Timeline for Enlarging a subdegree-finite "almost transitive" permutation group to a transitive one?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 30, 2018 at 15:52 | comment | added | YCor | Follow-up: mathoverflow.net/questions/311637 | |
| Sep 28, 2018 at 15:37 | vote | accept | Jens Bossaert | ||
| Sep 28, 2018 at 15:31 | comment | added | YCor | Ok, this is a reasonable question but should be asked as a separate post. | |
| Sep 28, 2018 at 15:29 | comment | added | Jens Bossaert | Yes, any $\tau\in\operatorname{Sym}(X)$ but $\tau\notin G$ (otherwise $\langle G,\tau\rangle=G$ will not be transitive). The only assumptions on $G$ are (1) finitely many orbits and (2) finite suborbits (orbits of point stabilisers). | |
| Sep 28, 2018 at 12:22 | comment | added | YCor | You mean $\tau\notin G$? or some additional requirement? Also, do you still assume that $G$ has finite stabilizers? | |
| Sep 28, 2018 at 11:40 | comment | added | Jens Bossaert | Thank you for the illuminating counterexamples! I was too optimistic thinking any $\tau$ would work. Might the title question still be salvable in the following way: for general $G$ with finitely many orbits and finite suborbits, can one always find an element $\tau$ (not necessarily of finite order) such that $\langle G,\tau\rangle$ is transitive with finite suborbits? | |
| Sep 27, 2018 at 23:29 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Sep 26, 2018 at 16:18 | history | answered | YCor | CC BY-SA 4.0 |