# Permutation group

let $G$ be a primitive permutation group of degree prime order $p$, such that for stabilizers $G_\alpha$ and $G_\beta$ in $G$, $G_\alpha\cap G_\beta=\{1\}$, prove $G$ is not simple.

• Is this homework? – Mariano Suárez-Álvarez Sep 19 '13 at 11:47
• Is $C_P$ with the obvious action not a counterexample to this? – Tobias Kildetoft Sep 19 '13 at 11:52
• How? please explain. – mehranian Sep 19 '13 at 11:56
• $C_P$ is a simple group, but with the usual action on $P$ elements, it is a primitive permutation group with all stabilizers trivial. – Tobias Kildetoft Sep 19 '13 at 11:58
• Well if the stabilizers are non-trivial this result is true, and follows because in a primitive permutation group, the stabilizers are maximum subgroups (and are of course all conjugate). – user641 Sep 19 '13 at 19:32

• If $[G:A]$ is finite then $[G:A\cap B]\le[G:A][G:B]$.
• If $G$ is a permutation primitive group on $\Omega$, then $|G|=p$ which $p$ is prime or $$\exists~\alpha\neq\beta\in\Omega,~~G=\langle G_{\alpha},G_{\beta}\rangle$$
• If $G$ is regular the claim is obvious, so think of $|G|$ and show that $|G|\leq p^2$.
• All $G_{\alpha}$s are conjugates since the action is transitive. And if $|\Omega|=p$ then $[G:G_{\alpha}]=p$ and so $p||G|$.
• And finally show that the $p-$sylow subgroup of $G$ is normal in the group.
• You are making this a lot more complicated than it needs to be, I think. The $G_\alpha$ are maximal, with trivial intersections. They each have index $p$. There are $p$ of them. So $|\cup G_\alpha| = |G|-p+1$, and thus the subgroup of order $p$ is normal (contains all the elements fixing no points). – user641 Sep 20 '13 at 19:28