Fix a finite set $S$ of places of $\mathbb Q$. Let $G\_{\mathbb Q,S}$ be the Galois group of the maximal extension of $\mathbb Q$ unramified outside S$. I [believe](https://mathoverflow.net/questions/63029/finitely-generated-galois-groups) that it is an open question whether this group is topologically finitely generated, that is, contains a dense, finitely generated subgroup. 



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> Is there a standard conjecture about whether $G\_{\mathbb Q,S}$ should be finitely generated or not?  
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>  Has anyone published an opinion?
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[JSE mentions](https://mathoverflow.net/questions/107817/could-unramified-galois-groups-satisfy-a-version-of-property-tau) a stronger conjecture: $|S|$ generators suffice; further, the existence of a special set of generators labeled by $S$ and inert at the labeling place. Is this refinement standard? (Note that a finite extension would suffice to disprove this hypothesis.)


Weakening the hypothesis, we could restrict the ramification at $S$ to be tame.



Are Galois groups of Q with restricted ramification supposed to be finitely generated?
I think you can look at page 532 of (the first version of) J. Neukirch, A. Schmidt, K. Wingberg, *Cohomology of Number Fields*, Springer, 1999. They explicitly write that we do not even know precisely what to conjecture for arbitrary number field. They also add "many mathematicians (including the authors) tend to think of $G\_S$ as being not finitely generated". In their notation, $G\_S$ is the Galois group of the maximal unramified outside of $S$ extension of a number field $K$ which has been fixed once and for all (and $S$ is finite). I do not know if the special case $K=\mathbb{Q}$ is somehow different, but I doubt it (after all, their $G\_S$ is a subgroup of your $G\_{\mathbb{Q},S'}$ for $S'=S\cap\mathrm{Ram}(K/\mathbb{Q})$...)


