Timeline for Is a $G$-invariant character $\theta$ of $H$ extendible to $G$?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2018 at 12:20 | vote | accept | C. Simon | ||
| Jun 22, 2018 at 12:18 | vote | accept | C. Simon | ||
| Jun 23, 2018 at 6:46 | |||||
| Jun 20, 2018 at 14:19 | answer | added | Geoff Robinson | timeline score: 12 | |
| Jun 20, 2018 at 11:37 | comment | added | kneidell | The order of $PSL(2,11)$ is, of course, $660$, and not $1320$. I can't edit for some reason.. anyway, it doesn't affect the (non-) argument. | |
| Jun 20, 2018 at 11:27 | comment | added | kneidell | This doesn't really answer your question, but it's what I had in mind- the assumption $\theta(1)=9$ implies that $\det(\rho_\theta(h))$ is a $3^k$ root of unity for some $k$. Here $\rho_\theta$ is a representation affording $\theta$. Using the defining relation of the cocycle $\beta\in H^2(G/H,\mathbb C)$ associated with $\theta$, it is possible to infer that $\beta$ must have order $3^k$ as well, but by the theorem is should also divide $|PSL(2,11)|=1320$, and hence is either $1$ or $3$. Maybe there's some reason why it can't be $3$? (IDK) | |
| Jun 20, 2018 at 11:19 | comment | added | C. Simon | @kneidell I know that Theorem 11.15. But I don't know how to use it. | |
| Jun 20, 2018 at 11:18 | comment | added | kneidell | Oh, wait.. I'm not sure what I suggested would work here. | |
| Jun 20, 2018 at 11:16 | comment | added | kneidell | @Simon: Ok, so I think you can use Theorem 11.15 of Isaac's book ''Character theory of finite groups'' to prove extendibility. Write down the defining relation of the cocycle associated to $\theta$, take determinants, and use what the theorem tells you about the order of this cocycle. | |
| Jun 20, 2018 at 11:13 | comment | added | C. Simon | @ kneidell $PSL(2,11)$ means the projective group of rank 1 over $F_{11}$. | |
| Jun 20, 2018 at 11:08 | comment | added | kneidell | Silly question- Does $PSL(2,11)$ mean the projective group of rank 1 over $\mathbb F_{11}$ or the projective group of rank $10$ over $\mathbb F_{2}$? | |
| Jun 20, 2018 at 8:50 | comment | added | Aurel | By Clifford theory, some multiple of $\theta$ extends to $G$. | |
| Jun 20, 2018 at 7:44 | history | edited | C. Simon | CC BY-SA 4.0 |
deleted 1 character in body
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| Jun 20, 2018 at 7:27 | comment | added | Dima Pasechnik | it would be very strange if this was true. look for counterexamples... | |
| Jun 20, 2018 at 3:32 | comment | added | C. Simon | @Venkataramana $\theta(1)$ is the degree of $\theta$. | |
| Jun 20, 2018 at 3:29 | comment | added | Venkataramana | I take it that $\theta (1)=9$ is a typo? | |
| Jun 20, 2018 at 3:27 | history | asked | C. Simon | CC BY-SA 4.0 |