Timeline for Does a non-simple perfect group always have a maximal subgroup whose derived subgroup has nontrivial core?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Apr 23, 2022 at 18:01 | history | bounty ended | Leyli Jafari | ||
| S Apr 23, 2022 at 18:01 | history | notice removed | Leyli Jafari | ||
| Apr 21, 2022 at 19:08 | vote | accept | Leyli Jafari | ||
| Apr 18, 2022 at 15:47 | answer | added | Richard Lyons | timeline score: 6 | |
| Apr 17, 2022 at 15:16 | comment | added | YCor | Does there exist a finite simple group $S$, a field of prime order $K$ and a nontrivial irreducible $KS$-module $V$ such that for every proper subgroup $M$ of $S$ one has $V^M\neq 0$? (if so, $S\ltimes V$ is an example for the question). Even the answer for $K=\mathbf{C}$ would be of interest (and probably enough), and could be checked with a computer for groups $S$ for which maximal subgroups are known and character tables are known for $S$ and its maximal subgroups (I have only checked $|S|=60$: there's no $V$ then). | |
| S Apr 17, 2022 at 14:42 | history | bounty started | Leyli Jafari | ||
| S Apr 17, 2022 at 14:42 | history | notice added | Leyli Jafari | Draw attention | |
| Apr 12, 2022 at 14:54 | comment | added | Richard Lyons | All you need for my comment is that $Z(G)\ne 1$. | |
| Apr 12, 2022 at 14:47 | comment | added | Richard Lyons | There are no counterexamples among quasisimple groups. If $G$ is quasisimple, let $p$ be a prime divisor of $Z(G)$, and $P$ a Sylow $p$-subgroup of $G$. Then by a transfer result, $P\cap Z(G)\le[P,P]$. Any maximal subgroup $M$ containing $P$ will then satisfy $P\cap Z(G)\le[M,M]$. | |
| Apr 12, 2022 at 9:01 | comment | added | Stefan Kohl♦ | @markvs I checked that the assertion is true for all non-simple perfect groups of order less than 1000000. | |
| Apr 11, 2022 at 14:53 | comment | added | markvs | @StefanKohl: For most of these groups the statement is obvious (I assume the "core" means normal core). | |
| Apr 11, 2022 at 14:31 | comment | added | Stefan Kohl♦ | @markvs More precisely, there are 444 perfect groups of order less than 100000. Of these, 31 are simple and 1 is trivial -- so in this range of orders, there are 412 nontrivial perfect groups which are not simple. | |
| Apr 11, 2022 at 9:57 | comment | added | markvs | There are not that many perfect groups of order less than $100000$. | |
| Apr 10, 2022 at 16:23 | history | asked | Leyli Jafari | CC BY-SA 4.0 |