While studying elliptic curves I was trying the following problem related with group theory:
Let $A$ be an abelian group of order $m^2$ with $|A[d]|=d^2$ for every $d\mid m$. Prove that $A\cong \mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$
$A[d]$ denote the set of points of $A$ with order dividing $d$.
If $m=p$ prime then either $A\cong \mathbb{Z}/p^2\mathbb{Z}$ or $A\cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, but $|A[p]|=p^2$ implies that there are $p^2$ elements of order $p$ and thus $A\cong \mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$, since in $\mathbb{Z}/p^2\mathbb{Z}$ there are $\varphi(p^2)=(p-1)p$ elements of order $p$. In a similar way, using the theorem of classification of finite abelian groups we can prove it for $m=p_1p_2$ with $p_1,p_2$ distinct prime numbers.
What about the general case? For any given $m$ we can solve it by checking cases, but if $m=p_1^{a_1}\dots p_n^{a_n}$ with $p_1,\dots,p_n$ there are many options that checking the cases one by one seems not a reasonable option.
My question is if there are some clever argument that solves the general case withouth checking the cases by hand.
Thanks for your help.
Since every finite abelian group is the direct product of its Sylow $p$-subgroups, it suffices to prove this when $m=p^a$ is a power of a prime. And when $m=p^a$, checking $d=p$ shows that the group is the direct product of two $p$-groups, while checking $d=p^a$ shows that the direct factors must both have order $p^a$ as needed.