1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.