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I come forth once again with a Dixon&Mortimer exercise (1.5.22), formulated as follows: let $G$ act faithfully and primitively (transitivity implicitly assumed) on finite set $A$ with $|A| \geqslant 2$ and fix $a \in A$. Assuming that $\mathrm{Stab}_{G}\ a$ has a non-trivial orbit of size $n$, show that any subgroup $H$ with $\{1_{G}\}<H\leqslant \mathrm{Stab}_{G}a$ also has a non-trivial orbit, of size at most $n$.

Provided the claim actually holds, any argument should rely on the existence of that one given non-trivial orbit of size $n$, say $B$ and show it is non-trivially partitioned into $H$-orbits, in other words that no such $H$ can fix $B\cup \{a\}$. Now, why should the primitivity of the action preclude this? It is not readily apparent that the $H$-fixed points would form a block in general (unless for the very particular case when $H$ is the stabilizer, but then the claim on the orbits follows trivially....).

It thus seems that what the exercise actually suggests is that any non-trivial orbit of the stabilizer is non-trivially partitioned into $H$-orbits. Perhaps there is something I am missing, which is why any clarifications are welcome!

P.S. Denoting for simplicity $\mathrm{Stab}_{G}\ a=E$, it is known that $E$ is maximal in $G$; as $E$ by hypothesis also has a non-trivial orbit, any orbit $X \in A/E$ will be a proper subset of $A$; defining $H=\mathrm{Stab}_{G}\ X$ for such an orbit, as $H<G$ (for otherwise equality would entail that $X$ is a nonempty proper sub-$G$-set of $A$, in contradiction with transitivity), $E \leqslant H$ by definition as $X=Ex$ for a certain $x \in A$ and maximality of $E$, we infer that $E=H$. Not that it seems to help too much..

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