Every non trivial discrete subgroup of R is cyclic.
Proof: It is known that every additive subgroup of R is either cyclic or dense in R, and that a non trivial subgroup H is cyclic when inf ($R_{\gt 0}\cap H)>0$. Let H be a non trivial subgroup of R. Suppose on the contrary that it is not cyclic. Hence it is dense in R. There exists h in H, $h\in (0,1)$. Since H is discrete, there exists an open interval (a,b), a>0, b<1, a>b such that ${h}=(a,b)\cap H$. By density (a,(a+h)/2) should contain an element of H but that's not possible, which is a contradiction.