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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.

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    $\begingroup$ The first thing that comes to mind is to tackle it via the contrapositive: assume that $A\cong \mathbb{Z}/m_1\mathbb{Z}\times\mathbb{Z}/m_2\mathbb{Z}$ with $m_1m_2=m^2$ but $m_1\neq m_2$ and show that there's some $d$ with $|A[d]|\neq d^2$. Given $m_1\neq m_2$ there's at least one prime $p$ that occurs to different powers in $m_1$ and $m_2$, and looking at $|A[p^n]|$ for some suitably chosen $n$ seems like a good starting point. $\endgroup$ Feb 7 at 18:22
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    $\begingroup$ I doubt that $A[d]$ is "the set of points of $A$ with order $d$". I expect it is the set of points with exponent $d$, that is, of order dividing $d$. Otherwise, no group satisfies the equation, so you can prove whatever you want with it. For example, the number of elements of order $2$ in $\mathbb{Z}_2\times\mathbb{Z}_2$ is three, not $4$. $\endgroup$ Feb 7 at 18:33
  • $\begingroup$ @ArturoMagidin yeah, you're right, sorry for the typo. $\endgroup$
    – Marcos
    Feb 7 at 19:24

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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.

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  • $\begingroup$ Thanks, short and elegant. $\endgroup$
    – Marcos
    Feb 7 at 19:24

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