 Solution

Suppose inductively that $G=H{\gamma }_i(G)$. In the quotient $G/{\gamma }_{i+1}(G)$, the image of ${\gamma }_i(G)$ is central. Hence this quotient is generated by the image of $H$ together with a central subgroup, so its commutator subgroup lies in the image of $H$. Pulling this back givesThe original assumption now gives $G=H{\gamma }_2(G)\le H{\gamma }_{i+1}(G)\le G$. Starting with $i=2$, induction proves $G=H{\gamma }_i(G)$ for every $i\ge 2$. These products are subgroups because each ${\gamma }_i(G)$ is normal; no normality of $H$ is needed.

If $G$ is finitely generated, every quotient, including its abelianization $G/{\gamma }_2(G)$, is finitely generated. Conversely, lift a finite generating set of that quotient to $g_1,\dots ,g_t\in G$, and set $H=⟨g_1,\dots ,g_t⟩$. Then $G=H{\gamma }_2(G)$. For a nilpotent group, ${\gamma }_{c+1}(G)=1$ for some $c$, and the established identity gives $G=H$. Thus