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Let $K$ be a number field, $S$ a finite set of primes of $K$ and $ G_{K,S} $ the Galois group of the maximal extension of $ K $ (inside a fixed algebraic closure of $K$) unramified outside $ S $. In Theorem 10.2.5 of Cohomology of Number Fields, it's proved that the group $ G_{K,S} $ is topologically generated by the conjugacy classes of finitely many elements, and hence $ G_{K,S} $ contains a proper dense normal subgroup if $ G_{K,S} $ is infinite. Note that if $ G_{K,S} $ is topologically finitely generated, then $G_{K,S}$ contains a countable dense subgroup.

Question: Is it true that the profinite group $ G_{K,S} $ contains a countable dense normal subgroup, at least at the level of conjecture?

I know that it's still an open problem whether $ G_{K,S} $ is topologically finitely generated. Any comments and reference would be highly appreciated.

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  • $\begingroup$ Yes: a metrizable compact space is indeed separable. $\endgroup$
    – YCor
    Commented Jun 29, 2023 at 19:26
  • $\begingroup$ @YCor Do you mean that any metrizable profinite group (=first countable profinite group) contains a countable dense normal subgroup? Maybe you missed the condition “NORMAL"? This conclusion shocks me, could you point out some references or say a few more words? $\endgroup$
    – Nobody
    Commented Jun 29, 2023 at 19:33
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    $\begingroup$ @R.vanDobbendeBruyn certainly you also want to discard those with an open abelian subgroup. $\endgroup$
    – YCor
    Commented Jun 29, 2023 at 20:57
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    $\begingroup$ For the question, I think it is enough to find a suitable infinite quotient with no such normal subgroup, and any $p$-adic analytic group with no abelian open subgroup should do the job. For instance, if there exists a local field of char. zero and a representation $G\to\mathrm{GL}_2(L)$ with open image, then we should be done. Whether there exists such a representation is maybe well-known to number theorists, but not to me. $\endgroup$
    – YCor
    Commented Jun 29, 2023 at 22:18
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    $\begingroup$ You're right, e.g. Serre's Open Image Theorem guarantees the existence of such a representation. So the answer to the question is no. $\endgroup$
    – Nobody
    Commented Jun 30, 2023 at 7:06

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