## non-continuous inverse Galois problem

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.

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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.

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Will: for Q1 are you saying/implying that the abstract commutator subgroup is closed? Why is this? Also surely the answer to Q2 is "yes", at least if you believe the inverse Galois problem. That said, this question seems very pathological to me. I can't really see any logic in considering a Galois group without its topology -- somehow the topology is what makes the Galois group worth studying. Do people really care about just the abstract group structure? – Kevin Buzzard Jun 4 at 22:28
@Kevin: from number theortical and Galois theory points of view the abstract group might not be interesting. However, there is an increasing intrest amongst group theoriests in the abstract structure of profinite groups. So yes it could be of some interest. You could ask Nik Nikolov a bit more about it. – Yiftach Barnea Jun 4 at 23:21
No profinite group can discontinuously surject to $\mathbb Z$. If so, choose a splitting and take the closure of the image. This is a commutative profinite group surjecting to $\mathbb Z$. In fact, it is a quotient of the free profinite group on one generator, $\hat {\mathbb Z}$, and Will shows this group doesn't map to $\mathbb Z$.....It is worth mentioning that on a finitely generated profinite group, the topology is intrinsic. – Ben Wieland Jun 5 at 0:02
@Will: This only shows the set of commutators is closed. Why should the commutator subgroup be closed? – Kevin Ventullo Jun 5 at 5:18
I'm pretty sure it falls out the Segal and Nikolov results that the abstract commutator subgroup is closed for finitely generated profinite groups. I believe it is false in general and that counterexamples can be found in the book of Ribes and Zalesskii – Benjamin Steinberg Jun 5 at 5:33
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 $G/H\simeq S$) .