Timeline for Classification of countable subgroups of the circle
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 17, 2011 at 17:23 | comment | added | Steve D | 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 $n\le \infty$. | |
| Nov 17, 2011 at 14:19 | vote | accept | Joanna K-P | ||
| Nov 17, 2011 at 14:18 | vote | accept | Joanna K-P | ||
| Nov 17, 2011 at 14:18 | |||||
| Nov 17, 2011 at 12:56 | comment | added | Joanna K-P | Thanks. Now I see that it is really a complex problem to try to classify these subgroups. Is there any reference to the fact that every countable torsion-free abelian group in embeddable in $\mathbb{R}/\mathbb{Z}$ or is it some folklore fact? | |
| Nov 16, 2011 at 21:01 | comment | added | Will Sawin | 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. | |
| Nov 16, 2011 at 16:28 | comment | added | Emil Jeřábek | 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. | |
| Nov 16, 2011 at 16:12 | comment | added | Joanna K-P | 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? | |
| Nov 16, 2011 at 14:33 | comment | added | Emil Jeřábek | 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 . | |
| Nov 16, 2011 at 14:01 | history | answered | Jonathan Kiehlmann | CC BY-SA 3.0 |