Timeline for Does solubility imply nilpotency?
Current License: CC BY-SA 4.0
12 events
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| Apr 5, 2021 at 14:43 | history | edited | Derek Holt | CC BY-SA 4.0 |
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| Apr 5, 2021 at 14:16 | comment | added | Geoff Robinson | @spin : Yes, thanks, that is the theorem I refer to, and there is a full reference to Kegel's paper in the accepted answer.. | |
| Apr 5, 2021 at 14:08 | comment | added | spin | @MeisamSoleimaniMalekan: See this question, for what I think Geoff is referring to: link | |
| Apr 5, 2021 at 14:06 | comment | added | MSMalekan | @GeoffRobinson I do not know about that theorem, could you please provide me a reference? | |
| Apr 5, 2021 at 13:32 | comment | added | Geoff Robinson | I think that condition 3 already implies that $G/G^{(n-1)}$ is nilpotent (applying a theorem of O. Kegel to $G/G^{(n-1)}$ with the automorphism induced by $\phi$). | |
| Apr 5, 2021 at 12:56 | history | edited | MSMalekan | CC BY-SA 4.0 |
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| Apr 5, 2021 at 12:46 | history | edited | MSMalekan | CC BY-SA 4.0 |
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| Apr 5, 2021 at 12:37 | comment | added | MSMalekan | @DerekHolt, Thank you, I think you are right. I I have to rewrite my assumptions more accurately. | |
| Apr 5, 2021 at 12:35 | comment | added | MSMalekan | @YiftachBarnea The motivation is giving a new version of Theorem 13 of this paper. | |
| Apr 5, 2021 at 11:57 | comment | added | Derek Holt | Isn't this false with $G=S_3$ and $\phi$ conjugation by an element of order $2$? | |
| Apr 5, 2021 at 11:54 | comment | added | Yiftach Barnea | Can you please explain the motivation behind this question and why do you think it is true? | |
| Apr 5, 2021 at 11:13 | history | asked | MSMalekan | CC BY-SA 4.0 |