Can free groups on different cardinals be isomorphic in ZF?

Question

For the free group on a finite number of generators, $F_n$, this is simple. It is enough to find a group that can be a quotient of one and not the other, so it is sufficient to find a group that can be generated by $n$ generators and no less. The group $\mathbb{Z}_2^n$ works. Similarly, $\mathbb{Z}_2^{(\aleph_0)}$ cannot be finitely generated, showing the countably generated free group is not isomorphic to a finitely generated free group.

With the axiom of choice, for any infinite cardinal $\kappa$, $|F_{\kappa}|=\kappa$, so cardinality sorts out all the other infinite free groups, showing $F_{\kappa}=F_{\lambda} \iff \kappa = \lambda$. However, this is clearly requires the axiom of choice. In ZF, this proves free groups on different well-orderable cardinals are diferent, but not all cardinals.

So I would like to know: is it consistent with ZF that free groups on two different cardinals can be isomorphic? And if so, which cardinals work, and what is the proof?

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My thinking:

The proof in ZFC doesn'tn work here, as it is possible for $\mathbb{Z}_2^{(\kappa)} \cong \mathbb{Z}_2^{(\lambda)}$ for $\kappa \ne \lambda$.

Free groups are much like direct sum vector spaces in that they don't need any choice to define them and get a generating set, so there's no obvious reason choice would be necessary for this. It doesn't seem like being given an isomorphism between free groups on different cardinals would allow you to prove anything choicy about those cardinals. But there also isn't any obvious approch to proving they're not isomorphic, and when there's no obvious proof for something simple, it seems to usually be independent.

I first thought that the universal property might be enough to distinguish them, but there are other algebraic categories where free objects on different cardinals are isomophic, even for finite cardinals (Quick proof that free objects on sets of different cardinality are not isomorphic?). As such, this seems a no-go. It also makes it look very likely that different cardinals can give isomorphic free groups.

I then thought that I could leverage the fact that the same vector space can have different sized bases: if I quoutient out the necessary relations to make it commutative and self inverse, $F_\kappa \to \mathbb{Z}_2^{(\kappa)}$. So this natural quotient can make free groups on different cardinals turn into isomorphic groups. It would be sufficient to prove that this quotient can't make non-isomorphic free groups become isomorphic. But this quotient can clearly make non-isomorphic groups isomorphic in general, and it may well be able to do so with free groups. And I don't know how to prove ths either way.

Just knowing about some of the oddities possible in cardinal arithmetic without choice doesn't seem to help either, e.g. I'm pretty sure different cardinals can generate free groups of the same cardinality, but they might not have the same group structure.

It seems like the methods I have knowledge of are not suffient.

Answer 1

This appears to still be an open problem, though I have not conducted an exhaustive search of the literature.

For simplicity, we call the statement "For all $S$, $R$, if $F_S \cong F_R$ then $|S| = |R|$" the Cardinal Theorem for Groups. Here, $F_S$ is the free group on $S$.

This answer cites some relevant papers. One paper from 1985 states on its second to last page that the problem is still open. This was after some substantial developments in forcing, so I doubt there are basic tools available today that will obviously and simply solve the problem. The paper is

Paul E. Howard, Subgroups of a Free Group and the Axiom of Choice. The Journal of Symbolic Logic, Vol. 50, No. 2 (Jun., 1985), pp. 458-467.

and it states that it is unknown whether one can prove "the cardinal number of a free generating set of a free group is determined by the group" without the axiom of choice. This is precisely the Cardinal Theorem (which Howard calls the CNG).

A paper from 2014 indicates that one can prove that free groups on different cardinals are not isomorphic from the Boolean Prime Ideal Theorem, which indicates that this proposition is strictly weaker than the axiom of choice. The author also proves that that the proposition "$\kappa$ is determined by $|F_\kappa|$" is equivalent to the axiom of choice. So an argument based purely on cardinality surely cannot work. That paper is

Philipp Kleppmann, Generating sets of free groups and the axiom of choice. Mathematical Logic Quarterly, Vol. 60, No. 3 (2014), 239–241.

Finally, note that for any field $F$, there is a model of ZFA in which one can find a vector space with two bases of differing cardinalities. This is exercise 10.5 in Jech's The Axiom of Choice. One can then use an embedding theorem to conclude there are models of ZF where this holds. So the analogous question for vector spaces has a known answer. I believe the methodology laid out in Jech's hints ought to apply to free Abelian groups as well, but I don't have time today to verify the details. However, I didn't see an immediate way of extending his idea to free groups.

Edit: after reviewing the literature some more, I note that Halpern proved in 1965 that the Boolean Prime Ideal Theorem (BPIT) implies that any two bases of a vector space have the same cardinality. That any two bases of a vector space have the same cardinality implies the Cardinal Theorem for Groups.

Halpern, J. (1966). Bases in vector spaces and the axiom of choice. Proceedings of the American Mathematical Society.

For fix any field $k$, and let $Vec$ be the category of vector spaces on it with linear maps. Then the forgetful functor $U_{Vec \to Grp} : Vec \to Grp$ has a left adjoint $F_{Grp \to Vec}$. We also have free/forgetful adjunctions $F_{Set \to Vec} \dashv U_{Vec \to Set}$ and $F_{Set \to Grp} \dashv U_{Grp \to Set}$ (note we previously denoted $F_{Set \to Grp}(S)$ by $F_S$). Adjunctions compose, so $F_{Grp \to Vec} \circ F_{Set \to Grp} \dashv U_{Grp \to Set} \circ U_{Vec \to Grp} = U_{Vec \to Set}$. By uniqueness of adjoints, we have $F_{Grp \to Vec} \circ F_{Set \to Grp} \cong F_{Set \to Vec}$.

Now suppose we have sets $S, R$ with $F_{Set \to Grp}(S) \cong F_{Set \to Grp}(R)$. Then we have $F_{Set \to Vec}(S) \cong F_{Grp \to Vec} (F_{Set \to Grp}(S)) \cong F_{Grp \to Vec} (F_{Set \to Grp}(R)) \cong F_{Set \to Vec}(S)$, and thus, $|S| = |R|$. So it really should have been clear in 1965 that the Cardinal Number Theorem for Groups follows from BPIT, and it was known not long after that that BPIT is strictly weaker than the full axiom of choice. It certainly shouldn't still have been an open problem in 1985, and definitely not in 2014. But I am too young to remember having to actually look through physical copies of journals, so perhaps I underestimate the difficulties of doing a proper literature review and making the right connections.

In fact, if you read through Kleppmann's 2014 paper and Halpern's 1965 paper, they both use the same combinatorial result which follows from BPIT. Kleppman even cites Halpern's proof of this theorem, contained in Halpern's 1965 paper, and remarks on the similarities between their proofs.