My understanding is that every countable Abelian $p$-group can be written as a direct sum of at most countably many finite cyclic groups and at most countably many copies of the Prüfer $p$-group.
Suppose that when such a direct-sum decomposition is written for a particular countable Abelian $p$-group $G$, there are $N$ copies of the Prüfer $p$-group in the decomposition, where $N$ can be $0$, finite, or infinite. Then my question is, does that mean that there are exactly $N$ subgroups of $G$ that are isomorphic to the Prüfer $p$-group? Or can there be more than $N$ such subgroups?
(in additives notations)
With $N=0$ then $$G=\prod_{n\ge 1}' \Bbb{Z}/p^{k_n} \Bbb{Z}$$ where $\prod'$ means we consider the 'sequences' $x_n\in \Bbb{Z}/p^{k_n} \Bbb{Z}$ with only finitely many terms $\ne 0$.
For $x,y\in G$, $x = p^m y$ implies that $x_n=0$ whenever $m\ge k_n$.
Thus there is no non-zero element $x\in G$ such that $\forall m, \exists y\in G,x=p^m y$ ie. there is no copy of $\Bbb{Z}[p^{-1}]/\Bbb{Z}$ in $G$.
With $N=1$ we are looking at $G\times \Bbb{Z}[p^{-1}]/\Bbb{Z}$ and it works more or less the same way.
With $N=2$ we are looking at $G\times \Bbb{Z}[p^{-1}]/\Bbb{Z}\times \Bbb{Z}[p^{-1}]/\Bbb{Z}$ and something different happens.