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: Let $H$ be a finite group. Is it always possible to find a non continuous surjective homomorphism $\rho:G\rightarrow H$?
If you think that removing the continuity assumption does make the inverse Galois problem any simpler then please provide some explanations.
Best Answer
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.