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?

arefinitely 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