I am self-studying group theory and in Exercise 10.38 of Rotman's An Introduction to the Theory of Groups, the exercise is to prove that a torsion abelian group is indecomposable if and only if it is isomorphic to a subgroup of the Prüfer group $\mathbb{Z}(p^\infty)$ for some prime $p$. So far I have:
Assume $G$ is indecomposable. Since $G$ is a torsion abelian group, it is the direct product of its $p$-primary components, but as it is indecomposable, it must itself be $p$-primary for some prime $p$. Imbed $G$ in a divisible abelian group $D$. Now the torsion subgroup of $D$, $tD$, is also divisible and contains $G$, and by the classification of divisible abelian groups, $tD$ is isomorphic to a direct sum of copies of $\mathbb{Q}$ and $\mathbb{Z}(p_i^\infty)$ for various primes $p_i$. But there can be no copies of $\mathbb{Q}$ as $tD$ is torsion, so there are only copies of the $\mathbb{Z}(p_i^\infty)$, and therefore $G \leq tD \cong \sum \mathbb{Z}(p_i^\infty)$ for some primes $p_i$. But since $G$ is $p$-primary, it cannot contain any elements with a nonzero component from any direct summand $A_i$ with $p_i \neq p$, for then that element's order would contain other factors than $p$. Therefore $G \leq \sum_{i \in I} \mathbb{Z}(p^\infty)$ for some indexing set $I$.
From here, there must be some way to "reduce" $|I|$ down to $1$, but I don't see how to show $I$ can even be finite. As far as I know, it isn't necessarily the case that $G$ lies in only one "coordinate" of $\sum_{i \in I} \mathbb{Z}(p^\infty)$, since there could be an isomorphic copy of $\mathbb{Z}(p^\infty)$ in the sum that is not one of these direct summands, and that could be where $G$ lies inside. I have considered using the Ulm invariants, but we have only talked about and proven results for $U\{n, G\}$ with $n$ finite and $G$ a direct sum of cyclic groups. I have also considered writing $G$ as a direct sum of a divisible group and a reduced group, and so by indecomposability it must be either divisible or reduced – if it is divisible, it must be isomorphic to the whole of $\mathbb{Z}(p^\infty)$ and we are done, but in the reduced case, I have not found any use of knowing that $G$ is reduced.
(The converse direction is much more straightforward: subgroups of $\mathbb{Z}(p^\infty)$ are either cyclic of order $p^n$ or are the whole of $\mathbb{Z}(p^\infty)$. The former case is indecomposable as a consequence of the classification of finite abelian groups, and the latter case is indecomposable since it is infinite and all proper subgroups are finite.)
Best Answer
If $G$ is divisible, then $G$ is a direct sum of copies of $\mathbb{Z}/(p^\infty)$ and thus isomorphic to $\mathbb{Z}/(p^\infty)$ by indecomposability. Otherwise, $G$ must have a cyclic direct summand (by Corollary 10.43 in Rotman) and so must be cyclic by indecomposability.