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.

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    $\begingroup$ 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? $\endgroup$ – Kevin Buzzard Jun 4 '12 at 22:28
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    $\begingroup$ @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. $\endgroup$ – Yiftach Barnea Jun 4 '12 at 23:21
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    $\begingroup$ 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. $\endgroup$ – Ben Wieland Jun 5 '12 at 0:02
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    $\begingroup$ @Will: This only shows the set of commutators is closed. Why should the commutator subgroup be closed? $\endgroup$ – Kevin Ventullo Jun 5 '12 at 5:18
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    $\begingroup$ 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 $\endgroup$ – Benjamin Steinberg Jun 5 '12 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 $G/H\simeq S$) . 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.

  • $\begingroup$ This is a nice answer. Compact groups have fixed points on trees. $\endgroup$ – Benjamin Steinberg Jun 5 '12 at 18:40
  • $\begingroup$ @Benjamin: some compact groups admit a "horocyclic" action, namely all elements (and thus all finitely generated subgroups) have fixed points, but there's no global fixed point. The criterion for the existence of such an action is that the group has countable cofinality, i.e. is union of a sequence of proper subgroups; some profinite groups satisfy this property as an abstract group (e.g. those infinite abelian, or infinite linear over a field) and some others don't. $\endgroup$ – YCor Jun 5 '12 at 19:36

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