Does anyone know if there is a classification of the subgroups of the real numbers taken under addition? If not can anyone point me in the directiong of any papers/materials which discuss properties of or interesting facts about these subgroups?
If you would like to classify the subgroups in the sense of Lebesugue measure, you may find the following facts helpful.
(1) Any measurable proper subgroup of the real line is of measure $0$.
(2) Any nonmeasurable subgroup $G$ of the real line charges fully everywhere, i.e., for any interval $I$, $m^{\ast}(G \cap I)=I$, where $m^{\ast}(\cdot)$ denotes the outer Lebesgue measure.
(3) Nonmeasurable subgroup of the real line exists.


1$\begingroup$ @Hans, for (1) use Steinhaus' interior point theorem (immediate); for (2) consider the "distrubution function" $F(x)=m^*([0,x]\cap G)$; for (3) construct a nonmeasurable additive functional whose kernel gives such a group. I think (1) and (3) should be available in some articles. $\endgroup$ – Syang Chen Feb 12 '12 at 2:46
Every torsionfree abelian group of cardinality at most $2^\omega$ is isomorphic to a subgroup of the reals. (To see this, note that any such group can be embedded in a divisible torsionfree group of the same cardinality, i.e., a vector space over $\mathbb Q$, which can in turn be embedded in any other vector space over $\mathbb Q$ of the same or greater dimension.) Since already the structure of rank 2 abelian groups is hopelessly complicated, you are not going to find any sensible classification.
I'm surprised no one has yet stated the most obvious fact (though I guess Xianghong's answer comes pretty close), which is that an additive subgroup of the reals is either of the form $a\mathbb{Z}$ or is dense in the real line (an obvious consequence from division with remainder).
The 2002 paper of Simon Thomas in JAMS provides a precise measure of "hopelessly complicated". As the rank of the groups increases, so does the complexity of the classification problem.

2$\begingroup$ And although this is in another direction of hopeless complication, adding a finite rank additive subgroup to the reals with the field structure allows one to define the integers. To see this, take the set of reals for which multiplication by $r$ is a map from $G \rightarrow G.$ Take the fraction field of this set. This is a finite degree extension of $\mathbb{Q}.$ Now results of J. Robinson give the definability of the integers. So, this is another instance of hopeless complication. $\endgroup$ – James Freitag Mar 29 '11 at 17:20
If you're interested in topological classification, then this might be useful: Farah and Solecki  Borel subgroups of Polish groups, Advances in Mathematics 199, 2006, 499541.
Among a lot of other things, one of their results shows that for any countable ordinals $\alpha \neq 2$ and $\beta \geq 2$, there are $\Pi_{\alpha}^{0}$complete and $\Sigma_{\beta}^{0}$complete additive subgroups of any uncountable polish group. For connected abelian polish groups, this was previously shown by Mauldin by refining a result of Klee.

$\begingroup$ Here are some concrete examples of such groups: mathoverflow.net/questions/34369/… $\endgroup$ – Ashutosh Mar 29 '11 at 17:44

$\begingroup$ Adding temporary comment just to update linked questions in the sidebar: On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational. $\endgroup$ – Martin Sleziak Mar 3 '18 at 7:53
For every real number $\alpha$ with $0<\alpha<1$, there is an uncountable subgroup of $\mathbb{R}$ with Hausdorff dimension $\alpha$: see e.g. https://perswww.kuleuven.be/~u0018768/artikels/actionsfreegroup.pdf
Slightly offtopic, a weird subgroup in two dimensions is constructed in this paper:
Ryuji Maehara. On a connected dense proper subgroup of ${\bf R}^2$ whose complement is connected . Proc. Amer. Math. Soc. 97 (1986) 556558. MR 840645.
That subgroup is constructed as a graph of a wild group homomorphism $f:\mathbf{R}\to \mathbf{R}$, so it is also a subgroup: $\{ (x,f(x)) \mid x\in \mathbf{R}\}$

1$\begingroup$ Clickable: Maehara  On a connected dense proper subgroup of $R^2$ whose complement is connected (MSN). $\endgroup$ – LSpice Jul 8 at 18:00

$\begingroup$ Thaanks. I'll get an opportunity to read again, hope it is free. $\endgroup$ – P Vanchinathan Jul 9 at 8:04
You may be interested in the following elementary proofs: