non-continuous inverse Galois problem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:55:43Z http://mathoverflow.net/feeds/question/98805 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98805/non-continuous-inverse-galois-problem non-continuous inverse Galois problem Hugo Chapdelaine 2012-06-04T19:47:43Z 2012-06-05T08:34:06Z <p>Let $G=Gal(\bar{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group over $\mathbf{Q}$. </p> <p>Q1: Is it possible to find a (necessarily non-closed) normal subgroup $K\leq G$ such that $G/K$ is free of infinite rank ?</p> <p>Q2: If the answer to Q1 is no then is it always possible to find a <strong>not necessarily</strong> continuous surjective homomorphism $\rho:G\rightarrow H$ where $H$ is an arbitrary finite group? </p> <p>If you think that removing the continuity assumption does make the inverse Galois problem any simpler then please provide some explanations why. </p> http://mathoverflow.net/questions/98805/non-continuous-inverse-galois-problem/98819#98819 Answer by Will Sawin for non-continuous inverse Galois problem Will Sawin 2012-06-04T22:16:22Z 2012-06-05T07:03:57Z <p>Q1: No.</p> <p>Suppose there were such a subgroup. Then there would certainly be a $K$ such that $G/K$ is $\mathbb Z$. $K$ would have to contain the commutator subgroup. The quotient of $G$ by the commutator subgroup is the idele class group, which in this case is $\prod_L \mathbb Z_l^{\times}$. (EDIT: This might not be true. There are better arguments in the comments.) Thus there must be a nontrivial map from some $\mathbb Z_l^{\times}$ to $\mathbb Z$. $\mathbb Z_l^{\times}$ has a finite index subgroup isomorphic to $\mathbb Z_l^{+}$, which must also has a nontrivial map to $\mathbb Z$. But $\mathbb Z_l^{+}$ is a $p$-divisible group for any $p\neq l$, and thus has no nontrivial maps to $\mathbb Z$.</p> <p>Q2: I don't know. It seems unlikely.</p> http://mathoverflow.net/questions/98805/non-continuous-inverse-galois-problem/98851#98851 Answer by Yves Cornulier for non-continuous inverse Galois problem Yves Cornulier 2012-06-05T08:34:06Z 2012-06-05T08:34:06Z <p>It's an old result by R. Alperin that a compact group has no abstract homomorphism onto $\mathbf{Z}$. [Compact groups acting on trees. Houston J. Math. 6 (1980), no. 4, 439--441.] So Q1 has a negative answer.</p> <p>Here $\mathbf{Z}$ cannot be replaced by any countable abelian group. For instance, the direct product $G$ of all finite perfect groups (up to isomorphism) has an infinite abstract abelianization. Indeed since $G$ is isomorphic to its countable power, if $G$ were perfect it would be uniformly perfect, and thus all finite groups would be together perfect with a uniform commutator width, which is false. So $G$ has nontrivial abstract abelianization, and again since $G$ is isomorphic to its countable power, its abstract abelianization is uncountable (and thus admits a countable infinite quotient). </p> <p>About Q2: I don't know any example of a profinite group $G$ and finite simple group $S$ such that $G$ admits $S$ as a quotient abstractly, but not as a quotient by any open subgroup (in all the examples I know with $G$ admitting $S$ as a quotient by a non-open subgroup, $G$ actually has <i>infinitely</i> many <i>open</i> subgroups $H$ with $G/H\simeq S$) .</p>