Solvable transitive groups of prime degree Is the following true ?

Every solvable transitive subgroup
  $G\subset\mathfrak{S}_p$ (the symmetric group on
  $p$ letters, where $p$ is a prime)
  contains a unique subgroup $C$ of
  order $p$ and is contained in the
  normaliser $N$ of $C$ in $\mathfrak{S}_p$.  The
  quotient $G/C$ is cyclic of order
  dividing $p-1$.  If $G$ is not cyclic,
  then it has exactly $p$ subgroups of
  index $p$.

I need such a result for an arithmetic application.  A reference or a short argument will be appreciated.
Addendum. For those interested in the arithmetic application, see http://arxiv.org/abs/1005.2016
 A: Happened to come across the following Satz in Huppert, Endliche Gruppen I, [S. 163].

A: A transitive subgroup $G$ of $S_p$ contains a Sylow $p$-subgroup $P$
having order $p$. If it has only the one, then $P$ is normal in $G$
and so $G$ lies in the normalizer $N$ of $P$ in $S_p$. This is the affine
linear group $\mathrm{AGL}(1,p)$ which is soluble. Thus $G$ is soluble.
Otherwise $G$ has more than one Sylow $p$-subgroup. By Sylow's theorems,
these $p$-subgroups are conjugate in $G$.
If $H$ is a nontrivial normal subgroup of $G$ then $H$ must be transitive
since $G$ is primitive (the orbits of $H$ form a partition invariant under the
action of $G$). So $H$ contains a Sylow $p$-subgroup $P$ of $G$. So
$H$ contains all the Sylow $p$-subgroups of $G$ (as they are conjugate under $G$).
Therefore $G$ cannot be soluble, as by repeatedly taking nontrivial normal subgroups
we always get groups with more than one Sylow $p$-subgroup.
A: This is exercise 7.2.12 of Robinson's Course in the Theory of Groups, page 195 in the first edition.
A transitive subgroup of prime degree is primitive, and primitive solvable groups have a regular normal subgroup that is complemented by a unique conjugacy class of maximal subgroups.  In particular, the Sylow p-subgroup C of order p is that regular normal subgroup, and the complement (being a permutation group) acts faithfully on it.  In other words, the centralizer of the subgroup C is C itself.  Hence G/C is a subgroup of Aut(C), so cyclic of order dividing p-1.
Since G is solvable it has a Sylow p-complement M, and by Hall's 1928 theorem, the number of
such Sylow p-complements is a divisor of p.  If it is 1, then M is normal, so M centralizes C, so M=1, and G=C is cyclic.
This is summarized by saying that G is a subgroup of AGL(1,p) containing the translation subgroup.  More generally every primitive solvable group is a subgroup of AGL(n,p) where p^n is the degree of the permutation action (but AGL(n,p) is no longer solvable itself).
