How can we find and categorize the subgroups of $\mathbb{R}$? $\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}\newcommand{\Z}{\Bbb Z}$
What are all the subgroups of R =  $(\R, +)$ and how can we categorize them? 
I started thinking about this question last night after looking at the structure of the cosets of $\R / \Q$ What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?. I did some searching on SO and google but didn't find anything giving a full categorization (or even a partial one) of the subgroups of $\R$. 
Here are the subgroups that I came up with so far:


*

*$\Z$ (there are no finite subgroups and $\Z$ is the universal smallest subgroup I think)

*n$\Z$ eg 2$\Z$ all even numbers

*a$\Z$ where a is any real number, including a in $\Q$ which "nest" nicely in each other

*$\Z$[a] - group generated by adding one real a to $\Z$

*n$\Z$[a] which equals $\Z$[na] and so is just a case of the one above

*Dyadic rationals  eg a numbers of the form a/2b or similar subgroups such as a/3b, a/2b7c etc

*$\Q$

*$\Q$[a]

*$\Q$[a in A] where A is a subset of $\R$ - could be finite, countable or uncountable. Group generated by adding all elements of A to $\Q$
eg  $\Q[\sqrt2]$


It is clear that the "n$\Z$ subgroups" n$\Z$ and m$\Z$ are related according to the gcd(n,m)
Also when H is a subgroup of R looking at the structure of the cosets of R / H. eg for H any of the Z subgroups we get R / H homomorphic to [0,1) or the circle. For H one of the Q subgroups it is more complex and I currently don't have ideas on the larger subgroup cosets
I am not clear how "big" a subgroup H can get before it becomes the whole of R. I do know that if it contains any interval then it is the whole of R. But what about H with dimension less than 1?
I am aware of one question on SO about the proper measurable subgroups of R having 0 measure Proper Measurable subgroups of $\mathbb R$, one on dense subgroups Subgroup of $\mathbb{R}$ either dense or has a least positive element? and one on the subgroups of Q How to find all subgroups of $(\mathbb{Q},+)$ but that is all my searching found so far.
Why is this question interesting? 1) there seem to be so many subgroups and they are related in many groupings 2) I think the subgroups related to the structure of the reals in some subtle ways 3) I know the result for the complete classification of all finite subgroups was a major result so wondering what has been done in this basic uncountable case.
If anyone has any insight, intuition, info, papers or theorems on subgroups of R and how they are interrelated that would be interesting. 
 A: The subgroups of $(\mathbb{R},+)$ are up to isomorphism the torsion-free abelian groups of rank $\alpha$ for every cardinal $\alpha\leq 2^{\aleph_0}$, because $(\mathbb{R},+)$ is isomorphic to $(\mathbb{Q}^{(2^{\aleph_0})},+)$ (weak direct product) as a $\mathbb{Q}$-vector space and thus also as a group. 
The torsion-free abelian groups of rank $2$ already haven't been classified yet and it seems difficult to do so (cf. The classification problem for torsion-free abelian groups of finite rank).
László Fuchs book "Infinite abelian group theory" contains a lot of interesting shizzle related to this.
A: If you're only interested in the additive structure of $\mathbb R$, then the best algebraic description of it is that it is a continuum-dimensional vector space over $\mathbb Q$. Your best way forward might then be to try to classify additive subgroups of $\mathbb Q$, and then look to commutative algebra for an answer to what the submodules of an infinite power of modules are.
A: We can show that subgroups of $\mathbb R$ are :
$x\mathbb Z$ $(x\in\mathbb R)$ or dense in $\mathbb R$.
$0=0\mathbb Z$. 
So we can suppose that $H$ is a subgroup of $\mathbb R$ not $0$. 
So, let $H^+=\{h\in H,h>0\}\neq\varnothing$ ($\exists h\in H$, if $h <0$, take $-h$). Now let $m=\inf H^+$. We'll show :
(i) if $m\neq 0$, $H=m\mathbb Z$. 
(ii) or $H$ is dense in $\mathbb R$.
(i) Use $\inf$ properties to show that $m\in H^+$, so $m\mathbb Z\subset H$. To show the converse inclusion : let $h\in H$, let $g=E\left(\frac{h}{m}\right)$, we easily show that $h-mg=0$. 
(ii) Now suppose $m=0$. Let $x<y$ in $H$, let $h\in H^+$ such as $0<h<y-x$. Let $n=E\left(\frac{x}{h}\right)$ then $x<hn+h\le x+h<y$. Let $g=h(n+1)$, so $x<g<y$ and $n+1\in\mathbb Z$, so $g\in H$ (subgroup...). So $g\in ]x,y[\cap H$. QED.
K .H.
