For an integer homology sphere $M$, one can show that non-trivial connected finite-sheeted covers of $M$ must have at least $5$ sheets. This is because, for a given $n$-sheeted cover $M'$, one can make $\pi_1(M)$ act on the set of cosets $\pi_1(M)/p_*\pi_1(M')$ to get a homomorphism to the symmetric group $S_n$. If there are $4$ or fewer cosets (i.e., sheets), then you can use the solvability of $S_n$ along with $\operatorname{Ab}(\pi_1(M))=H_1(M)=0$ to deduce $n=1$. The Poincaré homology $3$-sphere has index-$5$ subgroups, so this bound is tight.
Assume $M$ is a homology $3$-sphere coming from $1/q$ Dehn surgery on a knot in $S^3$, and let's assume $M$ is not $S^3$.
- How many sheets can the first non-trivial connected finite-sheeted cover have?
- Is there something efficient to compute that can give an upper bound for this?
Feel free to assume the covers are regular if it simplifies things.
(In other words and more generally: given a nontrivial perfect $3$-manifold group $G$, can one compute an $n$ such that there exists a proper subgroup $H$ with $[G:H]\leq n$? Residual finiteness implies that there is an $n$. While subgroup enumeration algorithms will eventually yield such a subgroup, I am essentially interested in a time bound. )