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
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).
- Locally cyclic iff subquotient of rationals
- Equivalence of definitions of locally cyclic periodic group
- Equivalence of definitions of locally cyclic torsion-free group
Given: A locally cyclic group .
To prove: For any non-identity elements , either both and have finite order or both have infinite order.
Proof: Consider the subgroup . Since is locally cyclic, there exists such that .
We consider two cases:
- has infinite order: Both and must have infinite order, since they are both finite powers of . We can think of them as nonzero elements of the group of integers, with identified with .
- has finite order: Both and must have finite order, and in fact, their orders must divide the order of . We can think of them as elements of the finite cyclic subgroup generated by .