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?
 A: 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.

A: 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 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:


*

*$K_L$ 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, 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...
