Let $A$ be an abelian group of order $N^r$ such that ${\rm Card}(A[D]) = D^r$ for all divisor $D$ of $N$.
Here $A[D]$ is the subgroup of elements $a$ such that $D\cdot a = 0$.
I need to show that under those conditions $A$ is isomorphic to $(\mathbb{Z}/{N\mathbb{Z}})^r$.
I can show that if the group has cardinal $p^r$ ($p$ prime) then it is isomorphic to $(\mathbb{Z}/{p\mathbb{Z}})^r$ because every non-zero element has order $p$. I don't know where to go from that.
Would you have a proof in simple terms, relying only on basic group theory results if possible?
Note: I found this problem in Arithmetic of elliptic curves, which is way above my level but for some reasons I'd like to be able to understand the structure of $E[n]$ on an elliptic curve $E$ (I'm interested in applications of elliptic curves to cryptography but I have very limited knowledge of group theory).