It is a well-known result that all profinite groups arise as the Galois group of *some* field extension.



> 
> What profinite groups are the absolute Galois group
>  $\mathrm{Gal}(\overline{K}|K)$ of some extension $K$ over
>  $\mathbb{Q}$?
> 
> 
> 


The answer is simple enough in the finite case:


* (Artin-Schreier) Only the trivial one and $C\_2$.


This might tempt us to think that absolute Galois groups are not that diverse. But 0-dimensional anabelian geometry shows us differently:


* (Neukirch-Uchida) There's as many different absolute Galois groups as non-isomorphic number fields.


The answer might still turn out to be boring, but I haven't seen this discussed anywhere. The closest is Szamuely in his book "Galois Groups and Fundamental Groups", where he seems delighted by the fact that the absolute Galois group of $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{q})$ are non-isomorphic for primes $p\neq q$.



Profinite groups as absolute Galois groups
This is a very good question which is a big open problem. There are a number of theorems, some of them easy and some very difficult, and also a number of conjectures, restricting the class of groups which may turn out to be absolute Galois groups. But it seems that nobody has any idea about how a precise description of the class of absolute Galois groups might look like.


One general point-of-view position on which the experts seem to agree is that the right object to deal with is not just the Galois group $G\_K=\operatorname{Gal}(\overline{K}|K)$ considered as an abstract profinite group, but the pair $(G\_K,\chi\_K)$, where $\chi\_K\colon G\_K\to \widehat{\mathbb Z}^\*$ is the cyclotomic character of the group $G\_K$ (describing its action in the group of roots of unity in $\overline K$). The question about being absolute Galois should be properly asked about pairs $(G,\chi)$, where $G$ is a profinite group and $\chi\colon G\to \widehat{\mathbb Z}^\*$ is a continuous homomorphism, rather than just about the groups $G$.


For example, here is an important and difficult theorem restricting the class of absolute Galois groups: for any field $K$, the cohomology algebra $H^\*(G\_K,\mathbb F\_2)$ is a quadratic algebra over the field $\mathbb F\_2$. This is (one of the formulations of) the Milnor conjecture, proven by Rost and Voevodsky. More generally, for any field $K$ and integer $m\ge 2$, the cohomology algebra $\bigoplus\_n H^n(G\_K,\mu\_m^{\otimes n})$ is quadratic, too,
where $\mu\_m$ denotes the group of $m$-roots of unity in $\overline K$ (so $G\_K$ acts in $\mu\_m^{\otimes n}$ by the character $\chi\_K^n$). This is the Bloch-Kato conjecture, also proven by Rost and Voevodsky.


Here is a quite elementary general theorem restricting the class of absolute Galois groups: for any field $K$ of at most countable transcendence degree over $\mathbb Q$ or $\mathbb F\_p$, the group $G\_K$ has a decreasing filtration $G\_K\supset G\_K^0\supset G\_K^1\supset G\_K^2\supset\dotsb$ by closed subgroups normal in $G\_K$ such that $G\_K=\varprojlim\_n G\_K/G\_K^n$ and $G\_K/G\_K^0$ is either the trivial group or $C\_2$, while $G\_K^n/G\_K^{n+1}$, $n\ge0$ are closed subgroups in free profinite groups. (Groups of the latter kind are called "projective profinite groups" or "profinite groups of cohomological dimension $\le1$".) One can get rid of the assumption of countability of the transcendence degree by considering filtrations indexed by well-ordered sets rather than just by the integers.


Here is a conjecture about Galois groups of arbitrary fields (called "the generalized/strengthened version of Bogomolov's freeness conjecture). For any field $K$, consider the field $L=K[\sqrt[\infty]K]$ obtained by adjoining to $K$ all the roots of all the polynomials $x^n-a$, where $n\ge2$ and $a\in K$. In particular, when the field $K$ contains all the roots of unity, the field $L$ is (the maximal purely inseparable extension of) the maximal abelian extension of $K$. Otherwise, you may want to call $L$ "the maximal radical extension of $K$". The conjecture claims that the absolute Galois group $G\_L=\operatorname{Gal}(\overline{L}|L)$ is a projective profinite group.


