# Index of derived subgroup in derived group

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?

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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.

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thank you very much for your detailed answer. –  randl Sep 20 '12 at 13:48
If I am not mistaken, for an infinite nilpotent group $G$ the derived group $G^\prime$ must have infinite index. So the general statement in your third sentence does not apply. How do things look if $G$ is nilpotent? –  randl Sep 20 '12 at 13:51
For nilpotent groups $|G':H'|$ should be finite, I think. –  Mark Sapir Sep 20 '12 at 14:11
I think so too, but cannot find nor develop a proof. –  randl Sep 20 '12 at 14:28
I included the proof in my answer. –  Mark Sapir Sep 20 '12 at 14:41

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 $G'$.

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Thank you for this answer Derek. –  randl Sep 20 '12 at 14:49
What does the $\otimes$ in your answer denote? –  randl Sep 20 '12 at 16:45
That's the tensor product of abelian groups. Think of them as additive groups, then you can regard them as modules over the integers and take their tensor product. If commutators are in the centre of the group, then the commutator map is bilinear, and so you get the induced map from the tensor product. –  Derek Holt Sep 20 '12 at 19:46
Excellent, I had not come across that idea up to this point. thanks. –  randl Sep 20 '12 at 22:24