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>



Solvable transitive groups of prime degree
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.


