Timeline for Operation of a p'-group on a set of p-power order and fix points
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 3, 2018 at 23:40 | vote | accept | Sven Wirsing | ||
| Sep 3, 2018 at 17:33 | answer | added | Frieder Ladisch | timeline score: 4 | |
| Sep 3, 2018 at 14:56 | comment | added | Geoff Robinson | Malcev had results of this nature too. | |
| Sep 3, 2018 at 13:28 | comment | added | Sven Wirsing | Thanks for both comments. I thought maybe ist related to the fact that the radical complements are conjugated under the p-group $1+rad(A)$ and that $G$ is also acting on these "conjugation" elements. But I can't see/find a general argumentation for this fact. | |
| Sep 3, 2018 at 13:00 | comment | added | Frieder Ladisch | In your $S_3$-example, there is a fixed point? In fact, when $G$ is a $q$-group ($q$ prime) acting on a set $M$, then the number of fixed points in $M$ under $G$ is congruent to the size of $M$ modulo $q$. So when $\lvert M \rvert$ is coprime to $q$, then fixed points exists. The situation is different when $\lvert G \rvert $ is divisible by at least two different primes. An example would be a group $G$ of order $6$ acting on a set with $5$ elements, with one orbit of length $2$ and one of length $3$. | |
| Sep 3, 2018 at 11:10 | history | asked | Sven Wirsing | CC BY-SA 4.0 |