Integer Powers of 2 under Multiplication form Infinite Abelian Group

Theorem

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

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

Then $\struct {S, \times}$ is an infinite abelian group.

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

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

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

Let $a, b \in S$.

Then:

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

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

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

Then we have:

$\ds a \times b^{-1}$

$=$

$\ds 2^{k_1} \times \dfrac 1 {2^{k_2} }$

$\ds $

$=$

$\ds 2^{k_1 - k_2}$

We have that:

$k_1 + k_2 \in \Z$

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

Hence by the One-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.

$\blacksquare$

1964: Walter Ledermann: Introduction to the Theory of Finite Groups (5th ed.) ... (previous) ... (next): Chapter $\text {I}$: The Group Concept: Examples: $(1) \ \text {(a)}$