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:

1. 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).
2. It is isomorphic to a subgroup of the Group of rational numbers (?).

Facts used

1. Local cyclicity is subgroup-closed
2. Torsion-freeness is subgroup-closed

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 $\varphi :G\to \mathbb {Q}$ 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: [SHOW MORE]
   $\langle g,h\rangle$ 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: [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.
3. $\varphi$ is a homomorphism: [SHOW MORE]
   We need to show that if $\varphi (a)=n_{1}/m_{1}$ and $\varphi (b)=n_{2}/m_{2}$, 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.
4. $\varphi$ is injective: [SHOW MORE]
   Suppose $\varphi (a)=\varphi (b)=n/m$. Then, $ma=n$ and $mb=ng$, so we get $ma=mb$, or $m(a-b)=0$. Since $m\neq 0$, $a-b$ 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.