Is there a classification of all countable subgroups of the circle $\mathbb{T} \simeq \mathbb{R}/\mathbb{Z}$?

It seems that there are quite a lot of them, e.g.:

  • cyclic subgroups $\{a^n\colon n\in\mathbb{Z}\}$
  • finite subgroups
  • subgroups of the form $\{k/2^n\colon k,n\in \mathbb{Z}\}$ or something similar
  • direct sums of the above...

Equivalently, is there a classification of all compact monothetic groups? (Each countable subgroup of the circle can be realized as a group of eigenvalues of some probability-preserving transformation and vice versa).

  • The question is equivalent to describing the countable additive subgroups of $\mathbb{R}$ containing 1: if $A$ is such a group, write $A=D\oplus C$, where $D$ is divisible and $C$ is reduced (= no non-zero divisible subgroup). $D$ is the easy part: a vector space over $\mathbb{Q}$, hence classified by dimension. $C$ is the tricky part: you should look at the rank, but my knowledge stops short (rank 1 is equivalent to being isomorphic to a subgroup of $\mathbb{Q}$). See "Infinite abelian groups" in en.wikipedia.org/wiki/Abelian_group to have an idea of the difficulty of the question. – Alain Valette Nov 16 '11 at 14:51
up vote 7 down vote accepted

"subgroups of the form $\{k/2n:k,n∈\mathbb{Z}\}$ or something similar [and] direct sums of the above..."

Are precisely the class of divisible subgroups. See http://en.wikipedia.org/wiki/Divisible_group for a full discussion and classification. Divisible groups are well-understood, and are always direct summands. As $\mathbb{R}/\mathbb{Z}$ is divisible, we can say a lot about its structure.

$\mathbb{Q}/\mathbb{Z}$ will be the torsion subgroup, which is the direct sum of all prufer groups. This contains all subgroups of the last three types which you list.

As $\mathbb{Q}/\mathbb{Z}$ is divisible, $\mathbb{R}/\mathbb{Z}$ is isomorphic to $\mathbb{R}/\mathbb{Q}\oplus \mathbb{Q}/\mathbb{Z}$.

But $\mathbb{R}/\mathbb{Q}$ is just a $\mathbb{Q}$ vector space. So it is a direct sum of copies of $\mathbb{Q}$.

This doesn't exactly answer your question, but it makes some points.

  • 2
    Right. In particular: (1) a countable subgroup of $\mathbb R/\mathbb Z$ is the same thing as a subgroup of $\mathbb Q^{(\omega)}\oplus(\mathbb Q/\mathbb Z)$, where $\mathbb Q^{(\omega)}$ denotes the direst sum of countable many copies of $\mathbb Q$; (2) every countable torsion-free abelian group is embeddable in $\mathbb R/\mathbb Z$. The latter means that asking for classification is hopeless, even the structure of rank 2 torsion-free groups has unmanageable complexity. Cf. the answers to mathoverflow.net/questions/59978 . – Emil Jeřábek Nov 16 '11 at 14:33
  • What do you mean by unmanageable complexity? Could you give some examples of "strange" rank 2 torsion-free subgroups of $\mathbb{R}/\mathbb{Z}$? Does it have much to do with the rich structure of divisible groups? – Joanna K-P Nov 16 '11 at 16:12
  • 2
    As noted in Juris Steprans’ answer to the other question, a precise explanation of the complexity of the classification problem is provided in this paper by Thomas: jstor.org/stable/827087 . As for the second question, no, divisible groups have a completely trivial structure: they are direct sums of copies of $\mathbb Q$ and the Prüfer groups $\mathbb Z(p^{\infty})$ (in your case, there can be at most one $\mathbb Z(p^\infty)$ summand for each $p$, if the group is to fit into $\mathbb R/\mathbb Z$). In contrast, nondivisible groups are a mess. – Emil Jeřábek Nov 16 '11 at 16:28
  • Notably, for each subset of the primes, there is a ring contained in $\mathbb Q$ where one can divide by exactly those primes. Now classify all modules over those rings. – Will Sawin Nov 16 '11 at 21:01
  • 2
    Every abelian group embeds in its injective hull, which is divisible. A countable torsion free abelian group then has injective hull $\mathbb{Q}^n$, where $n\le \infty$. – Steve D Nov 17 '11 at 17:23

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.