Let $ G $ be a solvable primitive permutation group. Why the degree of $ G $ is a prime power

Question

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 $?

Answer 1

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.