Timeline for Embedding f.g. groups in 2-generated groups
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7 at 9:52 | vote | accept | Sean Eberhard | ||
| Jul 27 at 21:28 | history | became hot network question | |||
| Jul 27 at 20:28 | answer | added | Moishe Kohan | timeline score: 6 | |
| Jul 27 at 18:21 | comment | added | AGenevois | A relevant example could be the free product $G_n:= \mathbb{Z}^2 \ast \mathbb{Z}^3 \ast \cdots \ast \mathbb{Z}^n$. A group $H$ quasi-isometric to $G_n$ decomposes as a graph of groups whose edge-groups are finite and whose vertex-groups are virtually $\mathbb{Z}^k$ for $2 \leq k \leq n$ (at least one factor for each $k$). A naive guess is that, for $n >> m$, $H$ is not $m$-generated. But there is something to prove here. | |
| Jul 27 at 16:45 | comment | added | ADL | It may be too restrictive a case to be useful, but any group of Euler characteristic $-1$, e.g. a 3-generator 1-relator group, cannot be embedded with finite index as a proper subgroup of any torsion-free group. | |
| Jul 27 at 16:22 | comment | added | Moishe Kohan | I am unsure how to prove a qi version, but I know how to prove a "virtually isomorphic" ("commensurable") version. It is a bit more complicated variation on the direct product argument. | |
| Jul 27 at 14:59 | comment | added | Sean Eberhard | @MoisheKohan Interesting, thanks! Does this also rule out quasi-isometry with a 2-gen group? | |
| Jul 27 at 14:50 | comment | added | Moishe Kohan | For 2-generated embeddings, a counter-example is the (2,3,7)-Coxeter group. For 100, I think you should take a direct product of 101 pairwise nonisomorphic triangle Coxeter groups. | |
| Jul 27 at 14:27 | answer | added | Lee Mosher | timeline score: 2 | |
| Jul 27 at 13:27 | history | asked | Sean Eberhard | CC BY-SA 4.0 |