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$\newcommand{\R}{\Bbb R}\newcommand{\Q}{\Bbb Q}\newcommand{\Z}{\Bbb Z}$ 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.

Community
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Michaela Light
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  • ...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 02 '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 02 '12 at 22:21
  • This may be helpful http://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals – Norbert Jun 02 '12 at 22:52
  • @benmachine - thanks - I added in dyadic rationals and link to article on them – Michaela Light Jun 02 '12 at 22:56
  • @Norbert thanks for that article - looks interesting, though I hope the task is not completely hopeless! – Michaela Light Jun 02 '12 at 22:56
  • There are more like dyadic rationals (I mean, just think of $3^b$ on the denominator, or possibly $2^a3^b$, or ...) I didn't mean so much "you forgot the dyadics!" as "you forgot some subgroups of the rationals, maybe you should understand those fully first" – Ben Millwood Jun 03 '12 at 08:47
  • @benmachine - thanks for clarifying, I will look into those more and I found a proof of the classification of subgroups of Q at http://answers.yahoo.com/question/index?qid=20080425160119AA19f8E which shows that they are all either finitely generated or of the type of the dyadic rationals. – Michaela Light Jun 03 '12 at 14:47
  • Oops, I meant "there are more subgroups that are similar to the dyadic rationals", rather than "there are more dyadic rationals". The original description you had of the dyadics as $a2^{-b}$ was correct. – Ben Millwood Jun 03 '12 at 18:25
  • I am thinking that the structure of the subgroups of R might be clearer if we look at the generators of them. For example Z = [1], nZ = [n], aZ = [a], Dyadic =[1/2^n n in N], Q = [1/p p prime] etc. Then it is clearer which subgroups are subgroups or supergroups of others. Perhaps this idea can be extended to infinite (uncountable?) sets of irrational generating elements... – Michaela Light Jun 03 '12 at 23:43
  • @benmachine - thanks for correction of definition of dyadic rationals - I updated the question text to reflect this. – Michaela Light Jun 03 '12 at 23:44

3 Answers3

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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.

Kris H.
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  • 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 02 '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 03 '12 at 08:50
  • 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 http://math.stackexchange.com/questions/152263/what-do-the-cosets-of-mathbbr-mathbbq-look-like about stripping away leading digits doesn't matter. – Michaela Light Jun 03 '12 at 23:50
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    $E(h/m)$ is whole part? thanks – user89940 Apr 04 '18 at 22:43
  • what mean the capital $E$, $E(h/n)$, thanks – user89940 Jul 28 '23 at 03:00
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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.

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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.

hmakholm left over Monica
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  • 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 02 '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. – hmakholm left over Monica Jun 03 '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). – hmakholm left over Monica Jun 03 '12 at 14:02
  • I am probably missing something here but is not the infinite set { 1/2n where n in N} a basis for the dyadic rationals of binary form ( 1/2a )? Thanks for the idea on Z-modules too! – Michaela Light Jun 03 '12 at 14:54
  • @MichaelSmith: not linearly independent, since $2\cdot1/2^n - 1/2^{n-1} = 0$. – Ben Millwood Jun 03 '12 at 18:23
  • @benmachine Oh, thanks for that example! That is interesting - we have a set that spans the subgroup but no subset is linearly independent! In fact we can remove any number of elements (finite or infinite) from this spanning set and so long as an infinite number of elements remain the subset still spans the subgroup... However I am not clear how to prove that no basis exists for dyadic subgroups of Q. I tried to get a contradiction from expressing 1/2^n as Z linear combination of basis elements but don't see that yet. – Michaela Light Jun 03 '12 at 23:34
  • @MichaelSmith: Any two elements of $\mathbb Q$ are linearly dependent over $\mathbb Z$, so a basis for any subgroup can contain at most one element. On the other hand, the set generated by such an element has bounded denominators-in-lowest-terms, so it cannot be the dyadic rationals. – hmakholm left over Monica Jun 04 '12 at 10:58