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

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).