Let $G$ be a group with finite index subgroup $H$. Let $G^\prime = [G,G]$ denote the derived subgroup of $G$.


Is it true that $|G:H|<\infty$ implies that $|G^\prime: H^\prime|<\infty$.


If this is not true in general, is it true for a large class of groups? say, finitely generated.


Thanks to Mark Sapir for providing a simple counter-example to the general statement below.


What about if $G$ is nilpotent for example?



Index of derived subgroup in derived group
Consider the infinite dihedral group $G=C\_2\*C\_2=\langle a,b \mid a^2=b^2=1\rangle$ where $C\_2$ is cyclic of order 2. It has a finite index (2) infinite cyclic group $H=\langle ab \rangle$ which is also (almost) the derived subgroup of $G$ ($G'=\langle (ab)^2\rangle$). Then $H'$ (trivial group) is of infinite index in $G'$. In general if $G$ has finite abelianization $G/G'$ and $H$ has infinite abelianization, then $|G':H'|$ is infinite. For more non-trivial examples, let $G$ be the fundamental group of a closed hyperbolic $3$-manifold. By Dunfield, Nathan M.; Thurston, Dylan P. A random tunnel number one 3-manifold does not fiber over the circle. Geom. Topol. 10 (2006), 2431–2499 then $G/G'$ is very rarely finite. On the other hand, by Agol's theorem all of them have finite index subgroups $H$ with $H/H'$ infinite. On the other hand, in the "higher rank" situation the answer is different. By the 
 [Margulis normal subgroup theorem](http://www.mathematik.uni-regensburg.de/loeh/seminars/normal_subgroup_thm.pdf) , normal subgroups of lattices $\Gamma$ in higher rank connected semi-simple Lie groups with finite center have finite indices or are finite, so $|\Gamma':H'|$ is finite whenever $H$ is of finite index in $\Gamma$. 


Here is the proof that $|G':H'|$ is finite for nilpotent (finitely generated) groups.Induction on the Hirsh length. For cyclic groups everything is fine. Let $C$ be an infinite cyclic group in the center of $G$ intersected with $H'$ (every normal subgroup of a nilpotent group intersects with the center non-trivially and we can assume that $H$ is torsion-free. Then $G/C$ has smaller Hirsh length than $G$, so for that group everything is fine. But the index $|(G/C)':(H/C)'|$ is the same as $|G':H'|$. The only missing case is when $H$ is Abelian. But in that case the center is of finite index in $G$, the derived subgroup is finite, and we are done also. 


