Is every finite group a quotient of the Grothendieck-Teichmuller group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:16:32Z http://mathoverflow.net/feeds/question/86874 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/86874/is-every-finite-group-a-quotient-of-the-grothendieck-teichmuller-group Is every finite group a quotient of the Grothendieck-Teichmuller group? Makhalan Duff 2012-01-28T01:33:23Z 2012-01-28T10:50:19Z <p>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?</p> http://mathoverflow.net/questions/86874/is-every-finite-group-a-quotient-of-the-grothendieck-teichmuller-group/86875#86875 Answer by Adrien for Is every finite group a quotient of the Grothendieck-Teichmuller group? Adrien 2012-01-28T01:53:54Z 2012-01-28T10:50:19Z <p>It's only a partial answer since this <a href="http://www.math.jussieu.fr/~leila/farb.pdf" rel="nofollow">survey</a> is already 5 years old, but it suggest that almost nothing is known about (non-abelian) finite quotients of $\widehat{GT}$ (question 1.7).</p> <p><strong>Edit:</strong> 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 <em>surjective</em> group morphism $\widehat{GT}\rightarrow \hat{\mathbb{Z}}^{\times}$. And the good news is that the composite</p> <p>$$G_{\mathbb{Q}} \hookrightarrow \widehat{GT} \rightarrow \hat{\mathbb{Z}}^{\times}$$</p> <p>is nothing but the <a href="http://en.wikipedia.org/wiki/Cyclotomic_character" rel="nofollow">cyclotomic character</a>.</p>