# Proof corollary of the theorem about supersolubility

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 ?

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.