A closed $n$-manifold $M$ is called an integral homology sphere if $H_*(M; \mathbb{Z}) \cong H_*(S^n; \mathbb{Z})$. Moreover, we say such an $M$ is non-trivial if $M$ is not homeomorphic to $S^n$.
I am interested in the following question:
Does every non-trivial integral homology sphere admit a finite connected covering space (other than itself)?
The first examples of non-trivial integral homology spheres occur in dimension three. The prime decomposition of such manifolds can only contain aspherical factors. It follows from Ian Agol's solution of the virtual positive first Betti number conjecture that the answer to the above question is yes in dimension three.
One can reformulate the above question in purely group-theoretic terms. Recall that a group $G$ is called superperfect if $H_1(G; \mathbb{Z}) = 0$ and $H_2(G; \mathbb{Z}) = 0$. It follows that the fundamental group of an integral homology sphere is a finitely presented superperfect group. Conversely, every finitely presented superperfect group arises as the fundamental group of an integral homology sphere by a result of Kervaire, see here. Therefore, the above question is equivalent to the following one:
Does every non-trivial finitely presented superperfect group contain a finite index subgroup (other than itself)?
My main interest is the case where the group is also torsion-free, so I'd be happy with an answer which could address this case.
Best Answer
The Higman group is finitely-presented, super-perfect (even acyclic), see here (see also Example 41 here), and contains no proper finite index subgroups. The standard presentation complex of the Higman group is aspherical and, hence, the group is torsion-free (it has cohomological dimension 2).