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 , 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.

It's true for finitely generated nilpotent groups. Let $G$ be such a group with lower central series $G=G_1 > G_2 > \cdots > G_{c+1}=1$, and let $H<G$ with $|G:H|$ finite. . By induction on the class $c$, we can assume that $|G/G_c: H'G_c|$ is finite, so it suffices to prove that $|G_c:H' \cap G_c|$ is finite.

The commutator map induces a surjective homomorphism $G/G' \otimes G_{c-1}/G_c \to G_c$. Since $|G/G':G'H'/G'|$ and $|G_{c-1}/G_c:(H' \cap G_{c-1})G_c/G_c|$ are both finite, the image of the restriction of this map to $G'H'/G' \otimes (H' \cap G_{c-1})G_c/G_c$, which is equal to $H' \cap G_c$, has finite index in $G_c$, and we are done.

It is not rue in general for infinitely generated nilpotent groups. For example, let $G$ be a covering group of an infinitely generated elementary abelian $p$-group, and let $H$ be a subgroup of $G$ of index $p$ that contains $G'$.

This is a old question but I think it is still worthwile to supply a reference to a much more general statement (taken from subsection 2.3.3 of "The Theory of Infinite Soluble Groups" by John C. Lennox and Derek J. S. Robinson).

Let $\theta(x_1,\dots, x_n)$ be a word in the variables $x_1,\dots, x_n$. If $H_1,\dots H_n$ are subgroups of a group $G$, let $\theta(H_1,\dots, H_n)$ be the subgroup generated by all $\theta(h_1,\dots, h_n)$ where $h_i\in H_i$. Then a result of Philip Hall says:

Theorem Let $G$ be a finitely generated nilpotent group with subgroups $H_i\le K_i$, $i=1,\dots, n$ such that each index $|H_i:K_i|=m_i$ is finite. Then for every $n$-variable word $\theta$ the index $|\theta(H_1,\dots, H_n):\theta(K_1,\dots, K_n)|$ is finite and it divides some power of $m_1 m_2 · · · m_n$.

In our case $n=1$ and the word $\theta$ is the commutator. Thus if $|G:H|=m$, then $|G^\prime: H^\prime|=m^r$ for some nonnegative interger $r$ (which could be $0$ as happens e.g. when $G$ is abelian).