# Locally cyclic implies periodic or torsion-free

The statement of this article contains an assertion of the form that for a certain kind of group, either every element has finite order (i.e., the group is a Periodic group (?)) or every non-identity element has infinite order (i.e., the group is a Torsion-free group (?) or aperiodic group). The actual statement in this case may be considerably stronger.
View other such statements

## Statement

A Locally cyclic group (?) (i.e., a group in which every finitely generated subgroup is cyclic) is either a Periodic group (?) (i.e., every element has finite order) or a Torsion-free group (?) (i.e., every non-identity element has infinite order).

## Proof

Given: A locally cyclic group $G$.

To prove: For any non-identity elements $g,h \in G$, either both $g$ and $h$ have finite order or both have infinite order.

Proof: Consider the subgroup $\langle g,h \rangle$. Since $G$ is locally cyclic, there exists $a \in G$ such that $\langle g,h \rangle = \langle a \rangle$.

We consider two cases:

1. $a$ has infinite order: Both $g$ and $h$ must have infinite order, since they are both finite powers of $a$. We can think of them as nonzero elements of the group of integers, with $a$ identified with $1$.
2. $a$ has finite order: Both $g$ and $h$ must have finite order, and in fact, their orders must divide the order of $a$. We can think of them as elements of the finite cyclic subgroup generated by $a$.