This does not answer your question exactly, but I would like to point out a Theorem of H. Wielandt of this general nature, which I find appealing.
Wielandt generalized the famous theorem of Frobenius as follows: let $G$ is a finite group and let $H$ be a proper subgroup of $G$. Suppose that $H$ has a normal subgroup $H_{0}$ such that $H \cap H^{g} \leq H_{0}$ for all $g \in G \backslash H$. Then there is a normal subgroup $G_{0}$ of $G$ such that $G = HG_{0}$ and $H \cap G_{0} = H_{0}.$ This generalizes Frobenius' theorem which is the case $H_{0} = 1$. The proof may be found in (Curtis and Reiner, Representation theory of finite groups and associative algebras, 1962).
A consequence of this is that if $G$ is a finite simple primitive permutation group acting on the set $\Omega$ with non-trivial point stabilizer $H = G_{\alpha}$, then $G_{\alpha}$ is generated by the two-points stabilizers $G_{\alpha \beta} : \beta \neq \alpha \in \Omega$ (here, $\alpha$ is fixed).
To see this, note that the group generated by the above two-point stabilizers is $H_{0} = \langle H \cap H^{g} : g \in G \backslash H \rangle$ ( as $G$ is simple and primitive, $H$ is maximal and $H = N_{G}(H)$). This is a normal subgroup of $H$. Then clearly $H \cap H^{g} \leq H_{0}$ for all $g \in G \backslash H$, so there is $G_{0} \lhd G$ with $G = HG_{0}$ and $H \cap G_{0} = H_{0}.$ If $H_{0}$ is proper in $H$, then $G_{0}$ is proper in $G$. Since $H$ is proper and $G = G_{0}H$, we see that $G_{0}$ is non-trivial. Hence the simplicity of $G$ is contradicted. Thus we must have $H_{0} = H.$