Numbers of form 1 + 2m over 1 + 2n form Infinite Abelian Group under Multiplication

Theorem
Let $S$ be the set of integers defined as:
 * $S = \left\{ {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}\right\}$

Then $\left({S, \times}\right)$ is an infinite abelian group.

Proof
Let $k \in \Z$.

Then $1 + 2 k \ne 0$.

Thus:
 * $\forall x \in S: x \in \Q_{\ne 0}$

Thus by definition of subset:
 * $S \subseteq \Q_{\ne 0}$

From Non-Zero Rational Numbers under Multiplication form Abelian Group:
 * $\left({\Q_{\ne 0}, \times}\right)$ is an infinite abelian group.

It is noted that $S$ is an infinite set and so trivially $S \ne \varnothing$.

Consider $a = \dfrac {1 + 2 m} {1 + 2 n} \in S$.

We have that:
 * $\dfrac {1 + 2 m} {1 + 2 n} \times \dfrac {1 + 2 n} {1 + 2 m} = 1$

and so $\dfrac {1 + 2 n} {1 + 2 m}$ is the inverse of $b \in \left({\Q_{>0}, \times}\right)$.

By inspection it can be seen that $b^{-1} \in S$.

Let $a, b \in S$.

Then:
 * $\exists m_1, n_1 \in \Z: a = \dfrac {1 + 2 m_1} {1 + 2 n_1}$
 * $\exists m_2, n_2 \in \Z: b = \dfrac {1 + 2 m_2} {1 + 2 n_2}$

Then we have:

Hence by the Two-Step Subgroup Test, $\left({S, \times}\right)$ is a subgroup of $\left({\Q_{>0}, \times}\right)$.

It has been established that $S$ is an infinite set.

Hence by definition $\left({S, \times}\right)$ is an infinite group.

Finally, from Subgroup of Abelian Group is Abelian, $\left({S, \times}\right)$ is an abelian group.