Transitive and Imprimitive group action with certain orbit lengths

Question

I am reading the book of D.E.Taylor the Geometry of Classical Groups and it is asked that:

Let $G$ act transitively on a set $\Omega$ of size $n$ and that $Stab_{G}(\omega)$ has orbits of lengths $1$, $k$ and $l$ on $\Omega$ for some $\omega \in \Omega$. If $G$ is imprimitive, show that either $k + 1$ or $l + 1$ divides $n$.

Here is what I have done so far:

We can easily see that $|\Omega|=n=1+k+l$. Now since $G$ acts imprimitively on $\Omega$, $Stab_{G}(\omega)$ is not a maximal subgroup of $G.$ Then say $H$ is a proper subgroup of $G$ properly containing $Stab_{G}(\omega).$ Then $Orb_{H}(\omega)$ is a block of imprimitivity for $\Omega.$ That is for all $g \in G$ either $g\cdot Orb_{H}(\omega)=Orb_{H}(\omega)$ or $g\cdot Orb_{H}(\omega) \cap Orb_{H}(\omega)= \emptyset. $ This is the point where I am stuck.

When we consider the action of $H$ on $\Omega$ how can we find the lengths of orbits of $H$ on $\Omega$? Specifically how the orbits change when we pick $h \in H \setminus Stab_{G}(\omega) $ ?

Answer 1

Hint. See if you can prove

• the $H$-orbits are unions of $\mathrm{Stab}_G(\omega)$-orbits
• $\omega$ is not fixed by $H$
• $H$ has more than one orbit

using $\mathrm{Stab}_G(\omega)\subsetneq H\subsetneq G$.