I have this problem:
Let $G$ be an abelian group of order $72$. Show that $G$ has exactly one subgroup of order $8$.
I've seen how to find all abelian groups (up to isomorphism) of order $n$, and I know that if $m$ divides the order of a finite abelian group $G$, then $G$ has a subgroup of order $m$.
I also know that $72 = 2^33^2$. However, I feel like I'm missing something to put it all together. Can someone give me a hint or tell me the theorem I need to apply here?