Let $\mathfrak{c}$ be the cardinality of the continuum. How much Choice, if any, is needed to prove that there are $2^{\mathfrak{c}}$ distinct (mutually nonisomorphic) torsion-free abelian groups of cardinality $\mathfrak{c}$? This can be proved with AC, but I suspect a much weaker form of Choice, or maybe none at all, is necessary. A fairly simple argument, not requiring any choice, is that there are at least as many as there are antichains in the power set of $\omega$, and the AC implies there are $2^{\mathfrak{c}}$ such antichains. Maybe an entirely different argument is possible.
Number of torsion-free abelian groups
Bruce Blackadar
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