Timeline for Finite simple groups: smallest nonsplit automorphic extensions
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3, 2013 at 22:35 | vote | accept | Anvita | ||
| Jun 3, 2013 at 14:34 | comment | added | Derek Holt | @Anvita: Yes, they are all nonsplit extensions, and so the answer to your new question is yes, there are such nonisomorphic nonpslit extensions. I have convinced myself that the extensions really are nonsplit in the case of a product of diagonal and field automorphisms of order $r$ of ${\rm PSL}_r(q^r)$, where $r|q-1$, but the calculations are a bit messy. | |
| Jun 3, 2013 at 1:04 | comment | added | Anvita | @Derek However, there are several ways to find a subgroup in $\langle\phi,\delta\rangle$ of order $r$ other than $\langle\phi\rangle$ and $\langle\delta\rangle$; namely, $r-1$ ways, and $\langle \phi\delta \rangle$ is only one of them. Do the other choices result in a nonsplit extension as well? Could there be nonisomorphic such minimal nonsplit extensions? | |
| Jun 3, 2013 at 1:00 | comment | added | Anvita | @Derek thank you. That is an interesting example! If I understand it right, it may be generalized to construct a minimal nonsplit extension with $|A|=r$ for an arbitrary prime $r$. If we take $S={\rm PSL}_r(q^r)$, where $q\equiv 1 (r)$, and define $\phi$ and $\delta$ to be a field and diagonal automorphisms of $S$, both of order $r$, then $S.A =\langle S, \phi\delta \rangle$ should be nonsplit. | |
| Jun 2, 2013 at 20:28 | history | answered | Derek Holt | CC BY-SA 3.0 |