Timeline for Finite groups with all irreducible representations one dimensional
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 23, 2017 at 8:36 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 24, 2017 at 8:02 | answer | added | user103474 | timeline score: 2 | |
| Jan 23, 2017 at 17:46 | comment | added | user103474 | @Geoff Robinson Ok, now everything is clear. Thank you very much! | |
| Jan 23, 2017 at 17:04 | comment | added | Geoff Robinson | It is the case that if all absolutely irreducible representations of $G$ are one-dimensional, then $G/O_{p}(G)$ is an Abelian group of order prime to $p$, and Schur-Zassenhaus may indeed be applied directly. To see this, note that $O_{p}(G)$ is in the kernel of all absolutely irreducible characteristic $p$ representations while on the other, if $K$ s the intrestection of the kernel of all such irreducible representations, then $K$ is a $p$-group and $G/K$ is Abelian, so $G$ has a normal Sylow $p$-subgroup, which is then contained in $K$ as it is a normal $p$-subgroup. | |
| Jan 23, 2017 at 15:09 | comment | added | Kevin Buzzard | If it doesn't then maybe just replace $G'$ by the subgroup of $G$ corresponding to the Sylow $p$-subgroup of $G/G'$. | |
| Jan 23, 2017 at 15:04 | history | edited | YCor |
edited tags
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| Jan 23, 2017 at 14:42 | comment | added | user103474 | @ Geoff Robinson I know that Schur-Zassenhaus implies that N is complemented but I don't understand why in our case the commutator subgroup $G'$ should have order coprime to the order of $G/G'$. | |
| Jan 23, 2017 at 14:27 | comment | added | Geoff Robinson | Read about the Schur-Zassenhaus theorem, which says that if $G$ is a finite group and $N \lhd G$ with ${\rm gcd}(|N|,[G:N]) =1,$ then $N$ is complemented in $G$. This was originally proved in the case that either $N$ or $G/N$ is solvable, which suffices for your situation ( given the Feit-Thompson theorem, it automatically follows that one of $N$ or $G/N$ is solvable when they have coprime orders, but that is not needed for your problem). | |
| Jan 23, 2017 at 14:11 | history | asked | user103474 | CC BY-SA 3.0 |