# If $G$ is solvable and acts such that $\mbox{fix}(g) \in \{0,p\}$ for $g \ne 1$, then maximal normal subgroups are Frobenius groups

Let $p$ be an odd prime and let $G$ be a solvable, transitive permutation group such that each nontrivial element fixes no points or exactly $p$ points on a set $\Omega$. Further suppose that for $g \notin N_G(G_{\alpha})$ we have $$G_{\alpha} \cap G_{\alpha}^g = 1$$ for some point stabilizer $G_{\alpha}$ and that $p$ is the smallest prime dividing the order of $G$.

Now let $M$ be a maximal normal subgroup of $G$, then by solvability $|G / M|$ is a prime $q$. Assume $M$ is intransitive, then it has $q$ orbits as the number of orbits has to divide $|G : M|$ and $M_{\alpha} = G_{\alpha}$ by maximality. Let $\Delta$ be the orbit of $M$ containing $\alpha$. Denote by $\overline \alpha$ the set of $p$ fixed points of $G_{\alpha}$.

Why does $M$ acts as a Frobenius group on its orbits if $\overline \alpha \not\subseteq \Delta$?

I hope you see it, for me it is not clear, but I guess it is just a simple observation.

Since $N_G(G_\alpha)$ acts transitively on $\overline{\alpha}$ and $p$ is prime, there must be an element $g \in N_G(G_\alpha)$ that induces a $p$-cycle on $\overline{\alpha}$. If we had $\alpha,\beta \in \Delta \cap \overline{\alpha}$, then some power of $g$ would map $\alpha$ to $\beta$ and then, since $\Delta$ is a block of imprimitiveity, $g$ would fix the set $\Delta$, which would imply $\overline{\alpha} \subseteq \Delta$, contrary to assumption.
So $\Delta \cap \overline{\alpha} = \{\alpha\}$ and the result follows.