Equivalence of definitions of locally cyclic torsion-free group

Statement
The following are equivalent for a group:


 * 1) It is both a fact about::locally cyclic group (i.e., every finitely generated subgroup is cyclic) and an fact about::torsion-free group (i.e., every non-identity element has infinite order).
 * 2) It is isomorphic to a subgroup of the fact about::group of rational numbers.

Facts used

 * 1) uses::Local cyclicity is subgroup-closed
 * 2) uses::Torsion-freeness is subgroup-closed

(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 $$\varphi:G \to \mathbb{Q}$$ defined as follows: $$\varphi(g) = 1$$. For any $$h \in G$$, find any nonzero integer $$m$$ such that (in additive notation) $$mh = ng$$ for some integer $$n$$. Then, set $$\varphi(h) = n/m$$.


 * 1) Such $$m$$ and $$n$$ exist: $$\langle g,h \rangle$$ is cyclic, so they are both multiples of some element $$a$$, so there exist integers $$m,n$$ such that $$ma = g$$ and $$na = h$$. These $$m,n$$ work in the above.
 * 2) $$\varphi$$ is well-defined, i.e., for different choices of $$(m,n)$$ that work, $$n/m$$ is the same: If not, then there exist $$m_1,n_1,m_2,n_2$$ with $$m_1n_2 \ne m_2n_1$$ but $$m_1h = n_1g$$ and $$m_2h = n_2g$$. Multiplying and subtracting, we obtain that $$(m_1n_2 - m_2n_1)g = 0$$, so $$g$$ has finite order, contradicting our assumption that the group is aperiodic. This completes the contradiction.
 * 3) $$\varphi$$ is a homomorphism: We need to show that if $$\varphi(a) = n_1/m_1$$ and $$\varphi(b) = n_2/m_2$$, then $$m_1m_2(a + b) = (n_1m_2 + n_2m_1)g$$. This is a straightforward algebraic manipulation. We can also check explicitly the conditions for identity and inverses.
 * 4) $$\varphi$$ is injective: Suppose $$\varphi(a) = \varphi(b) = n/m$$. Then, $$ma = n$$ and $$mb = ng$$, so we get $$ma = mb$$, or $$m(a - b) = 0$$. Since $$m \ne 0$$, $$a - b$$ 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.