A couple of months ago I was in Oberwolfach, where Tomer Schlank told me about the following conjecture:
Conjecture. (Boston-Markin) Let \(G\) be a finite group. Then there exists a Galois extension of \(\mathbb{Q}\) with Galois group \(G\) ramified over exactly \(d(G)\) primes, where \(d(G)\) is the minimal number of generators of \(G^{\text{ab}}\). (Here we violate convention and say that \(d(\{1\})=1\).)
It's not hard to see that \(d(G)\) is a lower bound for the minimal number of primes over which a \(G\)-extension ramifies, but I don't see much evidence for the conjecture (except for some results in the case of e.g. solvable groups). Here is a topological analogue (translating the conjecture through the usual primes-vs.-knots dictionary).
Question. Let \(G\) be a finite group and \(M\) a \(3\)-manifold. Does there exist a link \(L\) in \(M\) with \(d(G)\) components, and a \(G\)-cover of \(S^3\) branched only over \(L\)?
The case most analogous to the original conjecture is \(M=S^3\) (or in general, the case where \(M\) is simply connected), where one can see that \(d(G)\) is a lower bound for the number of components of a link with the desired properties. We weren't able to make much progress with \(S^3\), but Tomer, Brian Lawrence, and I were able to show that the answer is yes if \(G\) is a finite simple group and \(M\) is \(S^2\times S^1\), or indeed any surface bundle over a circle! (Of course such manifolds have infinite fundamental group, so the problem is strictly easier in this case.) If I have time, I'll sketch the argument in another post. Please let me know if you have any thoughts about this question!