12
$\begingroup$

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?

$\endgroup$
4
  • 5
    $\begingroup$ 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. $\endgroup$ 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. $\endgroup$
    – HJRW
    Jul 27, 2010 at 20:01
  • $\begingroup$ I know, Henry, that's precisely why I put the quote marks around it. $\endgroup$ Jul 27, 2010 at 21:19
  • $\begingroup$ Apologies, Victor. I was addressing your remark "This construction was generalized by Bieri". $\endgroup$
    – HJRW
    Jul 27, 2010 at 22:38

2 Answers 2

15
$\begingroup$

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...

$\endgroup$
3
  • $\begingroup$ I see that Victor has already mentioned my second example in a comment. $\endgroup$
    – HJRW
    Jul 27, 2010 at 19:57
  • $\begingroup$ 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.) $\endgroup$ Jul 27, 2010 at 21:23
  • $\begingroup$ Indeed - hence the fact that they are able to distinguish finite presentability from the more homological property $FP_2$. $\endgroup$
    – HJRW
    Jul 27, 2010 at 22:39
5
$\begingroup$

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.
$\endgroup$
2
  • $\begingroup$ Who is the "R" in BNSR? (I know BNS=Bieri, Neumann and Strebel.) $\endgroup$
    – user1729
    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. $\endgroup$
    – AGenevois
    Apr 20, 2018 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.