# If $G$ acts such that $\mbox{fix}(g) \in \{0,3\}$ for $g \ne 1$, and stabilizers are t.i. subgroups, then the Sylow $3$-subgroups have maximal class

Let $G$ be a transitive permutation group such that every nontrivial element fixing some point fixes exactly $3$ points. Also assume that for $g \notin N_G(G_{\alpha})$ we have $$G_{\alpha} \cap G_{\alpha}^g = 1$$ (i.e. they are trivial intersection (t.i.) subgroups) and that $G$ has a regular normal subgroup $R$ of odd order.

Suppose $C$ is the subgroup generated by all regular normal subgroups of $G$, and assume $R < C$ is properly contained in $C$.

If $9 \mid |\Omega|$, then the Sylow $3$-subgroups of $C$ have order at least $27$ and maximal class and they are also Sylow $3$-subgroups of $G$.

Why do the Sylow $3$-subgroups of $C$ have maximal class and are also Sylow $3$-subgroups of $G$?

I see that $|C| = |C_{\alpha}||\Omega| = |C_{\alpha}||R|$ and that $C / R$ must be a $3$-group (I can add details about that if wanted), but the above mentioned statements I do not see?

Remark: The condition that the point stabilizers are trivial intersection subgroups implies that $|N_G(G_{\alpha}) : G_{\alpha}| = 3$ and that all nontrivial elements from $G_{\alpha}$ have the same three fixed points.

• "Stabilizers are t.i. subgroups" means? (in title) – p Groups Jan 9 '16 at 3:41
• @pGroups A subgroup $U$ is a trivial intersection (t.i.) subgroup, if different conjugates have trivial intersection, i.e. $U \cap U^g = 1$ for $g \notin N_G(U)$. I added this definition to my question. – StefanH Jan 9 '16 at 10:35