# Normal subgroups of transitive group actions are primitive

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?

The statement is wrong. (If it were true, then every transitive group would be primitive, by taking the normal subgroup to be $G$ itself.)
First off, the condition for $\Delta$ to be a block is $\Delta^x=\Delta$ or $\Delta \cap \Delta^x=\varnothing$. As regards the problem, since $G$ is transitive, there is only one orbit of elements of $\Omega$. Thus given $\delta_1, \delta_2 \in \Delta$ we can find $g \in G$ such that $\delta_1^g=\delta_2$. As stated in the paper, the orbits of the normal subgroup $N$ for a system of blocks. The idea here is that the elements of $G/N$ act by permuting the blocks themselves, and that the elements of $N$ act by permuting the elements WITHIN the blocks. You can show that if the elements of $N$ do not act transitively WITHIN the blocks, then $G$ as a whole is not transitive.
• I corrected the equation $\Delta^x \cap \Delta = \emptyset$. That $N$ acts transitively on the blocks is clear to me, because these blocks are orbits for $N$. But why does that implies that each proper subset $\Delta$ of $\Omega$ with $|\Delta| > 1$ could not be a block? I have to find some $x \in N$ such that $\Delta^x \ne \Delta$ and $\Delta^x \cap \Delta \ne \emptyset$, I do not see that what you wrote helps in finding such an $x \in N$?? – StefanH Sep 7 '15 at 17:01