The absolute Galois group $G\_{\mathbb Q}=\text{Gal}(\bar{\mathbb Q}/\mathbb Q)$, as a profinite group, encodes a lot of things: the whole lattice of number fields (closed subgroups of finite index), Galois extensions (normal subgroups), abelian extensions etc. Is it possible to recognize *unramified* abelian extensions from the group-theoretic data? So,



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> Question: Given a closed subgroup $H\subset G\_{\mathbb Q}$ of finite index, is there an explicit group-theoretic way to recover the class group or the class number of the fixed field $K=\overline{\mathbb Q}^H$?
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(Because all continuous automorphisms of $G\_{\mathbb Q}$ are inner, in theory $K$ itself can be recovered from $H$ and its class group computed, but this is not what I'd like to call "group-theoretic" or "explicit".)



Are class numbers encoded in the absolute Galois group of ${\mathbb Q}$?
Dear Tim,


As you're probably aware, this is part of the 'anabelian' etcetera. 


It suffices to recover all intertia subgroups $I\_v\subset H$, because their union will then be a normal subgroup $N$ such that $H/N$ is the Galois group of the maximal extension of $K$ unramified everywhere. We can get the ideal class group then by (topological) abelianization. The fact that we can get all the decomposition groups $D\_v\subset G$ is Neukirch's theorem (together with Artin-Schreier at infinity). This says the maximal subgroups isomorphic to a local Galois group are exactly the decomposition groups. If you want to make this purely group-theoretic for the finite places, you invoke the theorem of Jannsen-Wingberg that lays out a presentation for all local Galois groups and consider maximal elements in the lattice of subgroups isomorphic to such an explicit presentation. Once you have the $D\_v$, there is a standard group-theoretic recipe for $I\_v$, which escapes me for the moment. But I'll get back to you with it, if you don't figure it out in the meanwhile.


Added:


OK, so here is the easy part. Now let $F$ be a finite extension of $\mathbb{Q}\_p$ and
$D=Gal(\bar{F}/F)$. We know that $D^{ab}$ fits into an exact sequence
$$0\rightarrow U\_F\rightarrow D^{ab}\rightarrow \hat{\mathbb{Z}}\rightarrow 0,$$
so we recover $p$ as the unique prime such that the topological $\mathbb{Z}\_p$-rank of
$D^{ab}$ is $r\_D\geq 2$. The order $q\_D$ of the residue field is 1 greater than the order of
the prime-to-$p$ torsion subgroup of $D^{ab}$. Also, we know
$r\_D=1+[F:\mathbb{Q}\_p]$. Now we apply the same reasoning to the subgroups of finite index in
$D$ to figure out those corresponding to unramified extensions. That is, consider
the subgroups $E$ of finite index such that $q^{r\_D-1}\_E=q\_D^{r\_E-1}$. Then the inertia subgroup
of $D$ is the intersection of all of these.


