Timeline for About the character degree of an extension
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 17, 2016 at 14:21 | vote | accept | BHZ | ||
| Aug 17, 2016 at 11:48 | answer | added | Frieder Ladisch | timeline score: 2 | |
| Aug 16, 2016 at 21:12 | vote | accept | BHZ | ||
| Aug 17, 2016 at 9:13 | |||||
| Aug 16, 2016 at 20:40 | vote | accept | BHZ | ||
| Aug 16, 2016 at 21:12 | |||||
| Aug 16, 2016 at 18:14 | answer | added | Geoff Robinson | timeline score: 2 | |
| Aug 16, 2016 at 16:05 | comment | added | Geoff Robinson | There is some work to do. Note that $H \lhd G$, and that $H$ is centralized by each Sylow $3$-subgroup of $G$ and each Sylow $5$-subgroup of $G$. | |
| Aug 16, 2016 at 15:38 | comment | added | BHZ | Thanks for your omments. As you hint that $ N $ has a cyclic normal subgroup $ H $ of order 15 and so irreducible character degrees of $ N $ is 1 or 2. Therefore $ G $ has irreducible character degrees as $ b $ or $2b $ where $ b $ is a divisor of 60. But how we can get that $ b $ can not be equal to 15? | |
| Aug 16, 2016 at 13:52 | comment | added | Geoff Robinson | A group of order $30$ has a unique Sylow $3$-subgroup and a unique Sylow $5$-subgroup, and all its irreducible characters have degree $1$ or $2$ by Ito's theorem. | |
| Aug 16, 2016 at 13:46 | comment | added | BHZ | I try but I could not prove this result since there exist two groups of order 30 such that their automorphisms have orders divisible by 5 or both 5 and 3. | |
| Aug 16, 2016 at 13:38 | comment | added | Geoff Robinson | Clifford's Theorem and Ito's theorem are the main tools to use here. | |
| Aug 16, 2016 at 12:24 | history | asked | BHZ | CC BY-SA 3.0 |