Distinguishing 3-manifolds by homologies of covers In a blog post on ldtopology, a recent arxiv posting of Lins-Lins is discussed. The main argument of that paper is difficult to algorithmically distinguish two 3-manifolds and to that end the authors provide the challenge of showing two specific manifolds are not homeomorphic. 
Nathan Dunfield implemented a computation of the index 6 subgroups of the fundamental groups of the two manifolds. In the comments of the aforementioned blog post, he showed this computation distinguishes these two sets of subgroups by abelianizing the subgroups. In particular, this indicates the two manifolds are not homeomorphic.   
My first question is, "Can this trick always be used?", or more specifically: 
Let $M_1$ and $M_2$ be non-homeomorphic (finite volume) hyperbolic 3-manifolds 
and let $S_{i,k}$ be the set of degree $k$ covers of $M_i$. Finally, let $H_{i,k}$ be the sets of integral first homology groups for the manifolds in $S_{i,k}$. 
Is there a known pair of non-homeomorphic (finite volume) hyperbolic 3-manifolds $M_1$ and $M_2$ such that for all $n$, the sets of $H_{1,n}$ and $H_{2,n}$ are identical, i.e. there is a bijection between them?
As a second question:
In the absence of a pair of hyperbolic examples, is there a known pair of 3-manifolds with non-isomorphic fundamental groups with this property?
 A: To atone for my comment above, I'm going to write out the Sol
example. A Sol lattice takes the form $\mathbb Z^2\rtimes \mathbb Z$,
where the action is hyperbolic. The rank of the abelianization is
always 1, though the amount of torsion varies. The profinite
completion is $\hat{\mathbb Z}^2\rtimes \hat{\mathbb Z}$.
We can consider the group $\mathbb Z^2$ together with the action
as a module over $\mathbb Z[t,t^{-1}]$. One invariant is the
characteristic polynomial $P(t)$. Fixing this, we can consider the
group as a module over $\mathbb Z[t,t^{-1}]/P(t)$. For hyperbolic
actions, this ring is an order in a real quadratic number field.
Moreover, every real quadratic number field has a unit, and thus
has an order of this form. We can give $\mathbb Z^2$ the structure
of a rank 1 projective module over this ring; and if the ring has
nontrivial class group, there are non-isomorphic ways to do this.
So this gives rise to two Sol lattices that are not isomorphic,
but are pretty similar, having the same characteristic polynomial.
The obstruction to their isomorphism is global, so when we take
profinite completions and are dealing with modules over local
rings, where all projectives are free, the groups become
isomorphic. Thus their finite index subgroups are in bijection
preserving the quotients of corresponding groups; and preserving
the abelianizations. The action on the finite quotient group on
the homology of the subgroup is trivial on the torsion-free part,
because it is always just $\mathbb Z$, and matches the
corresponding action in the profinite group, so does not
distinguish the two discrete groups.
Similarly, it seems to me that there are pretty good candidates for
hyperbolic three-manifolds with isomorphic profinite completions.
Then these would have a bijection between finite index subgroups
preserving the abelianizations and quotient groups, answering the
original question. However, the action of the finite quotients on
the abelianizations might distinguish them, although that is a
subtle invariant. The idea is to take a imaginary quadratic number
field with appropriate class group and form nonisomorphic
rank 2 projective modules. For the right choice of projective
modules, the automorphism groups are not isomorphic as group
schemes and, I think, as discrete groups. Over the number field
all projective modules are free, so the groups of rational
automorphisms are equal, and thus the lattices are commensurable.
Over the ring of integers in a local field, all projective modules
are free, so the congruence completions are isomorphic. The
congruence subgroup property does not apply to these groups, but
the congruence kernel is attached to the rational group, so pretty
much the same for the two groups. Compatibility between the
congruence completion and the congruence kernel seems to me to be
the big problem in checking this candidate. Or maybe you could use
two $(3,1)$ quadratic forms in the same genus.
