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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.

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This property is called being "locally cyclic", by the way. – Chris Eagle May 8 '12 at 22:21
up vote 7 down vote accepted

An element of $\mathbb Z_{p^\infty}$ is a $p^k$'th root of unity. So, if you have a collection of $p^{k_1},\ldots, p^{k_n}$'th roots of unity, then the subgroup they generate is a subgroup of the cyclic group of $p^{\operatorname{lcm}(k_1,\ldots, k_n)}$ roots of unity.

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Also, the very same argument works if we look at $\Bbb Z_{p^\infty}$ as the quotient additive group $\Bbb Z[1/p]/\Bbb Z$. If we have a collection $\{p_1/q_1,\cdots,p_k/q_k\}$ then each element is a multiple of $\operatorname{lcm}(q_1,\cdots,q_k)^{-1}$.

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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.

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