Timeline for Splitting of a finite group with no abelian subfactor in composition series
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 7, 2022 at 10:40 | vote | accept | User01 | ||
| Sep 14, 2022 at 18:07 | history | edited | YCor | CC BY-SA 4.0 |
edited to reflect current answer
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| Sep 14, 2022 at 8:03 | comment | added | Derek Holt | OK I will answer the more precise version of the question as formulated by YCor and spin. | |
| Sep 14, 2022 at 7:59 | answer | added | Derek Holt | timeline score: 9 | |
| Sep 14, 2022 at 6:42 | comment | added | spin | @YCor: I would be curious if an example is written down somewhere? If $X$ and $Y$ are non-abelian finite simple groups, by Schreier conjecture (outer automorphism groups solvable) every extension $1 \rightarrow X \rightarrow G \rightarrow Y \rightarrow 1$ is split. That would make me naively guess that every finite group with non-abelian Jordan-Hölder factors is constructed by iterated semidirect products, but seems that is wrong. | |
| Sep 14, 2022 at 6:03 | comment | added | YCor | There's a reasonable question of finding a finite group with no abelian Jordan-Hölder factor, which is not obtained from simple groups by iterating semidirect products (and hence wreath as well, since wreath product is obtained from semidirect products). I think I heard of a quite elaborate example by Joseph Ayoub. | |
| Sep 14, 2022 at 5:33 | comment | added | User01 | Yes, it is now clear to me. Thank you, @DerekHolt | |
| Sep 14, 2022 at 5:32 | history | edited | User01 | CC BY-SA 4.0 |
deleted 4 characters in body
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| Sep 13, 2022 at 16:44 | comment | added | Derek Holt | It is trivially true that every (non-trivial) group has a (non-trivial) solvable subgroup - you don't need to use the non-trivial result of Suzuki to prove that! | |
| Sep 13, 2022 at 16:04 | comment | added | Derek Holt | You need to make the question more precise. What exactly does "etc" mean? I predict that whatever definition of product you come up with, the answer to the question will be no. | |
| Sep 13, 2022 at 8:05 | review | Close votes | |||
| Sep 27, 2022 at 3:09 | |||||
| Sep 13, 2022 at 7:11 | answer | added | spin | timeline score: 3 | |
| Sep 13, 2022 at 6:23 | history | edited | User01 | CC BY-SA 4.0 |
added 37 characters in body
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| Sep 13, 2022 at 6:21 | comment | added | User01 | I should mentioned that $G$ is not a product (direct, semi direct, wreath product etc.) of non abelian simple group | |
| Sep 13, 2022 at 6:00 | comment | added | YCor | @JoachimKönig this is indeed the smallest counterexample, in the sense that every finite group with $\le 5$ Jordan-Hölder subquotients, all non-abelian, is a direct product of simple groups. | |
| Sep 13, 2022 at 5:59 | history | edited | YCor | CC BY-SA 4.0 |
fixed English
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| Sep 13, 2022 at 5:36 | comment | added | Joachim König | No. For example $G=A_5\wr A_5$ (wreath product). | |
| Sep 13, 2022 at 4:53 | history | asked | User01 | CC BY-SA 4.0 |