Timeline for Locally finite groups of finite rank and bounded exponent
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 28, 2023 at 18:33 | answer | added | YCor | timeline score: 3 | |
| Feb 4, 2013 at 23:23 | vote | accept | Kıvanç Ersoy | ||
| Feb 4, 2013 at 22:13 | answer | added | Ian Agol | timeline score: 6 | |
| Feb 4, 2013 at 21:54 | comment | added | Ian Agol | @Kivanc: Ok, I understand your question now - the group should be infinitely generated. The terminology "rank" has a different meaning in geometric group theory, denoting the minimal number of generators for the group, which led to my confusion. So you want an infinitely generated group, all of whose finitely generated subgroups are finite and of bounded exponent and (geometric) rank. | |
| Feb 4, 2013 at 21:04 | comment | added | Alireza Abdollahi | @Agol: Finite rank and locally finite finite do not imply finite; consider the quasicyclic $p$-group. The bounded exponent is necessary and now the positive solution of RBP comes into pay. | |
| Feb 4, 2013 at 21:02 | comment | added | Kıvanç Ersoy | @Agol : finite rank and locally finite does not imply finite, for example Prüfer $p$ groups... | |
| Feb 4, 2013 at 20:57 | comment | added | Kıvanç Ersoy | I didn't mean "proper subgroups are finite", I mean "every finitely generated subgroup is finite". So, Tarski monsters are not example. Thanks. | |
| Feb 4, 2013 at 20:55 | comment | added | Kıvanç Ersoy | yes, a group has finite rank $r$ if every finitely generated subgroup is generated by at most $r$ elements. | |
| Feb 4, 2013 at 20:50 | comment | added | Ian Agol | finite rank and locally finite implies finite, so no. You probably mean by locally finite that proper subgroups are finite. In this case, you can consider the Tarski Monsters. en.wikipedia.org/wiki/Tarski_monster | |
| Feb 4, 2013 at 20:46 | comment | added | Derek Holt | So the answer is no by the Restricted Burnside Problem. | |
| Feb 4, 2013 at 19:49 | comment | added | Alireza Abdollahi | @Ersoy: By a group $G$ is of finite rank you mean that there exists an integer $r$ such that every finitely generated subgroup of $G$ can be generated by at most $r$ elements. Am I right? | |
| Feb 4, 2013 at 19:24 | history | asked | Kıvanç Ersoy | CC BY-SA 3.0 |