Orbits of Normal Subgroups Form a Block System

Proposition

Let $G⩽\mathrm{Sym}(\Omega )$ be a group and let $N⊲G$ be a normal subgroup of $G$. Then the set of orbits of $N$ forms a block system of $\Omega$.

\begin{proof} Suppose $O_1,\cdots ,O_k$ are orbits of $N$. Then ${\bigsqcup }_{i=1}^k{O}_i=\Omega$ and $O_i\cap O_j=\varnothing$ for any $i\ne j$. For any $g\in G$, we have

$$(O_i^g{)}^N=(O_i^{gN{g}^{-1}}{)}^g=(O_i^N{)}^g=O_i^g,$$

and it yields $O_i^g$ is a union of $N$-orbits. For ${\omega }_1^g,{\omega }_2^g\in O_i^g$, there exists $n\in N$ such that ${\omega }_1^n={\omega }_2$. Then $({\omega }_1^g{)}^m={\omega }_2^g$ with $m=g^{-1}ng\in N$. So $N$ is transitive on $O_i^g$ and so $O_i^g=O_j$ for some $j\in \{1,\cdots ,k\}$. Therefore, the set of orbits of $N$ forms a block system. \end{proof}