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Consider the symmetric group $S_{p}$ where $p$ is a prime, then its $p$-Sylow subgroups are isomorphic to the cyclic group $C_{p}$. And it is clear that the normalizer of this cyclic group in the symmetric group is $N_{S_{p}}(C_{p})\cong C_{p}\rtimes F_{p}^{\times}$.

My question is about how to show that this normalizer is a maximal subgroup of this symmetric group. Is there a simple group theoretic argument?

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    $\begingroup$ On the contrary, I am pretty sure that the only known proof requires the classification of finite simple groups. $\endgroup$
    – Derek Holt
    Commented Aug 28, 2018 at 8:40
  • $\begingroup$ I misread the question. Sorry. $\endgroup$
    – Uri Bader
    Commented Aug 28, 2018 at 9:29
  • $\begingroup$ 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 $\mathrm{Sym}(F_p)$? $\endgroup$
    – YCor
    Commented Aug 28, 2018 at 14:12
  • $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Aug 28, 2018 at 15:40
  • $\begingroup$ 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). $\endgroup$
    – Peter Mueller
    Commented Sep 1, 2018 at 9:34

2 Answers 2

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It was proved by Burnside that a permutation group of prime degree must either be solvable, in which case it has a normal Sylow $p$-subgroup and is contained in the affine group, or it is $2$-transitive. See here for a recent straightforward proof of this.

The classification of the finite $2$-transitive groups was completed as an application of the classification of finite simple groups (see the wikipedia page or the discussion here for references) and they are listed for example in Section 7.7 of Dixon and Mortimer's book on permutation groups.

A look through the list shows that the only examples of prime degree are affine groups, $A_p$, $S_p$, groups $G$ with ${\rm PSL}(d,q) \le G \le {\rm P \Gamma L}(d,q)$ when $(q^d-1)/(q-1)$ is prime, and a few small examples (${\rm PSL}(2,7)$ and ${\rm PSL}(2,11)$ of degrees $7$ and $11$, $M_{11}$, and $M_{23}$).

When $(q^d-1)/(q-1)$ is prime $p$, an element of order $p$ in ${\rm PSL}(d,q)$ is a Singer cycle, and its normalizer in ${\rm P \Gamma L}(d,q)$ has order at most $pde$, where $q=r^e$ with $r$ prime, which is smaller than $p(p-1)$. The small examples are all contained in the alternating groups.

So the affine group of order $p(p-1)$ is maximal in $S_p$ whenever $p>3$.

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  • $\begingroup$ @GeoffRobinson That's true, but I mentioned Burnside's result because it definitely does not use CFSG. $\endgroup$
    – Derek Holt
    Commented Aug 28, 2018 at 16:24
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I can't answer the question, but here is a remark to indicate that it can't be expected to be entirely straightforward (without the classification of finite simple groups). Note that ( for $p >2$) the Sylow $p$-normalizer in $S_{p}$ contains a $(p-1)$-cycle, hence an odd permutation. The even permutations in that normalizer form a subgroup $E_{p}$ of order $\frac{p(p-1)}{2}$ of the alternating group $A_{p}.$ However, $E_{p}$ is not necessarily a maximal subgroup of $A_{p}.$ For example, we have $E_{7} < {\rm PSL}(2,7) < A_{7}$ and, even worse, $E_{11} < {\rm PSL}(2,11) < M_{11} < A_{11}$ (all indicated inclusions are strict).

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