Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
This property is called being "locally cyclic", by the way. –  Chris Eagle May 8 '12 at 22:21
add comment

3 Answers

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.

share|improve this answer
add comment

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 $\operatorname{lcm}(q_1,\cdots,q_k)^{-1}$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.