Let $ G $ be a solvable primitive permutation group. Why the degree of $ G $ is a prime power Let $ G $ be a solvable primitive permutation group. Why the 
degree of $ G $ is a prime power and $ G $ has a unique minimal normal 
subgroup? (8B.4 problem of Finite group theory by Issac) Is transitive this minimal normal subgroup of $ G $? 
 A: Allow me to give a reference, make the statement more precise, expand Derek's answer, and state a converse.
The theorem actually goes back to Galois himself.  A nice historical account can be found in 
Peter Neumann, The concept of primitivity in group theory
and the Second Memoir of Galois, Arch. Hist. Exact Sci. 60
(2006) 4, 379--429.
Let $G$ be a transitive subgroup of the group of permutations $\mathfrak{S}_\Omega$ of a finite set $\Omega$.  Then, as a $G$-set, $\Omega$ is isomorphic to the $G$-set $G/H$ for some subgroup $H$ of $G$, uniquely determined up to conjugation by an element of $G$.  The permutation group $G$ is said to be primitive if the subgroup $H$ is maximal in $G$ (in the sense that $H\neq G$ and the only subgroups of $G$ containing $H$ are $G$ and $H$).
Recall also the notion of a torsor for a group $\Gamma$.  It simply means a transitive (and hence nonempty) $\Gamma$-set for which stabilisers of points are trivial.  If $\Gamma$ is a vector space of dimension $n$ over a field $k$, then a $\Gamma$-torsor $E$ is also called an affine $n$-space over $k$, and $\Gamma$ is said to be the space of translations of $E$.
Here is how I like to formulate the statement :
Theorem (Galois). If $\,G$ is a solvable primitive subgroup
of $\,\mathfrak{S}_\Omega$, then there is a unique structure on
$\,\Omega$ of an affine $n$-space over $\mathbf{F}_l$ (for some prime $l$ and
some $n>0$) such that
$$
N\subset G\subset\mathrm{AGL}(\Omega)\subset\mathfrak{S}_\Omega,
$$
where $N$ (resp. $\mathrm{AGL}(\Omega)$) is the space of translations
(resp. the group of automorphisms or invertible affine maps) of the
affine space $\Omega$.
Proof.  Let $N$ be a
minimal normal subgroup of $G$.  Since $G$ is solvable, $N$ is a
vector $n$-space over $\mathbf{F}_l$ for some prime $l$ and some $n>0$.
Since $G$ is primitive, $N$ is transitive.  Since $N$ is commutative
and transitive, $\Omega$ is an $N$-torsor.  Finally, one checks that $G$
acts on $\Omega$ by affine maps (because the conjugation action of
$G/N$ on $N$ is by $\mathbf{F}_l$-linear maps, automatically).  It follows from this that the degree of $G$ is $l^n$.  
Conversely,
Theorem. Let $\Omega$ be an affine space over $\mathbf{F}_l$ of
dimension $n>0$ and let $N$ be its space of translations.  An
intermediate group $N\subset G\subset\mathrm{AGL}(\Omega)$ is solvable
(resp. primitive) if and only if $G/N$ is solvable (resp. the
$\mathbf{F}_l[G/N]$-module $N$ is simple).
Proof. The bit about solvability is clear.  Suppose that $G$ is
imprimitive, and let $(\Omega_i)_{i\in I}$ be a $G$-stable
partition of $\Omega$ (with $1<|I|<l^n$).  Since $G$ is transitive (even
$N$ is transitive), the parts $\Omega_i$ have the same cardinal $l^m$,
for some $m\in[1,n[$.  One checks that they are affine subspaces, all
parallel to each other.  Their common direction $M\subset N$ is a
$G$-stable subspace of dimension $m$, so $N$ is not simple as an
$\mathbf{F}_l[G/N]$-module.  Conversely, if the $\mathbf{F}_l[G/N]$-module $N$ is not simple, let $M\subset N$ be a $G$-stable subspace of some dimension $m\in[1,n[$.  The family of affine subspaces of $\Omega$ of direction $M$ is a $G$-stable partition of $\Omega$ (having $l^{n-m}$ parts), so $G$ is imprimitive. 
