Number of countable torsion-free groups Is it true that there exist $2^{\aleph_0}$ pairwise non-isomorphic torsion-free countable groups?
 A: Extended cw answer based on Jason Starr's comment.

The additive group of rationals admits $2^{\aleph_0}$ non-isomorphic subgroups.

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 $2^{\aleph_0}.$
