TITLE: ($\mathbb Q$,+) and $\mathbf{Z}_{p^\infty}$ has this property?
QUESTION [5 upvotes]: There is an exercise telling that every finite subset of group ($\mathbb Q$,+) or of group $\mathbf{Z}_{p^\infty}$ generates a cyclic group itself.
For the first group if $X= \left\{\frac{p_{1}}{q_{1}},\frac{p_{2}}{q_{2}},\ldots,\frac{p_{n}}{q_{n}}\right\} $ be a finite subset, then obviously $X\subseteq \langle\frac{1}{q_{1} q_{2}...q_{n}}\rangle$ and so $\langle X\rangle$ is cyclic iself.
Kindly asking about the second group. How to show that about $\mathbf{Z}_{p^\infty}$ ? Thanks.

REPLY [2 votes]: Any cyclic group (finite or infinite) has this property - a finite subset of elements generates a cyclic sub-group. 
Both $\mathbb Z_{p^\infty}$ and $\mathbb Q$ are unions of a nested sequence of cyclic groups, so any finite subset is necessarily in one of those subgroups, and hence they must generate a subgroup of a cyclic group, which is again a cyclic group.
In general, a group has this property if and only if it is the union of a directed set of cyclic groups.  Let $G$ be a group, and let $\mathcal P$ be the directed set of finite subsets of $G$, and, for $S\in \mathcal P$, let $G_S$ be defined as the subgroup generated by the elements of $S$.  If $G$ has this property, then each $G_S$ is a cyclic group, and $G$ is the union of this directed set of cyclic groups.