# Hyperbolic groups with infinitely generated commutator subgroups

I am trying to get a sense of how often the commutator subgroup $[G,G]$ of a (Gromov) hyperbolic group $G$ is infinitely generated.

Remarks:

1. $[G,G]$ is infinitely generated if is $G$ is free noncyclic, and of course $[G,G]$ is finitely generated when $G$ has finite abelianization.

2. There are examples when a hyperbolic group has an finitely generated, normal (infinite) subgroup (of infinite index), obtained via various versions of the Rips constructions, or Morse theory considerations of Bestvina-Brady and Brady, see here .

More generally, how often is the kernel of a surjection $G\to\mathbb Z$ infinitely generated (or infinitely presented)? Here is a specific:

Question. Suppose the abelianization of $G$ has rank $>1$, so that there are infinitely many surjections $G\to \mathbb Z$. Is it true that there are infinitely many surjections $G\to\mathbb Z$ whose kernel is not finitely generated?

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It is easy to see that a free group has infinitely generated normal subgroup with quotient ${\mathbb Z}$. There are many lattices in $SO(n,1)$ which map onto a free group (I think there are results by Lubotzky) which tell us that the lattices which are unit groups of quadratic forms over totally real number fields, have finite index subgroups with this property. Hence they also will have inf generated normal subgroups with ${\mathbb Z}$ quotient. –  Venkataramana Mar 6 '13 at 17:11
It is an old result (from "On one-relator groups having elements of finite order" by J. Fischer, A. Karrass and D. Solitar (1972)), but if $G=\langle X; R^n\rangle$, $n>1$, then the derived subgroup is an infinite free product of cyclic groups of order $n$. Also, $G$ is hyperbolic (and has one end). –  user6503 Mar 6 '13 at 17:16
@Igor: I do not know any other general ways for showing that normal subgroups are infinitely generated, since one has to avoid all of the Rips-type examples. But Aakumadula's idea helps to construct more groups: for any group mapping onto the free group $F_2$ of rank 2, its derived subgroup maps onto $[F_2,F_2]$ and so it's not f.g. In particular this is satisfied by all non-abelian subgroups of RAAGs (by Wise and Agol there are plenty of hyperbolic groups among them). –  Ashot Minasyan Mar 6 '13 at 17:44
@Igor: for your final question, your version of Rips's construction with Denis will give a counterexample. Let $1\to N\to G \to \mathbb{Z}^2\to 1$ be given, with $G$ hyperbolic, $N$ f.g. with property (T) (or simply with finite abelianization). Then the kernel of any homomorphism from $G$ to $\mathbb Z$ contains $N$, and so it is f.g. –  Ashot Minasyan Mar 6 '13 at 18:25
If a group has a nontrivial first $\ell^2$ Betti number then the only normal subgroup that are finitely generated subgroups are finite or finite index (Gaboriau). –  YCor Mar 6 '13 at 20:20

No, consider a hyperbolic 3-manifold $M$ with $b_1(M)=2$, and all faces of the Thurston norm fibered. Then there are only finitely many surjections to $\mathbb{Z}$ with infinitely generated kernel, corresponding to the (projective classes of the) vertices of the Thurston norm ball.
The question to determine, given an f.g. group $G$, which homomorphisms to abelian groups (esp. cyclic groups) have a f.g. kernel, is addressed in detail in the paper: Bieri, W. Neumann, Strebel, A geometric invariant of discrete groups, Invent.math. 90, 451477 (1987). An older reference is Neumann, W.D.: Normal subgroups with infinite cyclic quotient. Math. Sci. 4, 143-148 (1979). This is not specific to the setting of hyperbolic groups (and in particular does not supersede Agol's answer).