# Index of maximal subgroup of soluble group

Let $G$ is a soluble group. If $P_{G}(M) > M$, for any subgroup $M$ of prime power index in $G$, then every chief factor of $G$ has order $4$ or a prime. ( $P_{G}(M) = \langle g\in G \ \vert \ \langle g \rangle M = M \langle g \rangle \rangle$ ). This is a theorem. Now if $M$ a maximal subgroup of $G$ then $M$ has a prime power index in $G$, since $G$ is a soluble group. Why $M$ has index a prime or $4$ ?

Let $|G:M|=p^m$ for a prime $p$. Consider the permutation action of $G$ on the right cosets of $M$. The image of $G$ is a soluble primitive permutation group of degree $p^m$. A minimal normal subgroup $N$ of this image is transitive and abelian (by solubility), so it acts regularly and has order $p^m$. But the image is isomorphic to a quotient group $G/K$ of $G$, so $N= L/K$ is a chief factor of $G$, and hence $p^m=p$ or $4$.
• Is $G$ primitive permutation group, since $N$ is a nontrivial subgroup of $G$ ? – Soroush Oct 15 '15 at 10:02
• It is primitive because $M$ is a maximal subgroup of $G$. – Derek Holt Oct 15 '15 at 11:00
• i don't understand $G$ is primitive since $M$ is the maximal subgroup. – Soroush Oct 15 '15 at 11:03
• It is a standard result that an action of a group $G$ is primitive if and only if the stabilizer of a point in the action is a maximal subgroup of $G$. In the case of the action by multiplication on the cosets of a subgroup $M$, the stabiliser of the coset $M$ is $M$ itself, which you are assuming to be maximal. – Derek Holt Oct 15 '15 at 12:38
• My last comment was not quite accurate. I should have said that a transitive action of $G$ is primitive if and only if the stabilizer is a maximal subgroup. The action on cosets by right multiplication is transitive, so this result applies in your case. – Derek Holt Oct 15 '15 at 14:46