I am trying to answer the following:
Let $G$ be a finite group, $n$ a divisor of the order $|G|$ of $G$. If there are no subgroups of $G$ of order $n$, prove that $n$ has at least two distinct prime divisors.
I know that if $|n|= p$ for some prime $p$ , then by Cauchy's theorem $\exists g \in G$ such that $|g| = p$ and so the cyclic subgroup $\langle g \rangle$ has order $p$, a contradiction. I also know if $|n| = p^{m}$ , where $p^{m+1} \nmid |G|$, then by Sylow's theorem $\exists$ a subgroup of order $n$, a contradiction.
I am struggling in the case where $|n| = p^{l}$ and $p^{l+k} \mid |G|$ for some natural number $k$. In toy examples I am able to construct subgroups of order $p^{l}$, but don't know how to do it more generally (and thus settle that $n$ has more than one prime divisor).
Any hints would be appreciated!