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 ?