TITLE: Abelian groups whose subgroup lattice is chain
QUESTION [6 upvotes]: Let $G=Z(p^k)$ where $k=1,2,..,\infty$. The group $G$ has exactly one subgroup series.
Does exist an other infinite abelian group with this property?

REPLY [2 votes]: For any prime number $p$, the group of roots of unity whose order is a power of $p$ (the direct limit of cyclic groups of order a power of $p$) is an example of an infinite group with the mentioned property. (It is unclear to me which group is mentioned in the question, maybe it is this one?)
Since the property necessarily applies to all subgroups of a group $G$ that has it, and neither $\mathbf Z$ nor any group that non-trivally decomposes as a direct product has the property (so that $G$ is a torsion group, and the order of any element of $G$ must be a prime power), one easily shows that these examples are up to isomorphism the only ones.
Concretely: $\mathbf Z$ has uncomparable subgroups, for instance $2\mathbf Z$ and $3\mathbf Z$, while a finite cyclic group whose order has at least two different prime factors decomposes as a direct product by the (additive part of) the Chinese remainder theorem; neither of these groups has a unique chain of subgroups.
If $p$ is the unique prime factor of some nontrivial element of $G$, then the order of any element must be a power of $p$, lest we can find elements whose order has two distinct prime factors. Any two elements of order $p^k$ must generate the same subgroup of order $p^k$ of $G$ (otherwise those subgroups would be incomparable), and given that $G$ is infinite, it must have such a subgroup for all $k\in\mathbf N$. Then $G$ is the union of these subgroups.