# Let $A$ be abelian of order $m^2$ with $|A[d]|=d^2$ for every $d\mid m$. Prove $A\cong \Bbb{Z}/m\Bbb{Z}\times\Bbb{Z}/m\Bbb{Z}$

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.

• 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. Feb 7 at 18:22
• 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 $4$. Feb 7 at 18:33
• @ArturoMagidin yeah, you're right, sorry for the typo. Feb 7 at 19:24

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.

• Thanks, short and elegant. Feb 7 at 19:24