Why does a “block of imprimitivity” of an acting group generate a partition on the acted on set

group-actionsgroup-theory

Godsil&Royle Algebraic Graph Theory section 2.5 states (slightly paraphrased):

Let $G$ be a transitive group acting on a set $V$. A nonempty subset $S$ of $V$ is a block of imprimitivity for G if for any element $g\in G$, either $g(S) = S$ or $g(S)\cap S = \emptyset$. Because $G$ is transitive, it is clear that the translates of $S$ form a partition on $V$.

Assuming "translates of $S$" refers to all the $g(S)$, I understand why those cover $V$.

Question: But how does it follow that any two $g(S)$ and $h(S)$, if not identical, have an empty intersection?

Put another way: why is the situation in this diagram not possible?

overlapping g(S) and h(S)

The diagram shall describe $x_1,x_2\in S$ both mapped to some $x\notin S$ by $g$ and $h$ respectively where otherwise $g(S)\neq h(S)$.
I tried to derive a contradiction from this situation, but failed.

Best Answer

Consider $h^{-1}(g(S))$. On one hand, its intersection with $S$ is nonempty because $g(S)\cap h(S)$ is nonempty. On the other hand, it is not $S$: consider $y$ which is in $g(S)$ but not $h(S)$; then $h^{-1}(y)$ is not in $S$. (If such $y$ doesn't exist, then use the same reasoning with $g\leftrightarrow h$, which is possible unless $g(S)=h(S)$.)

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