This question is comes from the book GTM$163$ "Permutation groups" by John D.Dixon, page13, exercise1.5.5.
The exercise is :
Suppose that $G$ is a group acting transitively on a set $\Omega$ with at least two points, and that $\Delta$ is a nonempty subset of $\Omega$. Show that $\Delta$ is not a block $\iff$ for each pair of distinct points $\alpha,\beta$, there is $x\in G$ such that $\alpha,\beta$ exactly one lies in $\Delta^{x}$.
Recall:
A subset $\Delta\subseteq\Omega$ is called a block for $G$, if $\forall x\in G$, either $\Delta^{x}=\Delta$ or $\Delta^{x}\cap\Delta=\emptyset$. Where $$ \Delta^{x}=\{\delta^{x}\mid\delta\in\Delta\} $$
My question:
I know $\Leftarrow$) is obvious, but I cannot get the inverse. I would gracefu if you can give me an instruction.
Here is the correct formulation of the problem as pointed out in the errata off Dixon's webpage: Let $G$ act primitively on $\Omega$ and let $\Delta$ be a proper subset containing at least two distinct elements. Then for each pair of distinct points $\alpha , \beta \in \Delta$, there exists an $x \in G$, such that exactly one of either $\alpha \in \Delta^x$ or $\beta \in \Delta^x$.
Now, to prove this statement, we assume for contradiction that there exists a pair $\alpha,\beta \in \Delta$ such that no such $x \in G$ exists. We first consider the set $\Gamma := \bigcap\{\Delta^g \; | \; \alpha \in \Delta^g\}$. We claim that $\Gamma$ is a block of $G$ containing $\alpha$ (this is routine to verify from the definition of a block). Also, our assumption that no such $x$ exists implies that $\beta \in \Gamma$ as well, so $\Gamma$ is a block of a primitive group $G$ such that $\{\alpha,\beta\} \subseteq \Gamma \subseteq \Delta \subset \Omega$. This is impossible.