Timeline for Textbook source for finite group properties deducible from character table?
Current License: CC BY-SA 3.0
13 events
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| Oct 21, 2017 at 20:55 | comment | added | nkm | Ah, of course, because $p$-groups are solvable. Thanks a lot. | |
| Oct 21, 2017 at 20:40 | comment | added | Will Sawin | @NathanielMayer Yes, because the commutator subgroup of the $p$-group, the commutator of the commutator, etc. are all normal subgroups, and the quotient of each one by the next is abelian. | |
| Oct 21, 2017 at 20:13 | comment | added | nkm | Or, as @GeoffRobinson implies, does the stronger fact that all minimal normal subgroups of all quotients are p-groups imply solvability? I can't find references for these facts, and don't see how to determine that a subgroup is abelian from the character table. | |
| Oct 21, 2017 at 20:11 | comment | added | nkm | Very late here, but @SteveD does the existence of a normal series (all terms normal in G) with p-group quotients automatically imply G is solvable? Or do you need to know that the quotients are abelian as well? | |
| May 21, 2012 at 23:10 | comment | added | Geoff Robinson | @Jim: Some questions of this nature are difficult to prove or disprove. I'll keep a mental note at least! | |
| May 21, 2012 at 22:16 | vote | accept | Jim Humphreys | ||
| May 21, 2012 at 22:16 | comment | added | Jim Humphreys | @Geoff: Meanwhile I hope you are taking notes for your forthcoming definitive (but concise) text on finite group characters. Ideally all of this material would be set up as an extended exercise, but at the moment it's complicated to locate all details in the literature. | |
| May 21, 2012 at 20:15 | comment | added | Jim Humphreys | Thanks for the quick answer and the illuminating comments, which reinforce what I vaguely recalled should be true (and more). Probably I drew my own notes from Isaacs, since at the time few books went that far into character theory, but I was hoping for a better summary in later texts. I'll wait a bit longer before accepting what's here already as an answer to my questions 1 and 2. | |
| May 21, 2012 at 19:21 | history | edited | Will Sawin | CC BY-SA 3.0 |
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| May 21, 2012 at 19:17 | comment | added | Geoff Robinson | I think there is a typo in the first line- you mean "from the character table of $G$" I think. Solvability is similar. You can find all the minimal normal subgroups and their orders from the character table, and if $M$ ia any one of them, you can find the character table of $G/M$ from the character table of $G$. If $G$ is solvable, then all such $M$ should have prime power order. On the othe hand if all such $M$ have prime power order, youcan work inductively- if $G$ is not solvbale, you must find a normal subgroup $N$ so that $G/N$ has a minimal normal subgroup not of prime power order. | |
| May 21, 2012 at 19:16 | comment | added | Steve D | Sorry, I meant "prime power" order above; if you can find all prime order quotients, the group is supersolvable! | |
| May 21, 2012 at 19:15 | comment | added | Steve D | You can deduce solvability as well; you can easily find series of normal subgroups, then the order of their quotients needs to be prime. A good source for a lot of questions like this is Isaacs's Character Theory book; he doesn't discuss all these things in one place, but a lot of them are scattered throughout the text. | |
| May 21, 2012 at 19:08 | history | answered | Will Sawin | CC BY-SA 3.0 |