Modulus 1 Rational Argument Complex Numbers under Multiplication form Infinite Abelian Group

Theorem

Let $S$ be the set defined as:

$S = \set {\cos \theta + i \sin \theta: \theta \in \Q}$

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

Proof

By definition of polar form of complex numbers, the elements of $S$ are also elements of the circle group $\struct {K, \times}$:

$K = \set {z \in \C: \cmod z = 1}$

$S$ is infinite by construction.

Thus $S \subseteq C$ and trivially $S \ne \O$.

Let $a, b \in S$.

Then:

$a = \cos \theta_1 + i \sin \theta_1$

and:

$b = \cos \theta_2 + i \sin \theta_2$

for some $\theta_1, \theta_2 \in \Q$.

We have that:

\(\ds a \times b\)

\(=\)

\(\ds \paren {\cos \theta_1 + i \sin \theta_1} \paren {\cos \theta_2 + i \sin \theta_2}\)

\(\ds \)

\(=\)

\(\ds \map \cos {\theta_1 + \theta_2} + i \map \sin {\theta_1 + \theta_2}\)

\(\ds \)

\(\in\)

\(\ds S\)

and:

\(\ds b^{-1}\)

\(=\)

\(\ds \cos \theta_2 - i \sin \theta_2\)

\(\ds \)

\(\in\)

\(\ds S\)

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

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.

$\blacksquare$

Sources

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