Timeline for Which finite groups have no irreducible representations other than characters?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 13, 2015 at 6:51 | vote | accept | Pablo | ||
| Aug 4, 2015 at 20:27 | history | edited | Timothy Chow | CC BY-SA 3.0 |
fixed spelling of title
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| Aug 4, 2015 at 16:58 | history | edited | Pablo | CC BY-SA 3.0 |
added 1 character in body
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| Aug 4, 2015 at 13:17 | comment | added | Ehud Meir | I do not have the book near me so it is hard to answer. However, I posted an answer. | |
| Aug 4, 2015 at 13:16 | answer | added | Ehud Meir | timeline score: 12 | |
| Aug 4, 2015 at 13:09 | comment | added | Pablo | @EhudMeir Does this follow from proposition 25 on page 62 subsection 8.2 (semidrect products by an abelian group) of Serre's Linear representations of finite groups? If you can prove your claim then we have an answer to the question. | |
| Aug 4, 2015 at 13:00 | comment | added | Ehud Meir | Sorry for that. What I can say is that your example of $S_3$ generalizes: if you have a finite group $G$ which is a semidirect product of an abelian group $A$ with a $p$-group $P$ (where $A$ acts on $P$), and the ground field contains enough roots of unity ($exp(A)$ roots of unity) then all the irreducible representations of $G$ will be one dimensional. | |
| Aug 4, 2015 at 12:52 | comment | added | Pablo | @EhudMeir Dear Udi, I already know these things, and it is this different game that I want to play. Please note that your remark on trviality (which has an easy proof using orbits and actions) is a special case of my explanations above: the upper triangular matrices with $1$'s on the main diagonal form a $p$-Sylow subgroup of the general linear group, so every $p$-group is conjugate to a subgroup of it, and thus every representation of a $p$-group is simultaneously triangulizable showing that all the Jordan-Holder constituents are trivial as you claimed. | |
| Aug 4, 2015 at 12:42 | history | edited | Pablo | CC BY-SA 3.0 |
added 119 characters in body
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| Aug 4, 2015 at 12:42 | comment | added | Ehud Meir | If the characteristic of the field divides the order of the group then it is a completely different game. It is no longer true that every representations splits as the direct sum of irreducible representations. For example, if $G$ is a finite $p$ group, and the field $k$ has characteristic $p$, then $kG$ only has one irreducible representation, namely $k$ itself with the trivial action. | |
| Aug 4, 2015 at 12:37 | history | asked | Pablo | CC BY-SA 3.0 |