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..


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.