# Do these properties of a countable abelian group guarantee a Prüfer subgroup?

Suppose $$(G,+)$$ is a countable abelian group and $$p$$ is a prime number such that:

1. The subgroup $$pG$$ has finite index in $$G$$, and
2. For every $$n \in \mathbb{N}$$, $$G$$ contains an element of order $$p^n$$.

Must $$G$$ contain a subgroup isomorphic to the Prüfer $$p$$-group?

I have tried carrying out a pigeonhole-type argument as follows. Let $$x_n$$ be an element of order $$p^n$$. To produce a Prüfer subgroup, it suffices to show that infinitely many of the cyclic groups $$\langle x_n\rangle$$ are nested. Let $$G_n = \langle x_1, x_2, \dots, x_n\rangle$$. Since $$G_n$$ is a finite abelian group, it can be written in the form $$G_n = \bigoplus_{i=1}^{k_n}{\mathbb{Z}/p^{r_i}\mathbb{Z}}$$. If $$k_n$$ is bounded, then there must be an infinite sequence of nested cyclic groups and hence a Prüfer group. One many be inclined (as I was) to use condition (1) to show that $$k_n$$ is bounded, since $$[G_n : pG_n] = p^{k_n}$$. However, $$[G_n : pG_n]$$ is not bounded by $$[G : pG]$$, so this argument does not work. For example, the divisible group $$G = \bigoplus_{n=1}^{\infty}{\mathbb{Z}[p^{\infty}]}$$ contains $$\bigoplus_{n=1}^{\infty}{\mathbb{Z}/p^n\mathbb{Z}}$$ as a subgroup.

I don't see any immediate remedy to this argument, but I also don't see any simple candidates for counterexamples. My hope is that another approach can be used to find a Prüfer subgroup.

Yes, it must. And $$G$$ doesn't need to be countable.
Let $$H$$ be the $$p$$-primary component of the torsion subgroup of $$G$$. Then the natural map $$H/pH\to G/pG$$ is injective, so $$H$$ also satisfies (1), and clearly satisfies (2). So, replacing $$G$$ by the subgroup $$H$$, we shall assume that $$G$$ is a $$p$$-group.
Let $$X$$ be a finite subset of $$G$$ that generates $$G/pG$$, so $$G=\langle X\rangle + pG.$$
For some $$n$$, $$p^nx=0$$ for all $$x\in X$$, and so $$p^nG=p^n\langle X\rangle + p^{n+1}G=p^{n+1}G.$$
So $$p^nG$$ is divisible, and by (2) is nonzero. And a divisible abelian $$p$$-group is a direct sum of copies of the Prüfer $$p$$-group.