Maximality of normalizer of $p$-Sylow groups in the symmetric group $S_{p}$ 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?

 A: 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 $p>3$.
A: 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).
