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.



Do these properties of a countable abelian group guarantee 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.


