Timeline for Is there a universal countable group? (a countable group containing every countable group as a subgroup)
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2019 at 16:30 | comment | added | Maxime Ramzi | @Misha sorry for the late response; a countable group has only countably many finite subsets, hence countably many finitely generated subgroups | |
| Dec 15, 2013 at 21:02 | comment | added | Misha | Sorry for the late question about this answer, but why a f.g. group contains only countably many isomorphism classes of subgroups? | |
| Aug 30, 2010 at 18:27 | vote | accept | Joel David Hamkins | ||
| Jun 21, 2010 at 23:19 | comment | added | Simon Thomas | My brain isn't working too well as I've just arrived home from London ... but surely a routine amalgamation argument shows that if $CH$ holds, then there is a group of size $\omega_{1}$ which contains every group of size $\omega_{1}$? Of course, if $CH$ fails, then no group of size $\omega_{1}$ can include every finitely generated group. So we end up with some potentially interesting independence results. | |
| Jun 21, 2010 at 22:58 | comment | added | Joel David Hamkins | Thanks very much! Does it generalize to larger cardinals $\kappa$? | |
| Jun 21, 2010 at 22:03 | comment | added | HJRW | Oh, right. Scratch that. I wasn't thinking straight. | |
| Jun 21, 2010 at 21:58 | comment | added | Simon Thomas | Not quite ... an infinite 2-generator group often has uncountably many countable subgroups. So it seems necessary to prove that there are uncountably many finitely generated groups up to isomorphism. | |
| Jun 21, 2010 at 21:52 | comment | added | HJRW | Surely this is an immediate consequence of Higman, Neumann and Neumann's proof that you can embed any countable group in a 2-generator group? | |
| Jun 21, 2010 at 21:44 | history | answered | Simon Thomas | CC BY-SA 2.5 |