Does this Galois group determine the class ideal class group? Let $K$ be an algebraic number field, and consider the Galois group:
$G = Gal(\bar{\mathbb{Q}}, K)$.
Is knowing the Galois group $G$ alone, without other information on $K$, enough to determine the ideal class group of $K$? 
A user suggested that in short the answer is "yes", via class field theory, as a comment to my other post:
Does the abelianization of the Galois group determine the ideal class group?
In that post, I was looking at the wrong Galois group.
I don't require a long answer. It would not be fair to ask for a complete explanation, because it seems like a standard result in class field theory (which is on my list of subjects to learn about). However, a brief outline with a couple of references would be great.
 A: 
Edit: The answer below is incorrect. While it is true that, via class field theory, we can recover the class group as a quotient of $G^{ab}$, the problem, as @ThePiper points out, is that this quotient is by $\widehat{\mathcal O}_K^\times$, which $G^{ab}$ knows nothing about.
Given the whole of $G$, we would be able to recover $\widehat{\mathcal O}_K^\times=\prod_{v}\widehat{\mathcal O}_{K_v}^\times$ via class field theory if we could recover the inertia groups $I_v$ from $G$: by local class field theory, $I_v\cong {\mathcal O}_{K_v}^\times$.
It is possible to recover the inertia groups from $G$. However, the fact that we can do so is a key part of the Neukirch-Uchida theorem. 

The answer is yes. Let $G^{ab}$ denote the abelianisation of $G$ $-$ i.e. $G^{ab} = G/\overline{[G,G]}$. By global class field theory, we have a canonical isomorphism
$$K^\times\backslash\mathbb A_K^{\times}/\overline{(K_\infty^\times)^0}\cong G^{ab}.$$
Here, $\mathbb A_K^\times$ are the ideles of $K$, and $\overline{(K_\infty^\times)^0}$ is the closure of the identity connected component of $(K\otimes_\mathbb Q\mathbb R)^\times$ viewed as a subgroup of $\mathbb A_K^\times$.
This isomorphism gives a concrete connection to the class group of $K$: the class group of $K$ is canonically isomorphic to
$$K^\times\backslash\mathbb A_K^{\times}/\widehat{\mathcal O_K^\times} K_\infty^\times,$$
and is therefore a quotient of $G^{ab}$.
On the Galois side, this quotient of $G^{ab}$ cuts out a finite abelian extension of $K$ -- the Hilbert class field.
A: The answer is yes, but the explanation given by @Mathmo123 is incorrect. 
While it is true that class field theory gives an adelic description of $G^{ab}$,
it is not at all clear one can recover from this description what the maximal unramified extension of $K$ should be.
In order to compute the class group, one has to take the quotient of the  idele class group (or really the idele class group modulo a maximal connected component at the infinite place)
by $\widehat{\mathcal O_K^\times}$ - but this subgroup is given in terms of $K$, and the problem is exactly
about giving a description which depends only on $G$, and not on $K$.
This error is a fatal one - it turns out that the abelianization $G^{ab}$ of $G = G_K$ is not enough to determine the class group. For some references and examples of this phenomenon, see this paper (in particular, the last line of the first page):
http://www.math.ucsd.edu/~kedlaya/ants10/angelakis/paper.pdf
That said, the answer by Alex J Best in the comments gives a complete positive answer to the question; the entire group $G$ determines $K$ by the Neukirch-Uchida theorem, and then knowing $K$ determines the class group of $K$.
