Locally cyclic iff subquotient of rationals

Statement
The following are equivalent for a group:


 * 1) It is a locally cyclic group: every finitely generated subgroup of the group is cyclic.
 * 2) It is isomorphic to a subquotient (i.e., a quotient group of a subgroup) of the group of rational numbers.

Facts used

 * 1) uses::Locally cyclic implies periodic or aperiodic: In a locally cyclic group, either all the element have finite order, or all non-identity elements have infinite order.
 * 2) uses::Equivalence of definitions of locally cyclic aperiodic group: A group is locally cyclic and aperiodic iff it is isomorphic to a subgroup of the group of rational numbers.
 * 3) uses::Equivalence of definitions of locally cyclic periodic group: A locally cyclic periodic group is a restricted direct product of groups indexed by distinct primes $$p$$, where the group indexed by a prime $$p$$ is either cyclic of order a power of $$p$$ or a $$p$$-quasicyclic group.

Proof: Subquotient of rationals implies locally cyclic
To prove this, it suffices to note the following things:


 * 1) The group of rational numbers is locally cyclic.
 * 2) uses::Local cyclicity is subgroup-closed: Any subgroup of a locally cyclic group is locally cyclic.
 * 3) uses::Local cyclicity is quotient-closed: Any quotient of a locally cyclic group is locally cyclic.

Proof: locally cyclic implies subquotient of rationals
Given: A locally cyclic group $$G$$.

To prove: $$G$$ is isomorphic to a subquotient of the group of rational numbers.

Proof:


 * 1) By fact (1), $$G$$ is either periodic or aperiodic.
 * 2) If $$G$$ is aperiodic, it is isomorphic to a subgroup of the group of rational numbers by fact (2), and hence to a subquotient of the group of rational numbers. This completes the proof of the aperiodic case.
 * 3) If $$G$$ is periodic, then by fact (3), it is a restricted direct product of groups indexed by distinct primes $$p$$, where the group indexed by a prime $$p$$ is either cyclic of order a power of $$p$$ or a $$p$$-quasicyclic group. We argue that a restricted direct product of the form proved above is isomorphic to a subgroup of $$\mathbb{Q}/\mathbb{Z}$$. Indeed, each $$G_p$$ is isomorphic to a subgroup: when finite cyclic of order $$p^k$$, it is the set of elements of order $$p^k$$; when infinite, it is the set of all elements whose order is a power of $$p$$. The subgroup these generate inside $$\mathbb{Q}/\mathbb{Z}$$ is readily seen to be isomorphic to $$G$$.