# How much choice is required for a countably-infinite index subgroup of the real additive group?

The existence of such subgroups implies the existence of a non-measurable set; simply intersect each of the cosets with $$[0,1]$$. The results will all have equal outer measure, but their union will be all of $$[0,1]$$, and you only have countably many of them. As a result, they are non-measurable.

So, for example, in the Solovay model, these subgroups do not exist. However, with choice, you can let $$B$$ be a Hamel basis, then using that to generate a surjective homomorphism $$F$$ from $$\mathbb{R}$$ to $$\mathbb{Q}$$ (as in here) and consider all of the elements $$x$$ such that $$F(x)=0$$. This forms a subgroup under addition (since its a homomorphism) and then its cosets are those that contain all $$y$$ such that $$F(y)=q$$ for different rationals $$q$$; as a result, the group has countably infinite index.

Now, as shown in the question linked to above, the existence of such a subgroup in $$\text{ZF+DC}$$ is undecidable (models where all sets of reals are Borel for example). However, the existence of a Hamel basis is enough to prove the existence of one of these subgroups.

I believe that the existence of one of these subgroups is not equivalent to the existence of a Hamel basis under $$\text{ZF}$$. Perhaps it is possible for there to be a model in which a countably-infinite subgroup exists but its cosets do not correspond to different rational numbers as above, in which case theoretically the Hamel basis would be unnecessary.

Question: How much choice is required for there to be a subgroup of the real additive group with countably-infinite index?