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A subgroup $ K $ of $ G $ is called almost maximal if $ K $ is a maximal subgroup of $ G $ or the index of $ K $ in $ G $ is of prime power order.

This is the theorem in the article and proved : Suppose that $ H \unlhd G $ such that $ G/H $ is supersoluble. Suppose that there is an element $ y \in H $ such that $ H = \langle y \rangle L $ for any almost maximal subgroup $ L $ of $ H $; then $ G $ is supersoluble.

Now prove this corollary of above theorem: Suppose that $ H \unlhd G $ such that $ G/H $ is supersoluble. If all the Sylow subgroups of $ H $ are cyclic, then $ G $ is supersoluble. Particularly, if $ H $ is a square free subgroup, then $ G $ is supersoluble. How prove this corollary by above theorem ?

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I can prove the corollary directly, without using the theorem. Use induction on $|H|$ - result is clear if $|H|=1$. Let $N$ be a minimal normal subgroup of $G$ with $N \le H$. Then, since $H$ is soluble, $N$ is elementary abelian but, since all Sylow subgroups of $H$ are cyclic, $N$ must be cyclic. Now by induction applied to $H/N \unlhd G/N$, $G/N$ is supersoluble. Then $N$ cyclic implies $G$ is supersoluble.

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