# Dixon&Mortimer exercise involving size of orbits under subgroup of point stabilizer of given primitive permutation group

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 $$\{1_{G}\} also has a non-trivial orbit, of size at most $$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 $$H (for otherwise equality would entail that $$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..