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

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Theorem

Let $S$ be the set of integers defined as:

$S = \set {\dfrac {1 + 2 m} {1 + 2 n}: m, n \in \Z}$

Then $\struct {S, \times}$ 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 Infinite Abelian Group:

$\struct {\Q_{\ne 0}, \times}$ is an infinite abelian group.

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

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 $b \in \struct {\Q_{>0}, \times}$.

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:

\(\ds a \times b\) \(=\) \(\ds \dfrac {1 + 2 m_1} {1 + 2 n_1} \dfrac {1 + 2 m_2} {1 + 2 n_2}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {1 + 2 m_1} \paren {1 + 2 m_2} } {\paren {1 + 2 n_1} \paren {1 + 2 n_2} }\)
\(\ds \) \(=\) \(\ds \dfrac {1 + 2 m_1 + 2 m_2 + 4 m_1 m_2} {1 + 2 n_1 + 2 n_2 + 4 n_1 n_2}\)
\(\ds \) \(=\) \(\ds \dfrac {1 + 2 \paren {m_1 + m_2 + 2 m_1 m_2} } {1 + 2 \paren {n_1 + n_2 + 2 n_1 n_2} }\)
\(\ds \) \(=\) \(\ds \dfrac {1 + 2 m} {1 + 2 n}\) where $m = m_1 + m_2 + 2 m_1 m_2, n = n_1 + n_2 + 2 n_1 n_2$


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

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

Hence by definition $\struct {S, \times}$ is an infinite group.

Finally, from Subgroup of Abelian Group is Abelian, $\struct {S, \times}$ is an abelian group.

$\blacksquare$


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