Group generated by finitely many abelian normal subgroups is nilpotent of class at most equal to the number of subgroups

Statement

For a group

Suppose $G$ is a group and $H_1,H_2,\dots,H_n$ are abelian normal subgroups of $G$. Suppose $G$ is the join of subgroups $\langle H_1, H_2, \dots, H_n \rangle$. (Since they are all normal, this is equivalent to the product of subgroups $H_1H_2 \dots H_n$).

Then, $G$ is a nilpotent group of nilpotency class at most $n$.

For a subgroup

Suppose $G$ is a group and $H_1,H_2,\dots,H_n$ are abelian normal subgroups of $G$. Let $H$ be the subgroup of $G$ defined as the join of subgroups $\langle H_1, H_2, \dots, H_n \rangle$. (Since they are all normal, this is equivalent to the product of subgroups $H_1H_2 \dots H_n$).

Then, $H$ is a nilpotent group of nilpotency class at most $n$. In particular, it is a nilpotent normal subgroup of $G$.

Case of two subgroups

In the case that $n = 2$, we get a group of nilpotency class two (in the group formulation) or a class two normal subgroup (in the subgroup formulation).