Timeline for A finite group that splits and does not split
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 24, 2018 at 11:04 | vote | accept | Pablo | ||
| Mar 22, 2018 at 8:24 | answer | added | Derek Holt | timeline score: 6 | |
| Mar 21, 2018 at 11:11 | comment | added | Derek Holt | I am starting to suspect that there are no such examples, but I don't have time to think about it right now. Note that $C$ is a minimal normal subgroup of the semidirect product, so it must be either elementary abelian or a direct product of isomorphic nonabelian simple groups. | |
| Mar 21, 2018 at 9:01 | comment | added | Nick Gill | I should add, also, that my first thought was to look at the Dempwolff group.... But I don't have the wherewithal right now to follow this through and see if it yields the example you seek... | |
| Mar 21, 2018 at 8:52 | comment | added | Nick Gill | This is relevant -- mathoverflow.net/questions/163041/… -- although it doesn't consider the irreducibility condition. | |
| Mar 21, 2018 at 8:05 | comment | added | Pablo | @DerekHolt You are right, but I think my question is stated unambiguously. In an attempt to answer it, people are considering variants, which is fine I think. | |
| Mar 21, 2018 at 8:02 | comment | added | Derek Holt | It is apparent from the comments and from Glasby's answer that there is doubt about what you asking. | |
| Mar 21, 2018 at 7:58 | comment | added | Pablo | @DerekHolt What is unclear? | |
| Mar 21, 2018 at 7:56 | comment | added | Derek Holt | This question needs clarifying. | |
| Mar 21, 2018 at 6:09 | answer | added | Glasby | timeline score: 1 | |
| Mar 18, 2018 at 4:53 | comment | added | zibadawa timmy | @LSpice I would think even the case with $\ker(\tau)\cong C$ need not split for a semidirect product. That it does for a direct product is a non-trivial (and somewhat recent) theorem of Ayoub's. | |
| Mar 17, 2018 at 18:49 | comment | added | Dima Pasechnik | The interpretation I was thinking about was that you are asking whether it's always the case that a subgroup of the semidirect product isomorphic to $A$ must have $C$ as a complement. | |
| Mar 17, 2018 at 18:44 | comment | added | Pablo | Not requiring this condition. | |
| Mar 17, 2018 at 18:40 | comment | added | LSpice | Ah, so you're comparing the 'natural' sequence $1 \to C \to A \ltimes C \to A \to 1$ coming from the construction as a semi-direct product, which splits by assumption, with the sequence $1 \to \ker(\tau) \to A \ltimes C \xrightarrow\tau A \to 1$? Do you require that $\ker(\tau)$ be isomorphic to $C$? | |
| Mar 17, 2018 at 18:38 | comment | added | Jay Taylor | @LSpice Urgh, thanks for checking my stupidity. The comment is so stupid I'm just going to delete it. | |
| Mar 17, 2018 at 18:36 | comment | added | Pablo | Our group is a semidirect product, this is the splitting. | |
| Mar 17, 2018 at 18:35 | comment | added | LSpice | @JayTaylor, $x \mapsto (x, 1)$ is a splitting of your map. | |
| Mar 17, 2018 at 18:34 | comment | added | LSpice | What does the title mean? You explicitly ask for a situation in which a certain epimorphism doesn't split, but where's the "does split" part? (Probably it's implicit in some other statement, but I'd like to see it made explicit!) | |
| Mar 17, 2018 at 18:18 | comment | added | Pablo | @DimaPasechnik can you elaborate? | |
| Mar 17, 2018 at 18:16 | comment | added | Dima Pasechnik | iirc there are such examples with e.g. $A=Sp_6(2)$ acting on $2^6$. | |
| Mar 17, 2018 at 18:10 | history | asked | Pablo | CC BY-SA 3.0 |