Locally cyclic iff subquotient of rationals

This article gives a proof/explanation of the equivalence of multiple definitions for the term locally cyclic group
View a complete list of pages giving proofs of equivalence of definitions

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. 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. 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. 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. Local cyclicity is subgroup-closed: Any subgroup of a locally cyclic group is locally cyclic.
3. 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 $G$.