Basic Abstract Algebra - Subgroups of Abelian Group I'm trying to prove the following:

Let $G$ be an abelian group of order 72. Show that $G$ has exactly one subgroup of order 8.

I know by theorem that $G$ must have at least one subgroup of order 8, but I'm having trouble showing it has only one.
I know that it is necessary to use the prime decomposition theorem, but I'm having trouble understanding how finding the isomorphism classes of $G$ will help.
Thanks

By the classification theorem of finite abelian groups, the group is isomorphic to $G \times H$ where $G$ and $H$ are groups of orders $8$ and $9$, respectively.
From here, we can prove that $(g,h) \in G \times H$ has order dividing $8$ only if $h=\mathrm{id}_H$.  This implies that $G \times \{\mathrm{id}_H\}$ is the only subgroup of order $8$.
