What would it mean to understand this Galois group? You could mean several things.
You could mean trying to give the group in terms of some smallish generators and relations. This would be nice, and help to answer questions like the inverse Galois problem that Greg Muller mentioned, and having a certain family of "generating" Galois automorphisms would allow you to study questions about e.g. the representation theory in quite explicit terms. However, the Galois group is an uncountable profinite group, and so to give any short description in terms of generators and relations leads you into subtle issues about which topology you want to impose.
You could also ask for a coherent system of names for all Galois automorphisms, so that you can distinguish them and talk about them on an individual basis. One system of names comes from the dessins d'enfant that Ilya mentioned: associated to a Galois automorphism we have some associated data.
- We have its image under the cyclotomic character, which tells us how it acts on roots of unity. By the Kronecker-Weber theorem this tells us about the abelianization of the Galois group.
- We also have an element in the free profinite group on two generators, which (roughly speaking) tells us something about how abysmally acting on the coefficients of a power series fails to commute with analytic continuation.
These two names satisfy some relations, called the $2$-, $3$-, and $5$-cycle relation, which are conjectured to generate all relations (at least the last time I checked), but it is difficult to know whether they actually do so. If they do, then the Galois group is the so-called Grothendieck-Teichmüller group.
The problem with this perspective is that the names aren't very explicit (and we don't expect them to be: we may need the axiom of choice to show they exist, and there are only two Galois automorphisms of $\mathbb{C}$ that are measurable functions!) and it seems to be a difficult problem to determine whether the Grothendieck-Teichmuller group really is the whole thing. (Or it was the last time I checked.)
However, the cyclotomic character is a nice, and fairly canonical, name associated for Galois automorphisms. We could try to generalize this: there are Kummer characters telling us what a Galois automorphism does to the system of real positive roots of a positive rational number number (these determine a compatible system of roots of unity, or equivalent an element of the Tate module of the roots of unity). This points out one of the main difficulties, though: we had to make choices of roots of unity to act on, and if Galois theory taught us nothing else it is that different choices of roots of an irreducible polynomial should be viewed as indistinguishable. Different choices differ by conjugation in the Galois group.
This brings us to the point JSE was making: if we take the "symmetry" point of view seriously, we should only be interested in conjugacy-invariant information about the Galois group. Assigning names to elements or giving a presentation doesn't really mesh with the core philosophy.
So this brings us to how many people here have mentioned understanding the Galois group: you understand it by how it manifests, in terms of its representations (as permutations, or on dessins, or by representations, or by its cohomology), because this is how it's most useful. Then you can study arithmetic problems by applying knowledge about this. If I have two genus $0$ curves over $\mathbb{Q}$, what information distinguishes them? If I have two lifts of the same complex elliptic curve to $\mathbb{Q}$, are they the same? How can I get information about a reduction of an abelian variety mod $p$ in terms of the Galois action on its torsion points? Et cetera.
Best Answer
My comment, Wojowu's answer, YCor's comment, and Z. M's comment already contain everything we need. Let me provide a little more detail here. I will shift the indices by $1$ for reasons that will become apparent:
Definition. Set $K_0 = \mathbf Q$ and let $K_1 = \mathbf Q(\boldsymbol \mu)$ be the field obtained by adjoining the roots of unity $\boldsymbol \mu \subseteq \bar{\mathbf Q}$. Inductively define $K_{i+1}=K_i\big(\sqrt[\infty\ \ ]{K_i^\times}\big)$, and set $K_\infty = \underset{\substack{\longrightarrow \\ i}}{\operatorname{colim}} K_i$.
We claim that this is the extension we're after. We first introduce some notation:
Definition. Given a profinite group $G$, its (profinte) derived series is the transfinite chain of closed subgroups $$G = G^{(0)} \trianglerighteq G^{(1)} \trianglerighteq \cdots \trianglerighteq G^{(\alpha)} \trianglerighteq \cdots$$ defined by $G^{(\alpha+1)} = \overline{[G^{(\alpha)},G^{(\alpha)}]}$ and $G^{(\beta)} = \bigcap_{\alpha < \beta} G^{(\alpha)}$ for any limit ordinal $\beta$ (which is already closed as each $G^{(\alpha)}$ is closed). One could alter the notation to distinguish it from the abstract derived series, but I will never use the latter (the same goes for the Kronecker–Weber theorem: it computes the topological abelianisation, not the abstract one!). Note that for any continuous surjective homomorphism $G \to H$ of profinite groups, the image of $G^{(\alpha)}$ is $H^{(\alpha)}$.
Lemma. Let $G$ be a profinite group. Then $G^{(\omega + 1)} = G^{(\omega)}$, and this group is trivial if and only if $G$ is pro-soluble¹.
Proof. For any finite group $G$, the descending chain $G^{(i)}$ stabilises after finitely many steps, so $G^{(\omega + 1)} = G^{(\omega)}$. The same statement for profinite groups follows since any closed normal subgroup $H \trianglelefteq G$ is the intersection of the open normal subgroups $U \trianglelefteq G$ containing it. Similarly, $G^{(\omega)} = 1$ if and only if the same holds in every finite quotient $G/U$, i.e. if and only if all $G/U$ are soluble. $\square$
Let's denote $G^{(\omega)}$ by $G^{(\infty)}$. For $n \in \mathbf N \cup \{\infty\}$, we will say that $G$ is $n$-soluble if $G^{(n)} = 1$, and we write $G^{n\text{-}\!\operatorname{sol}} = G/G^{(n)}$ for its maximal $n$-soluble quotient (in which we omit $n$ if $n = \infty$). For instance, $G$ is $1$-soluble if and only if it is abelian, and $\infty$-soluble if and only if it is pro-soluble (equivalently, hypoabelian as profinite group).
Theorem. Let $\Gamma = \Gamma_{\mathbf Q}$ be the absolute Galois group of $\mathbf Q$.
Proof. Statement (1) is trivial for $n=0$, and is the Kronecker–Weber theorem for $n=1$. Statements (1) and (2) for finite $n \geq 2$ follow inductively by Kummer theory (see the corollary below). Finally, statement (1) for $n = \infty$ follows from the statement at finite levels, since $K_\infty = \bigcup_n K_n$ and $G^{(\infty)} = \bigcap_n G^{(n)}$. $\square$
Note also that the Galois group $\operatorname{Gal}(K_1/K_0)$ is isomorphic to $\operatorname{Aut}(\boldsymbol \mu) = \hat{\mathbf Z}^\times$. However, explicitly computing $\operatorname{Gal}(K_{n+1}/K_n)$ in a meaninful way is pretty hard, let alone saying anything about how the various pieces fit together.
Edit: After writing this answer, I became aware of the following two striking results:
Theorem (Iwasawa). The Galois group $\operatorname{Gal}(K_\infty/K_1) = \Gamma^{(1)}/\Gamma^{(\infty)}$ is a free pro-soluble group $\widehat{F_\omega}^{\operatorname{sol}}$ on countably infinitely many generators.
So we know that $\Gamma^{\operatorname{sol}}$ sits in a short exact sequence $$1 \to \widehat{F_\omega}^{\operatorname{sol}} \to \Gamma^{\operatorname{sol}} \to \hat{\mathbf Z}^\times \to 1.$$ I find it hard to imagine that this sequence splits as a semi-direct product (but I am more optimistic about the derived length $\leq 2$ situation).
Theorem (Shafarevich). Any finite soluble group $G$ occurs as a quotient of $\operatorname{Gal}(K_\infty/\mathbf Q) = \Gamma^{\operatorname{sol}}$.
A modern reference is Neukirch–Schmidt–Wingberg's Cohomology of number fields, Corollary 9.5.4 (Iwasawa) and Theorem 9.6.1 (Shafarevich). (This is a truly great book, but even at $>800$ pages it can be a bit terse at times.)
We used the following general result:
Lemma (Kummer theory). Let $m \in \mathbf Z_{>0}$, and $K$ be a field of characteristic not dividing $m$ that contains $\boldsymbol \mu_m$.
We avoid the notation $\widehat A$ for Pontryagin duals, since it clashes with the notation for profinite completions. (Note that Z. M's comment uses $(-)^\vee$ for a linear dual, which differs from my notation by a Tate twist.)
Because it's not very hard, let's include a proof.
Proof. For (2), by Pontryagin duality it suffices to show that the dual map \begin{align*} K^\times/(K^\times)^m &\to \operatorname{Hom}\!\big(\Gamma_K,\boldsymbol \mu_m\big) = \left(\Gamma_K^{\operatorname{ab}}/m\right)^\vee \\ a &\mapsto \left(\sigma \mapsto \frac{\sigma(\sqrt[m\ \ ]{a})}{\sqrt[m\ \ ]{a}}\right) \end{align*} is an isomorphism. Note that it is well-defined since any two $m$-th roots of $a$ differ (multiplicatively) by an element of $\boldsymbol \mu_m \subseteq K$, on which $\sigma$ acts as the identity. Since $\boldsymbol \mu_m \subseteq K$, the $\Gamma_K$-module $\boldsymbol \mu_m$ has trivial action, so $\operatorname{Hom}_{\operatorname{cont}}(\Gamma_K,\boldsymbol \mu_m) = H^1(K,\boldsymbol \mu_m)$. The Kummer sequence $$1 \to \boldsymbol \mu_m \to \mathbf G_m \stackrel{(-)^m}\to \mathbf G_m \to 1$$ and Hilbert's theorem 90 compute $K^\times/(K^\times)^m \stackrel\sim\to H^1(K,\boldsymbol \mu_m)$ via the map above. Now (1) follows since $\sigma \in \Gamma_K$ is in the kernel of $\Gamma_K \to \big(K^\times/(K^\times)^m\big)^\vee$ if and only if $\sigma$ fixes all $m$-th roots of elements in $K$. $\square$
Corollay. Let $K$ be a field of characteristic $0$ containing $\boldsymbol \mu$.
Proof. Take inverse limits over all $m \in \mathbf Z_{>0}$ in the lemma above, noting that the inverse limit pulls out of $\operatorname{Hom}(K^\times,-)$. $\square$
¹Linguistic footnote: soluble and solvable mean the same thing. I used to think that this is one of those BrE vs AmE things (for instance, my Oxford Advanced Learner's dictionary does not contain the word solvable at all). But I think some folks in the UK also use solvable, so it's not entirely clear to me.