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Consider $M^3_{pq}$, a torus bundle over $S^1$ with fundamental group the HNN extension generated by three generators $x,y,z$ satisfying the relations $\quad [x,y], \quad x^z = x^p \quad$ and $y^z = y^q$. By the exact homotopy sequence, $M^3_{pq}$ is aspherical.

Suppose there is a closed manifold $E^{n+3}$ that is the total space of a $T^n$ bundle over infinitely many $M^3_{pq}$ satisfying $p\neq q$. As $M_{pq}^3$ is aspherical, so is $E$. Hence, this can only be the case if $\pi_1 M_{pq}$ is a quotient of $\pi_1 E$ by the fundamental group $\pi_1T^n$; note that $\pi_1E$ and $n$ may not depend on $(p,q)$. It seems unlikely that one can obtain an infinite amount of mutually non-isomorphic, non-abelian groups as quotients from a finitely generated group by only identifying elements that commute. However, I have failed to prove this.

This begs the general question: How many exact sequences with non-abelian groups $H$
$$ 0 \rightarrow Z^n \rightarrow G \rightarrow H \rightarrow e $$ can a finitely presented group $G$ admit for fixed $n$? What are possible obstructions, aside from the kernel of $G\rightarrow H$ having to be abelian?

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You have a very specific requirements on the kernel which are not in your title. This is a very different question - of course, a free group has a huge number of quotients; I suggest editing your title. –  HJRW Jul 25 '12 at 10:17
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Also, if I'm not mistaken, in your example $E^{n+3}$ is aspherical. This should be important. My guess is that there are finitely generated $G$'s with infinitely many distinct quotients by fg free abelian subgroups - indeed, as a general rule (in fact I think it's a meta-theorem of Gromov), when someone says 'it seems unlikely that there is a finitely generated group with property P', they are almost invariably wrong. But it may well be that there are no examples of 'type F' (ie fundamental groups of compact aspherical complexes). –  HJRW Jul 25 '12 at 10:24
    
Thank you, Henry. I implicitly used that $E$ is aspherical to obtain the short exact sequence in the third paragraph, but I have pointed it out more clearly now. –  Malte Jul 25 '12 at 10:41
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@Malte: Your description of fundamental groups of 3-manifolds fibering over circle with torus fiber is way too limited, it works only for $|p|=|q|=1$. The right thing to do is to take some matrix $A$ in $GL(2,Z)$ and use it to define semidirect product of $Z^2$ and $Z$. –  Misha Jul 25 '12 at 17:26
    
Hmm... could you elaborate on that, Misha? Just to clarify: I am considering surface bundles that are obtained by taking the mapping torus of some diffeomorphism of the surface. In this case, the surface is a torus and the diffeomorphism is the composition of two Dehn twists. In general, I would get an element of $GL(2,Z)$. But where is the problem with having $p,q > 1$? –  Malte Jul 26 '12 at 21:31

1 Answer 1

Take a finitely presented non-Abelian perfect group $H$ (i.e. $H=[H,H]$). Take $G=H\times \mathbb{Z}$. Then you have infinitely (countably) many short exact sequences

$$1\to \mathbb{Z}\to G\to H\times \mathbb{Z}/m\mathbb{Z}\to 1.$$

On the other hand, for every finitely generated group $G$ there are at most countably many short exact sequences of the form

$$ 1\to \mathbb{Z}^n\to G\to H\to 1.$$

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I interpreted the question as asking for non-isomorphic quotients. –  HJRW Jul 25 '12 at 12:19
    
$H\times \mathbb{Z}/m\mathbb{Z}$ are all non-isomorphic because $H$ has trivial abelianization. –  Mark Sapir Jul 25 '12 at 12:21
    
Oh right, yes, apologies. I was fixing $m$ for some reason. That's quite trivial! –  HJRW Jul 25 '12 at 12:32
    
Perhaps it's more interesting if you also require the quotient to be of type F? –  HJRW Jul 25 '12 at 12:34
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One can get a central extension of $H$ with center $C=\mathbb{Z^2}$. Then choose maximal cyclic subgroups $N$ in $C$. The groups $H/N$ are most probably pairwise non-isomorphic (one needs to prove it!) and will be of type $F$ for a suitable $H$. –  Mark Sapir Jul 25 '12 at 12:41

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