# Without choice, can every homomorphism from a profinite group to a finite group be continuous?

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?

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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.) – Lior Bary-Soroker Sep 3 '12 at 11:00
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. – Asaf Karagila Sep 3 '12 at 13:10
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. – Will Sawin Sep 3 '12 at 15:09

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).
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 $H$.