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?

5$\begingroup$ No. F.p. groups $G$ with a f.g. but not f.p. subgroup are known as "incoherent". "BieriStallings 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 BestvinaBrady to higher finiteness properties. $\endgroup$– Victor ProtsakCommented Jul 27, 2010 at 19:44

2$\begingroup$ 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. $\endgroup$– HJRWCommented Jul 27, 2010 at 20:01

$\begingroup$ I know, Henry, that's precisely why I put the quote marks around it. $\endgroup$– Victor ProtsakCommented Jul 27, 2010 at 21:19

$\begingroup$ Apologies, Victor. I was addressing your remark "This construction was generalized by Bieri". $\endgroup$– HJRWCommented Jul 27, 2010 at 22:38
2 Answers
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:
 $K$ is finitely generated; and
 $K$ 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 rightangled 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:
 $K_L$ is finitely generated if and only if $L$ is connected; and
 $K_L$ is finitely presented if and only if $L$ is simply connected.
So just take $L$ to be your favourite connected, nonsimply 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...

$\begingroup$ I see that Victor has already mentioned my second example in a comment. $\endgroup$– HJRWCommented Jul 27, 2010 at 19:57

$\begingroup$ The beauty of BestvinaBrady 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.) $\endgroup$ Commented Jul 27, 2010 at 21:23

$\begingroup$ Indeed  hence the fact that they are able to distinguish finite presentability from the more homological property $FP_2$. $\endgroup$– HJRWCommented Jul 27, 2010 at 22:39
It worth noticing that the Sigmainvariants (also known as BNS or BNSRinvariants) 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 Sigmainvariant $\Sigma^k(G)$ 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 rightangled Artin groups (mentioned in Henry's answer).
For instance, all the Sigmainvariants are completely known for rightangled Artin groups and some Thompsonlike 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 Sigmainvariants.
 Bux & Gonzales, The BestvinaBrady construction revisited  Geometric computation of $\Sigma$invariants for rightangled 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, BNSRinvariants, and finiteness properties.

$\begingroup$ Who is the "R" in BNSR? (I know BNS=Bieri, Neumann and Strebel.) $\endgroup$– user1729Commented Apr 20, 2018 at 12:54

1$\begingroup$ 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. $\endgroup$ Commented Apr 20, 2018 at 15:01