Timeline for Is the solvable radical of a finite perfect group contained in the Schur multiplier of the quotient of the group modulo the solvable radical?
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7 events
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| May 20, 2020 at 19:46 | comment | added | YCor | The answer is positive if $N=Z(G)$, but this is somewhat immediate from the definition. I think that the question belongs on MathSE rather than here. | |
| May 20, 2020 at 19:45 | vote | accept | Leyli Jafari | ||
| May 20, 2020 at 19:45 | comment | added | Leyli Jafari | Thanks a lot! -- And can you perhaps also tell what would be the answer if the assume $N = {\rm Z}(G)$? | |
| May 20, 2020 at 17:35 | review | Close votes | |||
| May 24, 2020 at 21:55 | |||||
| May 20, 2020 at 17:07 | answer | added | Geoff Robinson | timeline score: 2 | |
| May 20, 2020 at 11:10 | comment | added | YCor | "Is contained", you mean "is isomorphic to a subgroup of"? Anyway, in many cases $N$ is non-abelian, so this sounds hopeless. Even with $N$ abelian, you have plenty of perfect semidirect products $G=N\rtimes S$ with $S$ simple non-abelian (for given $S$ you have such $G$ with $N$ arbitrary large). Classifying them for $N$ abelian of prime exponent is essentially classifying reps of $S$ in arbitrary finite fields. Classifying them for $N$ nilpotent is even much harder. So this sounds hopeless. | |
| May 20, 2020 at 10:45 | history | asked | Leyli Jafari | CC BY-SA 4.0 |