Equivalent Condition of Primitivity

Theorem

A transitive permutation group is primitive iff its some point stabilizer is a maximal subgroup.

\begin{proof} Let $G\le \mathrm{Sym}(\Omega )$ be a primitive permutation group. Assume that the point stabilizer $G_{\omega }$ is not a maximal subgroup. Then there exists a subgroup $H$ such that $G_{\omega }<H<G$. Let $B=H\omega$. We claim that $B$ is a block. If there is a $\sigma \in G$ such that $\sigma (B)\cap B\ne \varnothing$, then we have $h_1,h_2$ satisfying $\sigma h_2\omega =h_1\omega$ and so $h_1^{-1}\sigma h_2\in G_{\omega }<H$. It yields that $\sigma \in H$ and $\sigma (B)=B$, i.e., $B$ is a block. By the primitivity of $G$, either $B=\{\omega \}$ or $B=\Omega$. However, both of them are impossible.

Conversely, if $G$ is imprimitive, then there is a non-trivial block system $\mathcal{B}$. For any $\omega \in \Omega$, there is a $B\in \mathcal{B}$ such that $\omega \in B$. Since $G_{\omega }<G_B<G$, the group $G_{\omega }$ is not a maximal subgroup. \end{proof}