Let $G$ be a finite abelian group of order $p^{2k}$, where $p$ is a prime. Suppose, for every divisor $m$ of $p^k$, $G_m:= \{ x \in G : mx=0\}$ has order $m^2$. Then is it true that $G \cong \mathbb Z /p^k \mathbb Z \times \mathbb Z /p^k \mathbb Z $ ?
I can see that $G_{p^k}= G$, so every element of $G$ has order dividing $p^k$. Moreover, if $d$ annihilates all the elements of $G$, then since $G=G_d$ ,so $d^2=p^{2k}$, so $d=p^k$. So $G$ has annihilator $p^k \mathbb Z$ as $\mathbb Z$-module, so $G$ has an element of order $p^k$. But I am unable to proceed further. Please help.
The structure theorem for finite Abelian groups says that there exists $e_1, \cdots , e_n > 0$ such that
$$ G \cong \mathbb{Z}/p^{e_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p^{e_n}\mathbb{Z}\, . $$
The fact that $G_{p^k} = G$ implies that all elements have order at most $p^k$, so this means that all $e_i$ are at most $k$.
The fact that $|G_p| = p^2$ implies that there are only two factors (so $n = 2$). This is so because every factor $\mathbb{Z}/p^{e_i}\mathbb{Z}$ contributes a factor $p$ to $|G_p|$.