Quotient of subgroups by center. Let $H \leq G$. Let $Z_G$ denote the center $[G,G]$ the commutator subgroup. Assume $[G,G] \leq Z_G$ (i.e. nilpotent of class 2). Then $G/Z_G$ is abelian since $Z_G$ contains the commutator subgroup. There is a theorem that states that $H$ must also be nilpotent of class at most 2, hence $[H,H] \leq Z_H$ so $H/Z_H$ is also abelian.
A few questions:
1) Suppose $G/Z_G$ $\simeq$ $\mathbb{Z}^k$ and $H/Z_H$ $\simeq$ $\mathbb{Z}^t$
Is it always the case that $t \leq k$?
2) What if $Z_G$ and $Z_H$ are replaced by $[G,G]$ and $[H,H]$ respectively?
 A: The answer to (1) is "yes."  As Carnahan notes, you have $H\cap Z_G\subseteq Z_H$. Since $H/(Z_G\cap H)\cong HZ_G/Z_G$ is isomorphic to a subgroup of $G/Z_G$, then, $H/Z_H$ is a quotient of $H/(Z_G\cap H)$, hence isomorphic to a quotient of a subgroup of $G/Z_G$, so $\mathrm{rank}(H/Z_H) \leq \mathrm{rank}(G/Z_G)$. 
The answer to (2) is "no". Take $G$ to be the relatively free group of class $2$ and rank $k$ (isomorphic to $F_k/(F_k)_3$, where $F_k$ is the free group of rank $k$ and $(F_k)_3$ is the third term of the lower central series of $F_k$). Then $G^{\rm ab}\cong \mathbb{Z}^k$. Let $H=[G,G]$; then $H$ is free abelian of rank $\binom{k}{2}$. Pick $k\gt 3$ to get that the rank of $H/[H,H]$ is greater than the rang of $G/[G,G]$. If you want to exclude the case where $H$ is abelian itself, just add two of the original generators to the commutator subgroup so the abelianization is of rank $2+\binom{k-1}{2}$. 
For the modified version of (2) (see comments), it is still not true that $\mathrm{rank}(Z_H)\leq \mathrm{rank}(Z_G)$. Take $G$ as above; then $Z_G=[G,G]$ is free abelian of rank $\binom{k}{2}$. Take $H=\langle [G,G], x_1\rangle$, where $x_1$ is one of the free generators of $G$. Then $H$ is free abelian of rank $\binom{k}{2}+1$. 
