= 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 gives
$$
\gamma_2(G)\le H\gamma_{i+1}(G).
$$
The 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 $i\ge2$. 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 $g_1,\ldots,g_t\in G$, and set $H=\langle g_1,\ldots,g_t\rangle$. 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
$$
\boxed{G\text{ finitely generated}\iff G/\gamma_2(G)\text{ finitely generated}.}
$$
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