Is it true that there exist $2^{\aleph\_0}$ pairwise non-isomorphic torsion-free countable groups?



Number of countable torsion-free groups
Extended cw answer based on Jason Starr's comment.



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> The additive group of rationals admits $2^{\aleph\_0}$ non-isomorphic subgroups.
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Denote by $P\subset \mathbb{Z}\_{>0}$ the set of positive, integer primes. This is a countably infinite set by Euclid's proof of the infinitude of primes. The set $\mathcal{G}$ of saturated, multiplicatively closed subsets $S$ of $\mathbb{Z}$ is in bijection with the power set $\mathcal{P}(P)$ by the rule $S\mapsto S\cap P.$ 


For every saturated, multiplicatively closed subset $S$ of $\mathbb{Z},$ denote by $G\_S$ the fraction ring, $$G\_S =S^{-1}\mathbb{Z} \subset \mathbb{Q}.$$ This is a subring of the countably infinite ring $\mathbb{Q}$, thus also $G\_S$ is countably infinite. Moreover, the subset $$ \{p\in P |\ \forall x\in G\_S, \ \exists y\in G\_S, \ p\cdot y= x\}$$ equals $S\cap P.$ Thus, if $G\_S$ is isomorphic to $G\_T$ as Abelian groups, then $S$ equals $T$. Therefore, the collection of Abelian groups $G\_S$ is a system of pairwise non-isomorphic, countably infinite, torsion-free groups that are indexed by the set $\mathcal{G}$ with cardinality $2^{\aleph\_0}.$


