In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the axiom of choice: ultrafilters in the first answer, and "every vector space has a basis", in Milne's notes as referenced in the second answer, and used to compute the number of finite-index subgroups in the third answer.

Is it possible to prove the existence of a discontinuous homomorphism from a profinite group to a finite group without the axiom of choice? Instead is it consistent with ZF that there is none?

  • $\begingroup$ I'm not sure that this is going to help, but anyway, since the easiest profinite group that has a non-open finite index subgroup (assuming ZFC) is $G=(Z/2Z)^{\aleph_0}$, one should check there. I'm a complete ignorant in set-theory, so I do not even know whether a finite index subgroup of $G$ is necessarily isomorphic to $G$ without AC. (In ZFC it's trivial since they have the same dimension.) $\endgroup$ – Lior Bary-Soroker Sep 3 '12 at 11:00
  • $\begingroup$ I'm not 100% sure whether you expect an answer which exhibit a profinite group that provably has a discontinuous epimorphism onto a finite group; or a profinite group which provably has only continuous epimorphisms onto finite groups. $\endgroup$ – Asaf Karagila Sep 3 '12 at 13:10
  • $\begingroup$ A profinite group which provably has only continuous epimorphisms onto finite groups is easy to find: Just take $\mathbb Z_p$, or $\mathbb Z/p$, or the trivial group. I want either a profinite group that has discontinuous homomorphisms to finite groups, or a proof that all such homomorphisms are continuous in some model of ZF. $\endgroup$ – Will Sawin Sep 3 '12 at 15:09

Saharon Shelah (elaborating on Solovay) proved that it is equiconsistent with ZF that there exists a model of ZF+DC (DC=dependent choice, ) in which all sets of reals have the Baire property. And indeed all subsets of any Polish space have this property.

Now if $G$ is a second countable (=metrizable) profinite group, it is a Polish space (indeed a Cantor), and if $H$ is a non-open finite index subgroup, $H$ cannot have the Baire property, by a standard argument (it would have to be meager, which is absurd).

Hence it is consistent with ZF(+DC) that there is no discontinuous homomorphism from a second countable profinite group to a finite group.

On the other hand, I don't know if one can reduce the general case to the second countable one, e.g. by finding a countable intersection of open normal subgroups inside $H$.

  • $\begingroup$ The second countability property tends to be called 'countably based' in the profinite groups literature. $\endgroup$ – Colin Reid Sep 3 '12 at 13:51

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