotent by induction and we conclude at once that $G$ is nilpotent. $\square$

My question is on how exactly the induction works here, I am a little bit unsure about that. I do not think that $G$ in general is nilpotent just because it contains two normal nilpotent subgroups. So to spell out the induction statement with the induction parameter involved I think it should be

If $H, K$ are normal nilpotent subgroups of $G$, with $|G| = n$, then so is $HK$.

For $n = 1$ this is clear. Now if $n > 1$ and $HK < G$, then we apply the induction hypothesis on $U := HK$, as$|U| < n$ and $H, K$ are also normal and nilpotent in $U$ we have that $U = HK$ is nilpotent. In the other case, as $Z(G) \ne 1$ we have $|G / Z(G)| < n$, and the two subgroups $HZ(G) / Z(G)$ and $KZ(G) / Z(G)$ are nilpotent and normal in $G/Z(G)$, hence by induction hypothesis $HZ(G) / Z(G) \cdot KZ(G) / Z(G) = HKZ(G) / Z(G) = G/Z(G)$ is nilpotent, which gives nilpotency of $G$.

I am just unsure if I am overcomplicating things and want to know if my reasoning is correct?

Answer 1

In finite group theory, usually induction means that $G$ is the smallest group that does not satisfy the property. More precisely, if $G_1$ is a group and $|G_1| < |G|$ then $G_1$ satisfies this property and $G$ does not. In your question, assume that $G_1 = HK$. If $|G_1| < |G|$ then by induction $G_1$ satisfies the property. Therefore $HK$ is nilpotent and we are done. So we can assume that $G = G_1 = HK$. So as you mentioned, it does not mean that $G$ is necessarily nilpotent.

Answer 2

For some reason I find it irritating when people assume finiteness unnecessarily, even in books on finite groups. So here is a general proof, which is not exactly difficult.

From the commutator identities, it is not hard to prove that, for normal subgroups $A,B,C$ of a group $G$, $[AB,C] = [A,C][B,C]$ and $[A,BC]=[A,B][A,C]$.

Using this,and letting $\gamma_i(G)$ be the lower central series of $G$ (i.e. $\gamma_1(G)=G$, $\gamma_{n+1}(G) = [G,\gamma_n(G)]$), it is straightforward to prove by induction on $k$ that, if $A,B \unlhd G$, then $$\gamma_k(AB) \le \prod_{i+j=k} (\gamma_i(A) \cap \gamma_j(B))$$ (where we include $i=0$ or $j=0$ in the product, and define $\gamma_0(A)=\gamma_0(B)=G$).

So, if $A$ and $B$ are nilpotent of class $m$ and $n$ respectively , then $\gamma_{m+1}(A)=\gamma_{n+1}(N)=1$, so $\gamma_{m+n+1}(AB)=1$, and $AB$ is nilpotent of class at most $m+n$.