TITLE: Countable Dense Sub-Groups of the Reals...
QUESTION [5 upvotes]: Can countable dense additive subgroups of the reals be well-ordered up to isomorphism by inclusion?
If so, is $\mathbb{Q}$ the smallest (up to isomorphism) countable dense subgroup of the reals, and what is the second smallest (up to isomorphism)?

REPLY [5 votes]: This should be a comment to Joel's answer, but it got too long.  Joel has exhibited a good part, but not quite all, of the classification of non-trivial subgroups of the rationals (also known as the classification of rank-one, torsion-free, abelian groups), a classification which, if I remember correctly, goes back to Reinhold Baer in the 1930's.  For the record, here's the classification.  Let $s$ be a function from a subset $D$ of the set of primes into the non-negative integers.  Associated to $s$ is the group $G_s$ of those rational numbers expressible as $a/b$ with integers $a$ and $b$ such that, for each prime $p\in D$, the power of $p$ in the prime decomposition of $b$ is at most $p^{s(p)}$.  (Primes not in $D$ can occur arbitrarily often in denominators.)  Then the non-zero subgroups of $\mathbb{Q}$ that contain the integers are exactly these $G_s$'s.  (Note that, up to isomorphism, containing the integers is unimportant, as it can always be achieved by rescaling.)  Two of them, say $G_s$ and $G_t$, are isomorphic iff $s$ and $t$ have the same domain and agree at all but finitely many points in that domain.  All these groups are dense in the reals, except for those isomorphic to $\mathbb Z$, i.e., those of the form $G_s$ where the domain of $s$ consists of all the primes and $s(p)=0$ for all but finitely many $p$.  
Note that all of this concerns only groups of rank 1, i.e., those in which every two elements are linearly dependent over the rationals.  For groups of higher but still finite rank, things get more complicated --- in a precise sense: If I remember correctly, the complexity of the isomorphism problem for torsion-free abelian groups of rank $n$ is known to be strictly increasing (in the sense of Borel reducibility) as $n$ increases.