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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 $ ?

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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 $4$.

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  • $\begingroup$ Is $ G $ primitive permutation group, since $ N $ is a nontrivial subgroup of $ G $ ? $\endgroup$ – Soroush Oct 15 '15 at 10:02
  • $\begingroup$ It is primitive because $M$ is a maximal subgroup of $G$. $\endgroup$ – Derek Holt Oct 15 '15 at 11:00
  • $\begingroup$ i don't understand $ G $ is primitive since $ M $ is the maximal subgroup. $\endgroup$ – Soroush Oct 15 '15 at 11:03
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Oct 15 '15 at 12:38
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    $\begingroup$ 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. $\endgroup$ – Derek Holt Oct 15 '15 at 14:46

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