Quasiprimitive Rank 3 are Almost Simple

Proposition

Let $G$ be a quasiprimitive but not primitive group. Then $G$ is almost simple.

\begin{proof} By Imprimitive Rank 3 Group is a Wreath Product of two 2-Transitive Groups, group $G$ is isomorphic to $G_B^B:G^{\mathcal{B}}$, where $G_B^B$ and $G^{\mathcal{B}}$ are $2$-transitive groups. If $G^{\mathcal{B}}$ is affine, then there is $N⊲G^{\mathcal{B}}$ such that $N$ is regular on $\mathcal{B}$. Then the orbits of $N$ on $\Omega$ forms a different block system, which is impossible by Uniqueness of Block System of Imprimitive Rank 3 Group. So $G^{\mathcal{B}}$ is almost simple.

We claim that $G\cong G^{\mathcal{B}}$. Otherwise, there is a nontrivial normal subgroup $M$ such that $G^{\mathcal{B}}\cong G/M$. Since $M$ is the kernel of $G$ acting on $\mathcal{B}$, it is not transitive on $\Omega$, contradiction. Therefore, it yields that $G\cong G^{\mathcal{B}}$ is almost simple. \end{proof}

ref: On imprimitive rank 3 permutation groups, Lemma 3.3 and Corollary 3.4.