Equivalence of definitions of locally cyclic torsion-free group

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

Statement

The following are equivalent for a group:

It is both a Locally cyclic group (?) (i.e., every finitely generated subgroup is cyclic) and an Torsion-free group (?) (i.e., every non-identity element has infinite order).
It is isomorphic to a subgroup of the Group of rational numbers (?).

Facts used

Proof

(1) implies (2)

Given: A locally cyclic torsion-free group $G$ .

To prove: $G$ is isomorphic to a subgroup of the rationals.

Proof: Let $g$ be any non-identity element of $G$ . Consider the homomorphism $φ : G \to Q$ defined as follows: $φ (g) = 1$ . For any $h \in G$ , find any nonzero integer $m$ such that (in additive notation) $m h = n g$ for some integer $n$ . Then, set $φ (h) = n / m$ .

Such $m$ and $n$ exist: [SHOW MORE]
$⟨ g, h ⟩$ is cyclic, so they are both multiples of some element $a$ , so there exist integers $m, n$ such that $m a = g$ and $n a = h$ . These $m, n$ work in the above.
$φ$ is well-defined, i.e., for different choices of $(m, n)$ that work, $n / m$ is the same: [SHOW MORE]
If not, then there exist $m_{1}, n_{1}, m_{2}, n_{2}$ with $m_{1} n_{2} \neq m_{2} n_{1}$ but $m_{1} h = n_{1} g$ and $m_{2} h = n_{2} g$ . Multiplying and subtracting, we obtain that $(m_{1} n_{2} - m_{2} n_{1}) g = 0$ , so $g$ has finite order, contradicting our assumption that the group is aperiodic. This completes the contradiction.
$φ$ is a homomorphism: [SHOW MORE]
We need to show that if $φ (a) = n_{1} / m_{1}$ and $φ (b) = n_{2} / m_{2}$ , then $m_{1} m_{2} (a + b) = (n_{1} m_{2} + n_{2} m_{1}) g$ . This is a straightforward algebraic manipulation. We can also check explicitly the conditions for identity and inverses.
$φ$ is injective: [SHOW MORE]
Suppose $φ (a) = φ (b) = n / m$ . Then, $m a = n$ and $m b = n g$ , so we get $m a = m b$ , or $m (a - b) = 0$ . Since $m \neq 0$ , $a - b$ has finite order, forcing $a = b$ , thus proving injectivity.

(2) implies (1)

This follows from the fact that the group of rationals is both locally cyclic and torsion-free, and facts (1) and (2). We thus have an injective homomorphism from $G$ to the rationals, completing the proof.