Let $G$ be an abelian group of $p^{2n}$. Suppose for every $1 \le r \le n$, the subgroup $G_r=\{g \in G:g^{p^r}=e\}$ has order $p^{2r}$. Show that $G \cong \mathbb{Z}/p^n\mathbb{Z} \times \mathbb{Z}/p^n\mathbb{Z}$.
This is clearly a generalisation of the fact that a group of order $p^2$ is either a cyclic group or a product of two cyclic groups of order $p$. I am working on an exercise where $p^n$ is replaced by an arbitrary positive integer $M$, and I have decided to start with the case of $p$-groups (as the rest can be considered as an application of the Chinese remainder theorem). Here is how far I have gone.
We know that as an abelian $p$-group, $G$ can be written as $$ G \cong \mathbb{Z}/p^{r_1}\mathbb{Z} \times \cdots \times \mathbb{Z}/p^{r_s}\mathbb{Z} $$ where $0<r_1 \le \dots \le r_s$ and $r_1+\dots+r_s=2n$. We also see $G_n=G$. Therefore every element in $G$ has order divisible by $p^n$, which implies that $p^{r_i}|p^n$ for all $i$, i.e. $0<r_1\le \dots \le r_s \le n$. If we can show that $r_1 \ge n$ or even $r_{s-1} \ge n$ then we are done. I am thinking about showing that $r_1<n$ gives rise to a contradiction, but I have not figured out where to start. I should not only use the fact that $G_n=G$. Perhaps I need to look at $G_{r_1}$.
How can I push it further? Or does my approach have some problems that should be fixed first?
Thank you in advance.