Integer Powers of 2 under Multiplication form Infinite Abelian Group

From ProofWiki
Jump to: navigation, search


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

$S = \left\{ {2^k: k \in \Z}\right\}$

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


It can be seen by inspection that $S \subseteq \Q_{>0}$.

That is, all the elements of $S$ are strictly positive rational numbers.

From Strictly Positive Rational Numbers under Multiplication form Countably Infinite Abelian Group:

$\left({\Q_{>0}, \times}\right)$ is an infinite abelian group.

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

Let $a, b \in S$.


$\exists k_1 \in \Z: a = 2^{k_1}$
$\exists k_2 \in \Z: b = 2^{k_2}$

We have that:

$\dfrac 1 {2^{k_2} } 2^{k_2} = 1$

and so $\dfrac 1 {2^{k_2} } = 2^{-k_2} = b^{-1}$ is the inverse of $b \in \left({\Q_{>0}, \times}\right)$.

Then we have:

\(\displaystyle a \times b^{-1}\) \(=\) \(\displaystyle 2^{k_1} \times \dfrac 1 {2^{k_2} }\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 2^{k_1 - k_2}\) $\quad$ $\quad$

We have that:

$k_1 + k_2 \in \Z$

and so $a \times b^{-1} \in \S$.

Hence by the One-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.