The GrothendieckTeichmuller conjecture asserts that the absolute Galois group $Gal(\mathbb{Q})$ is isomorphic to the GrothendieckTeichmuller group. I was wondering, would this conjecture imply the Inverse Galois Problem? I.e. is every finite group a quotient of the GrothendieckTeichmuller group?
1 Answer
It's only a partial answer since this survey is already 5 years old, but it suggest that almost nothing is known about (nonabelian) 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.

2$\begingroup$ I have the feeling that their is very little known about $\widehat{GT}$ period! $\endgroup$ Jan 28, 2012 at 3:38

1$\begingroup$ I too think this problem sounds extremely hard. To give some indication  as far as I know (and I could certainly be wrong)  the only reason we know the map GThat > Zhat^* is surjective is that G_Q embeds in GThat! Without that, I don't think I even know how to show that the image is infinite! $\endgroup$– JSEJan 28, 2012 at 22:19