If a group $G$ acts on some set $\Omega$, then we have a natural action on $\Omega\times\Omega$ given by $(\alpha,\beta)^x := (\alpha^x, \beta^x)$. If $\Delta \subseteq \Omega\times\Omega$ is some orbit of this action on $\Omega\times\Omega$, then we could look at the graph given by vertex set $\Omega$ and edge set $\Delta$, this graph is called the orbital graph of $\Delta$. One characterisation of primitivity is that $G$ acts primitive iff for each orbital its corresponding graph is connected. If all orbital graphs are strongly connected (i.e. between any two vertexes there exists a directed path), then we say that $G$ acts strongly primitive.
I want to show:
Suppose that the group $G$ acts transitively on $\Omega$ and let $G_{\alpha}$ be a point stabilizer. Show that $G$ is strongly primitive if and only if there is no subset $T$ with the properties $G_{\alpha} \subset T \subset G$ and $TT \subseteq T$ (in other words, $G_{\alpha}$ is a maximal subsemigroup of $G$).
I am stuck with this, has anybody hints how to show this? One Lemma concerning strong primitivity I have at hand is:
Lemma: Let $G$ be a group acting primitively on $\Omega$.
(i) The orbital graph of some orbit $\Delta$ is strongly connected iff it contains a nontrivial directed cycle,
(ii) If for each pair $\alpha, \beta$ there exists $z \in G$ with a cycle of finite length containing both $\alpha$ and $\beta$, then $G$ is strongly primitive,
(iii) If $G$ is periodic (in particular finite), then $G$ is strongly primitive.
Maybe this might be of help?