This question refers to permutation groups, in particular the primitive ones, and block systems. Let $G$ be a finite group acting on a set $\Omega$, and consider some partition $\Delta_1 \cup \ldots \cup \Delta_k = \Omega$. For $\omega\in \Omega$ denote by $G_{\omega} = \{ g \in G : \omega^g = \omega \}$ the point stabilizer. Is it true that if $\alpha \in \Delta_1$, then for every $g \in G_{\alpha}$ we have $$ \Delta_1^g = \Delta_1 $$ i.e. the elements of $G_{\alpha}$ leave $\Delta_1$ invariant?
This property is used in the proof of the following assertion.
Let $G$ be a group acting transitively on $\Omega$ and $|\Omega| > 1$. If the point stabilizer $G_{\omega}$ for some $\omega \in \Omega$ is not a maximal subgroup, then $G$ is imprimitive.
Proof: Suppose for $\omega \in \Omega$ that $G_{\omega}$ is not maximal, first we have $G_{\omega} \ne G$, because $G$ acts transitively and $|\Omega| > 1$. Now suppose $H$ is some subgroup with $G_{\omega} \lneq H \lneq G$ and let $\Delta_1 \cup \ldots \cup \Delta_k$ be the orbit decomposition of $\Omega$ by $H$. We claim this is an $G$-invariant partition. To show this let $g \in G$ and $\alpha \in \Delta_1 \cap \Delta_1^g$. Let $\beta \in \Delta_1$ such that $\beta^g = \alpha$ and let $h \in H$ with $\alpha^h = \beta$. Then we have $\alpha^{hg} = \beta^g = \alpha$, therefore $hg \in G_{\alpha}$, and $G_{\alpha}$ is contained in the stabilizer of $\Delta_1$ (as $\alpha \in \Delta_1$). So that $hg, h$ are contained in the stabilizer of $\Delta_1$, and so $g$ is also in the stabilizer, meaning $\Delta_1^g = \Delta_1$.
So the above partition is $G$-invariant, we just have to show that it is not trivial. But if we would have $\Omega = \Delta_1$, then $H$ would be transitive and by this $|G : G_{\omega}| = |\Omega| = |H : H_{\omega}| = |H : G_{\omega}|$, which contradicts $G_{\omega} \nleq H \nleq G$. Also the blocks are not singleton, because then $H$ would fix every element, meaning $G_{\omega} = H$, which also contradicts $G_{\omega} \nleq H$. $\square$.
I hightlighted the part I do not understand, the rest is clear to me.
Perhaps it is just that the wording there is a bit confusing.
The set $\Delta_{1}$ was defined to be an $H$-orbit, which means that $H$ stabilises $\Delta_{1}$ (i.e., $H\leq G_{\Delta_{1}}$), and $H$ was chosen initially to contain the point stabiliser $G_{\alpha}$, so we get the chain $$G_{\alpha}\leq H\leq G_{\Delta_{1}}$$ of inclusions.