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?



Quotient of subgroups by center.
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 $\binom{k}{2}+1$. 


