## A finite group of composite order n having a subgroup of every order dividing n is not simple

Let $G$ be a finite group of composite order $n$ with the property that for all $k|n$, $G$ has a subgroup of order $k$. Prove that $G$ is not simple.

Let $p$ be the smallest prime dividing $n$, and write $n = pm$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. By Corollary 5 in the text, $H$ is normal in $G$.