Timeline for Finding groups of odd order without non-cyclic nilpotent quotients
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 7, 2012 at 9:14 | vote | accept | Tom De Medts | ||
| Jun 24, 2011 at 12:39 | answer | added | Max Horn | timeline score: 4 | |
| Mar 10, 2011 at 15:30 | comment | added | Jack Schmidt | I think your low number (1016) is just due to not having enough room for the p-groups to get out of control by order 2015. You have to waste a little space with that top k*q^n, and then maybe the chief factors below it are not order p, but rather p^2 or p^3 or something, but you still get the ridiculous growth of p-groups. | |
| Mar 10, 2011 at 15:25 | comment | added | Jack Schmidt | Usually when I do this, I have specific ideas of what sort of modules I want to allow. I think as posed, you need to find all modules (reducible, decomposable, etc.) up to a certain dimension. There are probably too many to make that a sane undertaking. If the Sylow p-subgroup is cyclic, it might be more reasonable. If you want the Fitting series to be a chief series then it should be relatively easy. | |
| Mar 10, 2011 at 15:24 | comment | added | Jack Schmidt | Look at the largest nilpotent quotient. It must be cyclic of order, say, k. The next chief factor below it must have order coprime to k, and so be of order q^n where n is the order of q mod k. Now you are just looking for downwards extensions of a fixed (solvable) group. In GAP you might use "Extensions" to find all the extensions for a given module. | |
| Mar 10, 2011 at 14:30 | history | edited | Tom De Medts | CC BY-SA 2.5 |
added 190 characters in body; edited title; added 11 characters in body; edited title
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| Mar 10, 2011 at 14:28 | comment | added | Tom De Medts | That's true; I guess the new question (edited) makes more sense. | |
| Mar 10, 2011 at 12:22 | comment | added | Tim Dokchitser | Just a remark: if $G$ is nilpotent of odd order then something like $G\times(C_7\rtimes C_3)$ will be in your list, so non-nilpotency is perhaps not really restrictive | |
| Mar 10, 2011 at 11:32 | history | asked | Tom De Medts | CC BY-SA 2.5 |