Circle Group is Infinite Abelian Group

Theorem
$\newcommand{\S}{\mathbb S}$Let $\S$ be the set of all complex numbers of unit modulus:


 * $\S = \left\{{z \in \C: \left|{z}\right| = 1}\right\}$.

Then $\left({\S, \cdot}\right)$ is an infinite abelian group under the operation of complex multiplication.

This is called the circle group.

Proof
We note that $\S \ne \varnothing$ as the identity element $1 + 0 i \in K$.

Since all elements of $\S$ have modulus $1$, they have the polar form:
 * $ \sin(\theta) + i\cos(\theta) = \exp(i\theta) \qquad (1)$

Conversely, if a complex number has the form $(1)$ it has modulus $1$.

So we have a bijection of sets. By Sine and Cosine are Periodic on Reals we have:
 * $ \S^1 = \left\{ \exp(2\pi i\theta) : \theta \in [0,1) \right\} $

Now we prove the hypothesis of the subgroup test using the property of the exponential function:
 * $\forall a, b \in \C: \exp(a + b) = \exp(a) \exp(b)$

We must show that if $x,y \in \S$ then $x\cdot y^{-1} \in \S$.

Let $x,y\in \S$ be arbitrary, say:


 * $x = \exp(2\pi i s)$
 * $y = \exp(2 \pi i t)$

Then:

So $y^{-1} = \exp(-2 \pi i t)$. We note that this lies in $\S$.

Now:

... which has precisely the form of an element of $\S$. We are done.