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Assuming the axiom DC($\omega_1$), is there a definition of the rank of a group ?

Another related question: assuming DC($\omega_1$), if we have two groups $A$ and $B$ of the same infinite rank, is there necessarily a surjective homomorphism from $A$ onto $B$?

Edit: in the second question , we assume $A$ and $B$ both free abelian.

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  • $\begingroup$ What is "not too big"? By rank do you mean minimal cardinality of generators? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 5, 2013 at 15:05
  • $\begingroup$ Yes by rank I mean the minimal cardinality of generators. $\endgroup$
    – user38200
    Commented Aug 6, 2013 at 1:13
  • $\begingroup$ Isn't the statement of the second paragraph plainly false? $\endgroup$
    – François G. Dorais
    Commented Aug 6, 2013 at 1:19
  • $\begingroup$ @Francois: I suppose the meaning is at least one way; otherwise yes. Take $A$ to be free abelian and $B$ free, both with the same set of generators. Should be false, I think. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 6, 2013 at 1:47
  • $\begingroup$ My guess is that both answers are negative. All sort of Lauchli-like constructions using larger supports should give you a counterexample. The second question seems to be true in $\sf ZF$, there is a bijection between the generators and it extends uniquely. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 6, 2013 at 3:14

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Without sitting to verify the details in full, here is a sketch of a proof:

Consider Lauchli's construction of a vector space with two bases of different cardinality, as outlined in Jech The Axiom of Choice in problem 10.5 (p. 149). The construction is to create two infinite sets which span isomorphic vector spaces, but have different cardinalities.

Replace "countable" by $\aleph_2$ and "finite" by $\aleph_1$. The result will satisfy $\sf DC_{\aleph_1}$, and the conclusion should still hold. Moreover, you can take the free abelian group generated by those sets instead of a vector space, and it would solve the problem.

If one prefers to do that by forcing instead of atoms, then it is also possible, although slightly more technical.

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    $\begingroup$ Asaf, for the second question, you seem to be assuming that the free abelian groups are free over the set of generators that you happen to pick. But the OP defines "rank" not in terms of the size of such a basis, but in terms of the size of a smallest generating set. Can you prove in ZF that if a free abelian group is generated by set X, the it has a basis (over which it is free) of at most the same size? $\endgroup$
    – Joel David Hamkins
    Commented Aug 6, 2013 at 4:47
  • $\begingroup$ @Joel: Ah. Silly me. I generally recall (although not in 100%) that free implying the existence of a basis is itself equivalent to the axiom of choice. This seems weaker, but probably unprovable as well. I'll remove it from the answer for now. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 6, 2013 at 4:52
  • $\begingroup$ But perhaps if such a group has a minimal size set of generators, then it has a basis? $\endgroup$
    – Joel David Hamkins
    Commented Aug 6, 2013 at 4:55
  • $\begingroup$ Joel, that is what I meant when I said weaker. I don't know th answer off-hand, nor I believe it is known generally. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 6, 2013 at 5:49

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