Exercise 4.2.14 (If $G$ has a subgroup of order $k$ for each divisor $k$ of $n$, then $G$ is not simple)

Proof. Let $p$ be the smallest prime dividing $|G|$. Since $n∕p$ is a divisor of $n$, there is by hypothesis some subgroup $H$ of $G$ of order $n∕p$, so $|G:H|=p$. In particular, $H$ is a proper subgroup of $G$.

According to Corollary 5, $H$ is normal in $G$.

Since $n$ is composite, $n\ne p$, therefore $H$ of order $n∕p$ is not trivial, so $H$ is a proper non trivial normal subgroup of $G$. This shows that $G$ is not simple.

If $G$ has a subgroup of order $k$ for each for each positive integer $k$ dividing $n$, then $G$ is not simple. □