Circle Group is Infinite Abelian Group

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


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

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

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

Since all $z \in \mathbb S$ have modulus $1$, they have, for some $\theta \in \left[{0 .. 2 \pi}\right)$, the polar form:


 * $z = \exp \left({i \theta}\right) = \cos \left({\theta}\right) + i \sin \left({\theta}\right)$

Conversely, if a complex number has such a polar form, it has modulus $1$.

Observe the following property of the exponential function:


 * $\forall a, b \in \C: \exp \left({a + b}\right) = \exp \left({a}\right) \exp \left({b}\right)$

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

Let $x, y \in \mathbb S$ be arbitrary. Choose suitable $s, t \in \left[{0 .. 2 \pi}\right)$ such that:


 * $x = \exp \left({i s}\right)$
 * $y = \exp \left({i t}\right)$

We compute:

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

Furthermore, we have:

We conclude that $xy \in \mathbb S$.

By the Two-Step Subgroup Test, $\mathbb S$ is a subgroup of $\C$ under complex multiplication.