Number of torsion-free abelian groups

Question

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.

Answer 1

You can construct antichains of size $\frak c$ without using choice, you can even have them to be antichains in a stronger sense of the word: i.e. every two have a finite intersection.

The one thing you'd want choice for this is to make sure your antichains are maximal. So we can't do that. But do we need that? No. Once you have one antichain of size $\frak c$, you have $2^\frak c$ of them automatically.

To see how you'd go about doing that, enumerate the rational numbers as $q_n$, then for every real number $r$ construct a rational sequence converging to it: $r_n$, the $n$th element in the sequence approaching $r$, would be the least indexed rational in the enumeration whose distance from $r$ is at most $\frac1n$.

Now look at $A_r=\{n\in\omega\mid\exists k: q_n=r_k\}$. Then $\{A_r\mid r\in\Bbb R\}$ would an antichain as wanted.

(Note that we can easily get $2^\frak c$ distinct antichains by partitioning the reals into $\omega$ intervals, and then replacing $\frac1n$ in the definition of $r_n$, by $1/f(k,n)$ when $r$ is in the $k$th interval, for any function $f\colon\omega\times\omega\to\omega$, satisfying that $\lim_{n\to\infty}f(k,n)=\infty$ for every $k$.)

Answer 2

Consider a set $\mathcal{C}$ of cardinal $\mathfrak{c}$ of infinite subsets of primes such that no $S\in \mathcal{C}$ is contained in a finite intersection of other elements of $\mathcal{C}$ (easy to produce: consider a bijection from primes to $\mathbf{Q}$ and consider the set of intervals of length 1 intersected with $\mathbf{Q}$).

For any $S\in \mathcal{C}$, consider $H_S=S^{-1}\mathbf{Z}=\mathbf{Z}[1/p:p\in S]$. This is a torsion-free abelian group of $\mathbf{Q}$-rank 1.

For any $\mathcal{A}\subset\mathcal{C}$, define $G_A=\bigoplus_{S\in \mathcal{A}}H_S$. This is a torsion-free abelian group of cardinal $\mathfrak{c}$.

If $x\in G_{\mathcal{A}}$ and $\mathcal{B}\subset\mathcal{A}$ is its finite support, then the set $P_x$ of primes $p$ such that $x$ is $p$-divisible (in the sense that $x\in\bigcap_n p^nG_{\mathcal{A}}$) is $\bigcap\mathcal{B}=\{p:\forall S\in\mathcal{B}:p\in S\}$. In particular, if $x$ is a nonzero element of $H_S$, $P_x=S$. Conversely, if $S\in\mathcal{C}\smallsetminus\mathcal{A}$, then $S$ is not of the form $\bigcap\mathcal{B}$ for any finite subset $\mathcal{B}$ of $\mathcal{C}$. Thus the set of $S\in\mathcal{C}$ such that there exists $x\in G_{\mathcal{A}}$ such that $P_x=S$ is precisely $\mathcal{A}$. This proves that the $G_{\mathcal{A}}$ are pairwise non-isomorphic when $\mathcal{A}$ ranges over subsets of $\mathcal{C}$, and they are $2^\mathfrak{c}$ in number.