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).

  • $\begingroup$ 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. $\endgroup$ Nov 16, 2011 at 14:51

1 Answer 1


"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
    $\begingroup$ 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 . $\endgroup$ Nov 16, 2011 at 14:33
  • $\begingroup$ 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? $\endgroup$
    – Joanna K-P
    Nov 16, 2011 at 16:12
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 16, 2011 at 16:28
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Nov 16, 2011 at 21:01
  • 2
    $\begingroup$ 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 $n\le \infty$. $\endgroup$
    – Steve D
    Nov 17, 2011 at 17:23

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