Suppose that $G$ is a finite Abelian group that has exactly one subgroup for each divisor of |$G|$. Show that $G$ is cyclic.
What I have so far:
By the Fundamental Theorem of Finite Abelian Groups, we may write $G$ as $G=Z_{n_1}\oplus\dots\oplus Z_{n_k}$ for a set of $k$ integers $n_1$ through $n_k$ that are prime.
If $m$ divides the order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$.
I am not sure what else I need to know.