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 24, 2018 at 12:11 | comment | added | Geoff Robinson | We have constructed a central extension of $G$ by a finite cyclic $3$-group. This is $\{ (\omega^{i}, M_{g}): g \in G :0 \leq i \leq 3^{m}-1)$ where $omega$ is some primitive $3^{m}$-th of unity. It has a normal subgroup isomorphic to $\sigma(H),$ which is $\{(1,\sigma(h)): h \in H \}.$ When we factor out the latter normal subgroup, we get a central extension of ${\rm PSL}(2,11)$ by a cyclic group of order $3^{m}.$ | |
| Jun 24, 2018 at 12:04 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
more explanation
|
| Jun 24, 2018 at 11:29 | comment | added | C. Simon | Now, I understand that $\alpha(x,y)$ gives a $2$-cocycle for $G$. I don't understand that $\alpha(x,y)$ gives a $2$-cocycle for $G/H\cong PSL(2,11)$. I hope you give me the reason. Thank you, again. | |
| Jun 24, 2018 at 6:50 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Revision of some text
|
| Jun 23, 2018 at 12:38 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
Corrected a typo
|
| Jun 23, 2018 at 7:37 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
clarified
|
| Jun 23, 2018 at 6:46 | comment | added | C. Simon | Maybe there is a problem. Let $t,s\in T$. However, I am not sure $st\in T$. And, if we consider a linear character $\theta\in Irr(H)$, this proof should be effective. It means that every $G$-invariant linear character is extendible to $G$. | |
| Jun 22, 2018 at 12:18 | vote | accept | C. Simon | ||
| Jun 23, 2018 at 6:46 | |||||
| Jun 22, 2018 at 9:22 | comment | added | C. Simon | Now, I understand the proof. I think that we don’t have to prove " $M_t$ is a root of unity multiple of $σ(t^{660})$ and that each$ M_t$ has finite multiplicative order." Can you give me some reasons? @Geoff Robinson | |
| Jun 21, 2018 at 9:09 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo
|
| Jun 21, 2018 at 8:28 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo
|
| Jun 21, 2018 at 7:40 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Clarifications
|
| Jun 21, 2018 at 3:43 | comment | added | C. Simon | Firstly, Thanks. But, I don't understand: (1) why $M_t$ is a scalar; (2) why could we multiply each $M_t$ by a suitable scalar in paragraph two. | |
| Jun 20, 2018 at 15:23 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo and layout, one correction of numbers
|
| Jun 20, 2018 at 14:19 | history | answered | Geoff Robinson | CC BY-SA 4.0 |