The Grothendieck-Teichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the Grothendieck-Teichmuller group. I was wondering, would this conjecture imply the Inverse Galois Problem? I.e. is every finite group a quotient of the Grothendieck-Teichmuller group?
[Math] Is every finite group a quotient of the Grothendieck-Teichmuller group
ag.algebraic-geometrygalois-theorynt.number-theory
Related Solutions
V.G. Drinfeld: On quasi-Hopf algebras and on a group that is closely connected with $Gal(\overline{\mathbb{Q}}/\mathbb{Q}$), Algebra i Analiz, 2:4 (1990), 149–181
I am not saying it is easy to read but it is definitely well-motivated. By the way, the point of view on $\widehat{GT}$ might also be a bit different from what you expect.
In the very same spirit, but less technical, there is an excellent account of Drinfled's theory by Dror Bar-Natan: On Associators and the Grothendieck-Teichmuller Group I, Selecta Mathematica NS 4 (1998) 183-212 ( http://arxiv.org/abs/q-alg/9606021).
Let me now explain broadly how this works.
Consider the collection $\mathcal B_*=(\mathcal B_n)_n$ consisting of groupoids of parenthesised braids. It is an operad in groupoids (by using the so-called cabling operations).
A more algebroic way to look at it is to consider the free braided monoidal category with one generator $\bullet$, and consider the groupoid of isomorphisms in this category. Objects in this category are actually parenthesisations of $\bullet^{\otimes n}$... so if we fix $n$ we recover $\mathcal B_n$.
Now we would like to define the group $GT$ as the automorphism group of this operad in groupoids $\mathcal B_*$... the main problem is that it is rather small.
The main point is then, for any field $k$, to consider $k$-prounipotent completions $\mathcal B_n(k)$ of $\mathcal B_n$. Now $\widehat{GT}(k)$ is the automorphism group of the operad in $k$-groupoids $\mathcal B_*(k)$.
Using the version of Mac-Lane coherence Theorem for braided monoidal categories, one can prove that such automorphisms are determined by their action on $\mathcal B_3(k)$, and that the relations we have to impose are in $\mathcal B_3(k)$ and $\mathcal{B}_4(k)$.
The relation in $\mathcal B_3(k)$ is the hexagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_2(k)$ to $\mathcal B_3(k)$. The relation in $\mathcal B_4(k)$ is the pentagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_3(k)$ to $\mathcal B_4(k)$.
There are no such elements -- the intersection of the derived series of a free group is trivial. In fact, even more is true -- the intersection of the lower central series of a free group is trivial. This is a theorem of Magnus, and by now there are many proofs. The classical one is in the final chapter of Magnus-Karass-Solitar's book on combinatorial group theory.
By the way, a topological proof of this fact (lifting curves to covers to resolve self-intersections, etc) is contained in my paper "On the self-intersections of curves deep in the lower central series of a surface group" with Justin Malestein.
EDIT : I see that you really want finite solvable quotients, not general solvable quotients. It is still true. Fixing a prime $p$, there is a ``mod $p$ lower central series'' of a group whose quotients are $p$-groups (so finite nilpotent if the group is finitely generated). For a free group, Zassenhaus proved in his paper
H. Zassenhaus, Ein Verfahren, jeder endlichen p-Gruppe eine Lie-Ring mit der Charakteristik p zuzuordnen, Abh. Math. Sem. Hamburg Univ. 13 (1939), 200-207.
that the intersection of the mod $p$ lower central series of a free group is trivial. This can also be deduced from the paper I mentioned with Justin Malestein, at least for the prime $2$ (one of the proofs we give actually yields regular covers whose order is a power of $2$).
Best Answer
It's only a partial answer since this survey is already 5 years old, but it suggest that almost nothing is known about (non-abelian) finite quotients of $\widehat{GT}$ (question 1.7).
Edit: I should maybe recall what happen in the abelian case, and why it's encouraging: elements of $\widehat{GT}$ are pairs $(f,\lambda)$ where $f$ is in the derived subgroup of $\hat{F}_2$, and $\lambda \in \hat{\mathbb{Z}}^{\times}$, satisfying some complicated equations. It turns out that the set theoretic map $(f,\lambda) \mapsto \lambda$ induces a surjective group morphism $\widehat{GT}\rightarrow \hat{\mathbb{Z}}^{\times}$. And the good news is that the composite
$$G_{\mathbb{Q}} \hookrightarrow \widehat{GT} \rightarrow \hat{\mathbb{Z}}^{\times}$$
is nothing but the cyclotomic character.