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.


1 Answer 1


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.


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