Timeline for Universal group on $\kappa$ elements
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 8, 2019 at 15:34 | comment | added | Asaf Karagila♦ | @Paul: That was the bit I was missing, infinitely presented. Thanks! | |
| Oct 8, 2019 at 15:32 | comment | added | user35370 | @Asaf You can have infinitely presented, finitely generated, groups. You can arrange for them to be nonisomorphic too (for example by some small cancellation theory, although there are other constructions). Also there are embedding theorems where you can arrange that for any countably group there is an embedding into a 2-generated group. | |
| Oct 8, 2019 at 14:37 | comment | added | YCor | @AsafKaragila Yes, I am, it's no contradiction. Continuum many pairwise non-isomorphic 2-generated groups. And since you talk about formulas, continuum pairwise non-elementary equivalent 2-generated groups. Precisely for every set $J$ of primes there exists a 2-generated group $G$ in which for $p$ prime, the formula $\exists g\neq 1:g^p=1$ holds in $G$ iff $p\in J$. | |
| Oct 8, 2019 at 13:55 | comment | added | Asaf Karagila♦ | @YCor: Since there are only countably many formulas in the language of groups, I'm not sure how you'd make this calculation. Are you sure? | |
| Oct 8, 2019 at 13:54 | comment | added | YCor | @AsafKaragila yes this is what I wrote. | |
| Oct 8, 2019 at 13:36 | comment | added | Asaf Karagila♦ | @YCor: Continuum of finitely generated groups? | |
| Oct 8, 2019 at 8:09 | comment | added | Gabe Conant | Quoting from page 3 of Banach spaces and groups - order properties and universal models by Shelah and Usvyatsov: "Note that every AEC with amalgamation has a universal model in every regular $\lambda$ satisfying $\lambda=\lambda^{<\lambda}$ or $\lambda=\mu^+$ and $2^{<\mu}\leq\lambda$." So it seems one could take $\lambda=\beth_\omega^+$. | |
| Oct 8, 2019 at 7:47 | comment | added | YCor | Jónsson (1956) proved that under GCH, the answer is yes for every uncountable $\kappa$, see Theorem 3.1 therein. However to get the result for a single $\kappa$ probably does not require GCH, see notably Theorem 2.9 (unfortunately I can't find a definition for $\mathfrak{n}_\mu$ used there, so I'm not sure of meaning of the assumption). | |
| Oct 8, 2019 at 7:24 | comment | added | YCor | For $\kappa=2^{\aleph_0}$, I think this is an open question. Actually, concerning the group $G=\mathrm{Sym}(\aleph_0)/\mathrm{fin}$, the question whether every group of continuum cardinal embeds into $G$ is open, and is also open in ZFC+CH; in (ZFC + $2^{\aleph_0}=\aleph_2$) it consistently has a negative answer. | |
| Oct 8, 2019 at 7:18 | comment | added | YCor | Just for context, the answer is no for $\kappa<2^{\aleph_0}$, using that there exists continuum many non-isomorphic finitely generated groups, and that each countable group embeds in one of them. | |
| Oct 8, 2019 at 6:51 | history | asked | Dominic van der Zypen | CC BY-SA 4.0 |