Primitive permutation group

Remarks

Condition 2 says that every block $B$ of $G$ has $\abs{B} = 1$ or $B = \Omega$. If $\abs{\Omega} \ge 2$ and $G$ is transitive, then $G$ is primitive if and only if the stabiliser $G_\omega$ of a point $\omega$ is a maximal subgroup of $G$.

Example: a transitive group $G$ of prime degree $p$. For a transitive group, all parts of a block system have the same size, which divides $p$, so each block system is trivial. For instance, the cyclic group generated by $(1\,2\,3\,4\,5)$ is primitive on $\set{1, \dots, 5}$. Every $2$-transitive group is primitive (see multiply transitive group action).

Non-example: the eight symmetries of a square acting on its corners $1, 2, 3, 4$ in cyclic order. The action is transitive, but the partition $\set{\set{1, 3}, \set{2, 4}}$ into the two diagonals is a nontrivial block system, so clause 2 fails.

If $G$ acts primitively and faithfully, then every normal subgroup $N \ne 1$ of $G$ is transitive on $\Omega$, because the orbits of $N$ form a block system.