Primitive solvable group acts on a set iff the set has prime power size

Statement
The following are equivalent for a natural number $$n$$:


 * 1) $$n$$ is a fact about::prime power.
 * 2) There exists a solvable group $$G$$ with a fact about::primitive group action on a set of size $$n$$.
 * 3) There exists a solvable group $$G$$ with a fact about::doubly set-transitive group action on a set of size $$n$$.
 * 4) There exists a solvable group $$G$$ with a fact about::doubly transitive group action on a set of size $$n$$.

(Note that since solvability is quotient-closed, assuming that the action in any of the above cases is faithful does not alter the strength of the statement).

Related facts

 * Primitive implies Fitting-free or elementary abelian Fitting subgroup
 * Classification of solvable transitive subgroups of symmetric group of prime degree
 * Primitive solvable group acts on a set iff the set has prime power size

Facts used

 * 1) uses::Doubly transitive implies doubly set-transitive
 * 2) uses::Doubly set-transitive implies primitive
 * 3) uses::Primitive implies innately transitive: In a primitive group, there is a transitive minimal normal subgroup.
 * 4) uses::Minimal normal implies elementary abelian in finite solvable
 * 5) uses::Fundamental theorem of group actions

(1) implies (4)
This follows from an explicit construction: consider the field $$F$$ of order $$n$$, and consider the general affine group $$GA(1,F)$$, acting on the set $$F$$. This is doubly transitive.

(4) implies (3) implies (2)
This follows from facts (1) and (2).

(2) implies (1)
As remarked earlier, we can assume that the action is faithful, and hence, $$G$$ is a primitive group. Also, since the set is finite, $$G$$ is finite.

By facts (3) and (4), $$G$$ has a transitive minimal normal subgroup that is also elementary abelian. Thus, there is an elementary abelian group acting transitively on the set of size $$n$$. By fact (5), $$n$$ must divide the order of the elementary abelian group. Since elementary abelian groups have prime power order, $$n$$ must be a prime power.