Timeline for Does every group that satisfies the maximal permutizer condition then satisfy the permutizer condition?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2015 at 11:11 | answer | added | Derek Holt | timeline score: 3 | |
| May 26, 2015 at 23:43 | comment | added | Marty Isaacs | I suspect that all groups satisfying the maximal permutizer condition must be solvable. I think I could prove that if I knew that no nonabelian simple group satisfied the condition. If a simple group satisfies the maximal permutizer condition, then every maximal subgroup is complemented by a cyclic group, and that seems unlikely. In fact, a subgroup of even index in a simple group cannot be complemented, but unfortunately, a simple group can fail to have a maximal subgroup of even index. A_7 is a counterexample. | |
| Apr 24, 2015 at 12:54 | comment | added | Jim Humphreys | @Simin: In your definition of "permutizer condition", I guess you should require $H$ to be a proper subgroup of $G$. This is already understood in the notion of "maximal subgroup". (By the way, is there a reason for including the tag 'algebraic-groups'?) | |
| Apr 24, 2015 at 11:35 | comment | added | Geoff Robinson | @DerekHolt : Yes, you are right I was mistaken. Thanks for pointing it out. | |
| Apr 24, 2015 at 10:09 | history | edited | Nick Gill | CC BY-SA 3.0 |
Corrected some spelling and English.
|
| Apr 24, 2015 at 9:28 | review | Close votes | |||
| Apr 24, 2015 at 15:30 | |||||
| Apr 24, 2015 at 9:19 | comment | added | Derek Holt | @GeoffRobinson I don't think that's right. $S_3$ satisfies the maximal permutizer condition. | |
| Apr 24, 2015 at 9:03 | comment | added | Simin Eftekhari | The converse is also true? | |
| Apr 24, 2015 at 7:40 | review | First posts | |||
| Apr 24, 2015 at 7:42 | |||||
| Apr 24, 2015 at 7:36 | history | asked | Simin Eftekhari | CC BY-SA 3.0 |