I know every subgroup of a finitely generated nilpotent group is also finitely generated. Can we bound the number of generators of that subgroup?
In other words, Let $G$ be a nilpotent group of nilpotency class $k$ and generated by $r$ elements. Let $H$ be an arbitrary subgroup of $G$. Can we find a bound on the number of generators of $H$ in terms of $k$ and $r$?
Any reference on this question?
Yes. Every such group $G$ is a quotient of the free class $k$ nilpotent group $F_r/\gamma_{k+1}(F_r)$, where $F_r$ is an $r$-generator free group, and $\gamma_i = \gamma_i(F_r)$ is its lower central series.
Each $\gamma_i/\gamma_{i+1}$ is a free abelian group of rank $r_i$, say. Since, for $i>1$, $\gamma_i/\gamma_{i+1}$ can be generated by commutators of generators of $\gamma_1/\gamma_2$ and $\gamma_{i-1}/\gamma_i$, it is easy to see that $r_i \le r^i$. In fact there is an exact formula for $r_i$ due to Witt, which you can find in Chapter 11 of Marshall Hall's book on Group Theory.
It is clear that any subgroup of $G$ can be generated by at most $\sum_{i=1}^kr_i$ elements, so this provides an upper bound. Putting $j= \lfloor k/2 \rfloor$, the group $\gamma_j/\gamma_{k+1}$ is free abelian of rank $\sum_{i=j}^k$, and that might be the largest possible number of generators of a subgroup.