Timeline for Are generalized symmetric groups maximal finite groups (in a certain sense)?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2022 at 15:12 | comment | added | Geoff Robinson | Yes, it works for all n. | |
| May 21, 2022 at 13:31 | vote | accept | Jonas Anderson | ||
| May 21, 2022 at 12:50 | comment | added | Jonas Anderson | Thanks, that makes sense. For clairification this argument works for all $n$, correct? | |
| May 21, 2022 at 10:22 | comment | added | Geoff Robinson | When $m >6$, $M(m,n)$ has a natural diagonal subgroup $T$ of order $m^{n}$ , generated by $n$ elements of order $m$, each with $m-1$ eigenvalues $1$. Hence the group $A$ in the answer contains $T$, and necessarily needs $n$ generators. Then the continuation as in the answer should be clear. | |
| May 21, 2022 at 10:18 | comment | added | Geoff Robinson | The 1962 book of Curtis and Reiner has a section "On theorems of Frobenius, Schur and Burnside" (or something like that), and al the Character Theory book by Isaacs. The upshot is that in a finite subgroup of $U(n,\mathbb{C})$, any two elements which have all their eigenvalues close enough together on the unit circle commute. | |
| May 21, 2022 at 3:45 | comment | added | Jonas Anderson | Can you give me some more details and perhaps a link? This is exactly the type of answer I was looking for, I just don't fully understand it yet. | |
| May 20, 2022 at 16:46 | history | answered | Geoff Robinson | CC BY-SA 4.0 |