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.
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.
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 $G/H\simeq S$) . Edit: there's indeed an example: Nikolov and Segal in "Remarks on profinite groups...", preprint 2013 (arxiv link) have an example of a profinite group with $\mathbf{Z}/p\mathbf{Z}$ as a quotient as an abstract group, but not as a topological quotient. The idea is to find large perfect groups $G_n$ for which the proportion $p_n$ of central elements of order $p$ is big, in the sense that $p_n^{-k}\ll |G_n|$ for all $k$; (e.g., consider central extensions of $\text{SL}_2(\mathbf{F}_p)\ltimes(\mathbf{F}_p^2)^n$ by $\mathbf{F}_p^{n(n+1)/2}$, so $|G_n|\sim p^{3+2n+n(n+1)/2}$ while $p_n^{-1}\simeq p^{3+2n}$), then in the product $G=\prod G_n$, denoting $S=\{x^p[y,z]:x,y,z\in G\}$, the set of products of $k$ elements of $S$ is a closed subset with empty interior and therefore (using Baire) $S$ does not generate $G$, so if $N=\langle S\rangle$ then $G/N$ is an uncountable $p$-elementary abelian group and thus $G$ admits $\mathbf{Z}/p\mathbf{Z}$ as an abstract quotient although $G$ is topologically perfect (replace $\mathbf{F}_p$ by $\mathbf{F}_{p^2}$ if $p=2,3$). Still, I don't know any example with $S$ nonabelian.
