Uniqueness of Block System of Imprimitive Rank 3 Group

Lemma

Assume that $G⩽\mathrm{Sym}(\Omega )$ is a finite imprimitive rank $3$ group. Then $G$ has a unique non-trivial proper block system $\mathcal{B}$.

\begin{proof} Let $\alpha \in \Omega$, and let $n=|\Omega |$. Then $G_{\alpha }$ has three orbits: $\{\alpha \}$, ${\Gamma }_{\alpha }$ and ${\Delta }_{\alpha }$. We claim that a non-trivial proper block containing $\alpha$ is either $\{\alpha \}\cup {\Gamma }_{\alpha }$ or $\{\alpha \}\cup {\Delta }_{\alpha }$. Let $B$ be a block containing $\alpha$ such that $B\ne \{\alpha \}$ or $\Omega$. Then $B^{G_{\alpha }}=B$ as $\alpha \in B^{G_{\alpha }}$. It follows that $B$ is a union of some orbits of $G_{\alpha }$. Thus, either $B=\{\alpha \}\cup {\Gamma }_{\alpha }$ or $B=\{\alpha \}\cup {\Delta }_{\alpha }$.

Assume that $|{\Gamma }_{\alpha }|=k$ and $|{\Delta }_{\alpha }|=\ell$. Then $n=1+k+\ell$. We may assume that $k⩽\ell$. If $\{\alpha \}\cup {\Delta }_{\alpha }$ is a block, then $1+k+\ell$ is divided by $\ell +1$, which is impossible. Hence, $\{\alpha \}\cup {\Gamma }_{\alpha }$ is the unique non-trivial proper block containing $\alpha$. \end{proof}

Remark. This also tells us that $|{\Gamma }_{\alpha }|\ne |{\Delta }_{\alpha }|$ if $G$ is imprimitive.

ref: Finite permutation groups of rank 3