Timeline for Maximal subgroups of $S_{\Bbb N}$ and $A_{\Bbb N}$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2023 at 15:58 | comment | added | Najib Idrissi | @HenrikRüping Why couldn't a maximal subgroup be isomorphic to $S_{\mathbb{N}}$? In $\mathbb{Z}$, all the maximal subgroups ($p\mathbb{Z}$ for $p$ prime) are isomorphic to $\mathbb{Z}$, aren't they? | |
| Jul 24, 2023 at 14:25 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
added 47 characters in body
|
| Jul 24, 2023 at 14:08 | comment | added | Martin Seysen | According to en.wikipedia.org/wiki/Maximal_subgroup "a maximal subgroup H of a group G is a proper subgroup, such that ..." | |
| Jul 24, 2023 at 13:57 | comment | added | HenrikRüping | But then for $S_\mathbb{N}$ this would only be the whole group. Any other subgroup cannot be maximal in the poset of subgroups isomorphic to $S_\mathbb{N}$. Maybe we only have to look at proper subgroups ? | |
| Jul 24, 2023 at 13:42 | comment | added | Martin Seysen | Can the 2nd sentence in the question be formulated as follows: I'd like to describe all maximal subgroups of $S_{\mathbb{N}}$ (and also of $A_{\mathbb{N}}$) that are isomorphic to $S_{\mathbb{N}}$ . | |
| Jul 24, 2023 at 13:33 | answer | added | Sean Eberhard | timeline score: 9 | |
| Jul 24, 2023 at 13:33 | history | edited | Sean Eberhard | CC BY-SA 4.0 |
spelling
|
| Jul 24, 2023 at 12:30 | comment | added | Wojowu | What do you mean with "of type $S_{\Bbb N}$"? | |
| Jul 24, 2023 at 11:26 | history | asked | Анатолий Вершик | CC BY-SA 4.0 |