non-continuous inverse Galois problem - MathOverflow

non-continuous inverse Galois problem

2012-06-04T19:47:43Z

Let $G=Gal(\bar{\mathbf{Q}}/\mathbf{Q})$ be the absolute Galois group over $\mathbf{Q}$.

Q1: Is it possible to find a (necessarily non-closed) normal subgroup $K\leq G$ such that $G/K$ is free of infinite rank ?

Q2: If the answer to Q1 is no then is it always possible to find a not necessarily continuous surjective homomorphism $\rho:G\rightarrow H$ where $H$ is an arbitrary finite group?

If you think that removing the continuity assumption does make the inverse Galois problem any simpler then please provide some explanations why.

Answer by Will Sawin for non-continuous inverse Galois problem

2012-06-04T22:16:22Z

Q1: No.

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 $\mathbb Z$.

Q2: I don't know. It seems unlikely.

Answer by Yves Cornulier for non-continuous inverse Galois problem

2012-06-05T08:34:06Z

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.

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).

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 infinitely many open subgroups $H$ with $G/H\simeq S$) .