# Finitely generated subgroups with infinite cyclic quotient

Suppose that $G$ is a finitely presented group and $H$ is a finitely generated normal subgroup such that $G/H$ is infinite cyclic. Is it true that $H$ is finitely presented?

• No. F.p. groups $G$ with a f.g. but not f.p. subgroup are known as "incoherent". "Bieri-Stallings example": $G=F_2\times F_2$ and $H$ is the kernel of the map to $\mathbb{Z}$ sending each of the 2+2 free generators of the factors to 1. This construction was generalized by Bieri and further by Bestvina-Brady to higher finiteness properties. – Victor Protsak Jul 27 '10 at 19:44
• Victor, a very slight historical nitpick: the example studied by Stallings is one level up, namely the corresponding subgroup of $F_2\times F_2\times F_2$ (which gives an example of a fp group with infinitely generated $H_3$); Bieri then generalised this construction for an direct product of any number of free groups. Presumably this subgroup of $F_2\times F_2$ was known before Stallings's paper. – HJRW Jul 27 '10 at 20:01
• I know, Henry, that's precisely why I put the quote marks around it. – Victor Protsak Jul 27 '10 at 21:19
• Apologies, Victor. I was addressing your remark "This construction was generalized by Bieri". – HJRW Jul 27 '10 at 22:38

No. Ollivier & Wise's version of the Rips Construction gives, for any finitely presented group $Q$, a finitely presented group $G$ of cohomological dimension 2 and a surjection $G\to Q$ such that the kernel $K$ satisfies:

1. $K$ is finitely generated; and
2. $K$ has Kazhdan's property T, in particular $K$ has at most one end.

Now it follows from Theorem 5.3 of a paper of Bieri that $K$ is only finitely presented if $Q$ is finite.

Note: In my original answer, I only mentioned the unadulterated Rips Construction. Using Ollivier and Wise's version is overkill, but it makes the application of Bieri's theorem cleaner.

I should also mention another, famous and beautiful (though I suppose less general) counterexample. In its simplest cases this example is more elementary.

Given a flag complex $L$, Bestvina & Brady consider the corresponding right-angled Artin group $A_L$ and the kernel $K_L$ of the map $A_L\to\mathbb{Z}$ that sends each generator to $1$. They prove:

1. $K_L$ is finitely generated if and only if $L$ is connected; and
2. $K_L$ is finitely presented if and only if $L$ is simply connected.

So just take $L$ to be your favourite connected, non-simply connected flag complex to construct a counterexample. The square graph with four vertices and four edges is a good choice for $L$, in which case $A_L$ is just the direct product of two copies of the free group on two generators. In this simple case, it's easy to see that $K_L$ is finitely generated; one should be able to prove (though I haven't tried) that $K_L$ is not finitely presented by messing around with some spectral sequences...

• I see that Victor has already mentioned my second example in a comment. – HJRW Jul 27 '10 at 19:57
• The beauty of Bestvina-Brady approach via Morse theory is that there is no need to mess around with spectral sequences: you can just see infinitely many relations! (See Geoghegan's GTM 243 book for a recent exposition.) – Victor Protsak Jul 27 '10 at 21:23
• Indeed - hence the fact that they are able to distinguish finite presentability from the more homological property $FP_2$. – HJRW Jul 27 '10 at 22:39

It worth noticing that the Sigma-invariants (also known as BNS- or BNSR-invariants) provide a strategy to answer such a question.

Given a group $G$ of type $F_n$ and an integer $k \leq n$, the $k$-th Sigma-invariant $\Sigma^k(G)$ is a (complicated) subset of the character sphere $$S(G):= \left( \mathrm{Hom}(G,\mathbb{R}) \backslash \{ 0 \} \right) / \text{positive scaling}$$ of $G$. Notice that the class of every non trivial morphism $G \to \mathbb{Z}$ is an element of $S(G)$.

Theorem: The kernel of $\chi : G \twoheadrightarrow \mathbb{Z}$ is of type $F_k$ if and only if $[\chi], [-\chi] \in \Sigma^k(G)$.

So the case $k=2$ corresponds to $\mathrm{ker}(\chi)$ finitely presented. Unfortunately, these invariants are usually quite difficult to compute. Nevertheless, there exists a useful method, based on Bestvina and Brady's Morse theory, extending their work on right-angled Artin groups (mentioned in Henry's answer).

For instance, all the Sigma-invariants are completely known for right-angled Artin groups and some Thompson-like groups. An application I really like is:

Theorem: Any finitely presented normal subgroup of Thompson's group $F$ is of type $F_{\infty}$.

A few references on the subjet:

• Strebel, Notes on the Sigma-invariants.
• Bux & Gonzales, The Bestvina-Brady construction revisited - Geometric computation of $\Sigma$-invariants for right-angled Artin groups.
• Witzel & Zaremsky, The $\Sigma$-invariants of Thompson's group $F$ via Morse theory.
• Zaremsky, On the $\Sigma$-invariants of generalized Thompson groups and Houghton groups.
• Zaremsky, Symmetric automorphisms of free groups, BNSR-invariants, and finiteness properties.
• Who is the "R" in BNSR? (I know BNS=Bieri, Neumann and Strebel.) – user1729 Apr 20 '18 at 12:54
• It seems that the invariant $\Sigma^1$ was introduced by R. Bieri, W. D. Neumann and R. Strebel (in their paper untitled A geometric invariant of discrete groups, 1987), and next the invariants $\Sigma^2, \Sigma^3, \ldots$ were introduced by R. Bieri and B. Renz (in their paper untitled Valuations on free resolutions and higher geometric invariants of groups, 1988). So I guess that the R in BSNR stands for Renz. – AGenevois Apr 20 '18 at 15:01