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?



Finitely generated subgroups with infinite cyclic quotient
No. [Ollivier & Wise's version](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=ollivier&s5=wise&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) 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](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=bieri%252C%2520r%2a&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=39&mx-pid=390078) that $K$ is only finitely presented if $Q$ is finite.


**Note:** In my original answer, I only mentioned [the unadulterated Rips Construction](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&review_format=html&s4=rips%252C%2520e%2a&s5=&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq&r=18&mx-pid=642423). 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](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=AUCN&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=bestvina&s5=brady&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq) 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...


