If $G$ acts transitive on $\Omega$, a subset $\Delta \subseteq \Omega$ is called a block if for each $x \in G$ we have either $\Delta^x = \Delta$ or $\Delta^x \cap \Delta = \emptyset$. The singletons and the whole of $\Omega$ are always blocks, called the trivial blocks. If $G$ has no non-trivial blocks, it is called primitive. I want to prove the following:
Let $G$ be a transitive permutation group acting on $\Omega$. Then every non-trivial normal subgroup of $G$ is primitive.
This is mentioned here as Propoosition 1.1. (c) and seems to follow from the fact that the orbits of a normal subgroup form a system of blocks. I tried to prove it on my own, but I am stuck. So any hints?
Best Answer
The statement is wrong. (If it were true, then every transitive group would be primitive, by taking the normal subgroup to be $G$ itself.)
What Peter Cameron surely intended is "every non-trivial normal subgroup of a primitive group is transitive", a well-known and important fact.
(This intention is also clear from the next sentence, defining quasiprimitive groups.)