# 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.

• ...and that is why I am not going to try and edit any question long enough ever again (well not ever, but for a while...) – Asaf Karagila Jun 2 '12 at 22:11
• Some of the subgroups of $\mathbb Q$ are missing from your list, so I'd go work through the answer of that one first (e.g. the dyadic rationals, off the top of my head) – Ben Millwood Jun 2 '12 at 22:21
• This may be helpful mathoverflow.net/questions/59978/… – Norbert Jun 2 '12 at 22:52
• @benmachine - thanks - I added in dyadic rationals and link to article on them – Michaela Light Jun 2 '12 at 22:56
• @Norbert thanks for that article - looks interesting, though I hope the task is not completely hopeless! – Michaela Light Jun 2 '12 at 22:56

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.

• That is useful because it shows that there are no subgroups between the Z ones and the Q ones. However it doesn't say anything about the structure beyond Q once the subgroups have gotten dense in R. I am wondering how "many" subgroups there are between Q and the full group of R... – Michaela Light Jun 2 '12 at 23:18
• Sure there are, there are dense subgroups that don't contain $\mathbb Q$ (I said this in my other comment, just saying it here for completeness) – Ben Millwood Jun 3 '12 at 8:50
• Yes that is true. And the point I am trying to make is that I think there is a infinite hierarchy of dense subgroups between Q or these other non-Q subgroups and the full group R. And I am wondering how to classify those... My intuition says that this is somehow related to the diagonal proof of the uncountablity of the reals and the point that @rschwieb made on the related question math.stackexchange.com/questions/152263/… about stripping away leading digits doesn't matter. – Michaela Light Jun 3 '12 at 23:50
• $E(h/m)$ is whole part? thanks – user89940 Apr 4 '18 at 22:43

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.

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.

• That is a useful way to look at the problem. I know that R has a Hamel basis over Q and also any subgroup H that contains Q would also have a Hamel basis. Then we can make other subgroups by picking arbitrary subsets of these bases. I did some google searching on submodules of infinite power modules without any luck so you have any links for articles on that I would appreciate it. – Michaela Light Jun 2 '12 at 23:22
• It's not quite that simple. For example if you take $H$ to be $\mathbb Q + \pi \mathbb Z$, then $H$ does not have a Hamel basis over $\mathbb Q$, because it is not even a vector space over $\mathbb Q$ (it contains $\pi$ but not $\pi/2$). To classify additive subgroups, you'll need to look at them as $\mathbb Z$-modules, not $\mathbb Q$-vector spaces. And some of these submodules do not have bases at all -- for example, the dyadic rationals. – Henning Makholm Jun 3 '12 at 14:00
• (And unfortunately my own commutative algebra knowledge is too weak to do more than point at some of the problems you (probably) need to solve. I have no good idea of their solution). – Henning Makholm Jun 3 '12 at 14:02
• I am probably missing something here but is not the infinite set { 1/2<sup>n</sup> where n in N} a basis for the dyadic rationals of binary form ( 1/2<sup>a</sup> )? Thanks for the idea on Z-modules too! – Michaela Light Jun 3 '12 at 14:54
• @MichaelSmith: not linearly independent, since $2\cdot1/2^n - 1/2^{n-1} = 0$. – Ben Millwood Jun 3 '12 at 18:23