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Sep 1, 2018 at 9:34 comment added Peter Mueller Peter Neumann proved and surveyed some results on groups of prime degree. For instance a group properly containing $\text{AGL}_1(\mathbb F_p)$ is $3$-transitive, and as of 1973 it was an open problem (raised by Wielandt) whether the group is $S_p$. It wasn't even known whether the group need to be $4$-transitive. See Neumann's survey `Transitive permutation groups of prime degree' in Proc. 2nd Internat. Conf. Theory of Groups, Canberra 1973, Lect. Notes Math. 372, 520-535 (1974).
Aug 28, 2018 at 15:40 comment added YCor One immediate remark: any intermediate subgroup $Aff(F_p)\le G\le Sym(F_p)$ has a simple normal subgroup, which is non-abelian as soon as $Aff(F_p)<G$ (so $Aff(F_p)$ is maximal solvable). Indeed, let $N$ be a minimal nontrivial normal subgroup of $G$. Since $N\neq 1$, by primitivity, $N$ has to be transitive, and in particular $p$ divides $|N|$. Also, as a minimal normal subgroup, $N\simeq S^k$ for some simple group $S$. Since $p^2$ does not divide $|S_p|$, we have $k=1$, so $N$ is simple. So either $N=F_p$ and hence $G$ is contained in its normaliser $Aff(F_p)$, or $N$ is non-abelian simple.
Aug 28, 2018 at 15:26 answer added Derek Holt timeline score: 9
Aug 28, 2018 at 14:47 answer added Geoff Robinson timeline score: 6
Aug 28, 2018 at 14:12 comment added YCor Restatement of the question: let $\mathrm{Aff}(F_p)$ be the group (of order $p(p-1)$ of affine automorphisms of the 1-dimensional line $F_p$, $p$ a given prime. How to show that $\mathrm{Aff}(F_p)$ is a maximal subgroup in the permutation group $\mathrm{Sym}(F_p)$?
Aug 28, 2018 at 9:29 comment added Uri Bader I misread the question. Sorry.
Aug 28, 2018 at 8:40 comment added Derek Holt On the contrary, I am pretty sure that the only known proof requires the classification of finite simple groups.
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Aug 28, 2018 at 8:01
Aug 28, 2018 at 7:53 history asked Ling CC BY-SA 4.0