Galois group of $\overline{\mathbb{F}_{p}}$ gives arithmetical information for finite fields $K/\mathbb{F}_{p}$? Let $\mathbb{F}_{p}$ be the field with $p$ elements and $\overline{\mathbb{F}_{p}}$ be its algebraic closure.
For some reason, we want to understand the structure of the Galois group of such an extension.
We then find that we have an isomorphism of topological groups $G(\overline{\mathbb{F}_{p}}/\mathbb{F}_{p})\cong\hat{\mathbb{Z}}$.
My question is:
The description of the group $G(\overline{\mathbb{F}_{p}}/\mathbb{F}_{p})$ certainly gives the "subfield lattice" of $\overline{\mathbb{F}_{p}}$ and this is nice.
On the other hand, such a subfield lattice can be found very easily without using this isomorphism, and in fact most books even use this fact to establish the isomorphism.
In view of this comments, why does one want to describe such an absolute Galois group and why it is useful to know that, for instance, $\hat{\mathbb{Z}}\cong\prod_{p}\mathbb{Z}_{p}$, where $\mathbb{Z}_{p}$ denotes the $p$-adic integers?
Do we get any arithmetical information about finite fields $K/\mathbb{F}_{p}$? Thanks a lot.
 A: I think you have things a bit backwards here. The prevalence of $G_{\mathbb F_p}:=\mathrm{Gal}(\overline{\mathbb F}_p/\mathbb F_p)$ in number theory is not as a gadget to study the finite extensions of $\mathbb F_p$, since, as you've said, the extensions of $\mathbb F_p$ are well understood. Rather, it is to enable us to use what we know about the extensions of $\mathbb F_p$ to understand the extensions of $\mathbb Q_p$. In particular, there is a natural isomorphism
$$\mathrm{Gal}(\mathbb Q_p^{\text{nr}}/\mathbb Q_p)\cong \mathrm{Gal}(\overline{\mathbb F}_p/\mathbb F_p),$$
where $\mathbb Q_p^{\text{nr}}$ denotes the maximal unramified extension of $\mathbb Q_p$. 
Let $G_{\mathbb Q_p} = \mathrm{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$. If $\sigma \in G_{\mathbb Q_p}$, then, identifying $\overline{\mathbb F}_p$ with the residue field of $\overline{\mathbb Z}_p$, we obtain an action of $\sigma$ on $\overline{\mathbb F}_p$. In particular, there is a map
$$G_{\mathbb Q_p}\to G_{\mathbb F_p},$$
which can be shown to be surjective (using Hensel's lemma). Define the inertia group $I_p$ to be the kernel of this map. Then $\mathbb Q_p^{\text{nr}}$ can be identified with the fixed field $\overline{\mathbb Q}_p^{I_p}$. 
We have a short exact sequence
$$1\to I_p\to G_{\mathbb Q_p}\to G_{\mathbb F_p}\to 1.$$
Hence, in order to understand $G_{\mathbb Q_p}$, we would do well to understand $I_p$ and $G_{\mathbb F_p}$. Here, $G_{\mathbb F_p}$ tells us about the unramified extensions of $\mathbb Q_p$, and $I_p$ tells us about the ramified ones.
Knowing that $G_{\mathbb F_p}\cong \widehat{\mathbb Z}\cong\prod_l\mathbb Z_l$ means that we completely understand $G_{\mathbb F_p}$.
A: In afterthought, let me give a striking illustration of what I called the "functorial" approach in my previous comment. Let $K/ \mathbf Q_p$ be a given extension of degree $N$, not containing a primitive $p$-th root of 1. Suppose somebody comes up with a finite $p$-group $G$ and asks you to construct a Galois extension $L/K$ with group $G$. After some hard work, you succeed in doing it, and the guy comes back with a second $p$-group asking the same question. Toiling even harder, you begin to suspect that there could be no such extension. Then the guy comes back with a third $p$-group, and this time you draw your gun, I mean your "functorial" gun, which is a theorem of Safarevic : "The Galois group over $K$ of the maximal pro-$p$-extension of $K$ is pro-$p$-free with $N + 1$ generators". This means that any (resp. no) $p$-group on less (resp. strictly more) than $N + 1$ generators can be realized as the group of a Galois extension of $K$.
The point here is that of course the proof of Sh.'s theorem does not (cannot) rely on explicit constructions. As explained e.g. in H. Koch's "Galois theory of $p$-extensions", it builds on Galois cohomology. To give a quick survey : for a pro-$p$-group $\mathcal G$, the minimal number of generators (resp. of relations) of $\mathcal G$ equals the $\mathbf F_p$-dimension of $H^1(\mathcal G,\mathbf F_p)$ (resp. of $H^2$). If $\mathcal G$ is the maximal pro-$p$-group introduced above, the first dimension is easily computed from that of $K^*/K^{*p}$, which even gives a system of generators consisting of a lift of the Frobenius automorphism (see @Mathmo123's answer) and of $N$ automorphisms in the inertia subgroup.The second dimension is shown to be $0$ by adding $p$-th roots of 1 to $K$ in order to  use the Brauer group, then going back to $K$ applying cohomological restriction.
I hope this example will convince you of the utility, but also the necessity of the functorial approach.
