Locally cyclic implies periodic or torsion-free

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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.
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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).

Related facts

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.