Let $p$ be prime integer. I've read that every subgroup of $(\mathbb{Q}_p/\mathbb{Z}_p)^2$ has the form
\begin{equation} (\mathbb{Q}_p/\mathbb{Z}_p)^e\times U \end{equation}
with $0\le e \le 2$ and $U$ a finite group. My question is: why?
Some ideas:
$\mathbb{Q}_p/\mathbb{Z}_p$ is a $p$-divisible group that has only finite subgroups. One idea could be: consider $H$ a subgroup of $(\mathbb{Q}_p/\mathbb{Z}_p)^2$ and consider $D$ its maximal divisible subgroup. If $D\cong (\mathbb{Q}_p/\mathbb{Z}_p)^e$ and has finite index in $H$, then we're done since divisible subgroups are always direct summands. But I don't know if (or why) the hypotheses are true.