Timeline for Examples of extensions of non-solvable groups by one another
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2018 at 13:55 | vote | accept | Gro-Tsen | ||
| Jun 2, 2018 at 10:59 | comment | added | Derek Holt | I was in any case being overcomplicated. There is a nonsplit extension of $C_2 \times A_5$ by itself, giving $C_4 \times A_5^2$. | |
| Jun 2, 2018 at 10:43 | comment | added | YCor | @DerekHolt the central product might split. I think you take $N$ to be the central product and the left arrow in the exact sequence being given by the first factor. Say that the center $Z=\{1,s\}$ is cyclic of order 2, so $G$ is the quotient of $N\times N$ by $\{(1,1),(s,s)\}$. Then the diagonal yields a semidirect product decomposition (similar to a semidirect decomposition $SO(4)=SU(2)\rtimes SO(3)$). | |
| Jun 2, 2018 at 9:47 | comment | added | Derek Holt | Of course is you allow $N$ and $Q$ to have abelian composition factors then it is much easier to find examples. For example you could take a central product of two (nonisomorphic) quasisimple groups with a common nontrivial centre. | |
| Jun 2, 2018 at 9:39 | comment | added | YCor | PS Ayoub's example seem to exactly coincide with Derek's one. After embedding $Out(S)$ into a nonabelian simple group $K$, he uses an identification between non-abelian 1-cohomology pointed sets $H^1(Out(S),Out(S))=H^1(K,Out(S)^{K/Out(S)})$. But I've failed to understand details. (PS $S$ is not the same here and in Derek's post.) | |
| Jun 2, 2018 at 9:28 | answer | added | Derek Holt | timeline score: 10 | |
| Jun 2, 2018 at 9:14 | comment | added | YCor | Joseph Ayoub once gave me an (somewhat involved) example of a non-split exact sequence of finite groups without abelian JH factors, relying on the existence of a nonabelain finite simple group $S$ for which $Aut(S)\to Out(S)$ does not split. | |
| Jun 2, 2018 at 6:23 | comment | added | YCor | To make the question precise you might require that $N,Q$ both have no abelian Jordan-Hölder factor. | |
| Jun 2, 2018 at 0:53 | history | edited | Gro-Tsen | CC BY-SA 4.0 |
give at least one example (essentially the only one I can think of)
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| Jun 2, 2018 at 0:39 | history | asked | Gro-Tsen | CC BY-SA 4.0 |