Silverman AEC: exercise 3.30 Supposing A is finite abelian with #A = $N^r$ and for every $d | n$ we know #$A[D] = D^r$, with $A[D]$ the subgroup with all elements of order $D$. How do you prove that $A \simeq Z_N^r$ ?
I undestand that the structure theorem for finite abelian groups is required, but how can it be applied to solve the problem? As I'm an undergraduate student, a simple explanation would help me a lot.
 A: First, show that it suffices to prove the result for $p$-primary components, for each prime $p|N$. That is, it suffices to show that under your conditions, $A[p^n] \cong (\mathbb{Z}/p^n\mathbb{Z})^r$ for every $p|N$, where $p^n||N$.
Then, look at $p$-primary components and see what it is possible, and what is not. For example, suppose $\# A = (p^2)^2$ and $\#A[p]=p^2$, and $\#A[p^2]=p^4$. Since $A$ is abelian and has size $p^4$, there are only a few possibilities:
$$\mathbb{Z}/p^4\mathbb{Z},\ \text{ or }\  \mathbb{Z}/p^3\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z},\ \text{ or } \ \mathbb{Z}/p^2\mathbb{Z}\times \mathbb{Z}/p^2\mathbb{Z}.$$
What possibility satisfies the hypotheses and why? That should give you the clue about how to finish the proof. (Hint: count elements of order $p$ in each case.)
A: *

*Fundamental theorem of finite abelian groups $$ N = \prod_p p^{n_p}, \qquad A \cong \prod_p A[p^\infty], \qquad A[p^\infty] \cong \prod_{i=1}^{j_p} \Bbb{Z/p^{e_{p,i}}Z}$$

*Note $\# A[p] =p^r$ tell us that $j_p = r$. 
So the number of elements whose order divides $p^k$ is $$\# A[p^k] = 
 \prod_{i=1}^r p^{\min(k,e_{p,i})}$$ 
We are told that $\# A[p^{n_p}] = p^{rn_p}$ thus $e_{p,i} \ge n_p$ and since $\#A[p^\infty] = p^{rn_p}$ then $e_{p,i} = n_p$ and $$A[p^\infty] \cong (\Bbb{Z/p^{n_p}Z})^r, \qquad A \cong \prod_p (\Bbb{Z/p^{n_p}Z})^r \cong (\Bbb{Z}/ N \Bbb{Z})^r$$
