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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?

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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
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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

1 Answer 1

up vote 11 down vote accepted

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...

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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

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