Circle Group is Infinite Abelian Group

Theorem
Let $K$ be the set of all complex numbers of unit modulus:
 * $K = \left\{{z \in \C: \left|{z}\right| = 1}\right\}$.

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

This is called the circle group.

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


 * We now show that $z, w \in K \implies z w \in K$:


 * Next we see that $z \in K \implies z^{-1} \in K$:


 * $\displaystyle \left|{z}\right| = 1 \implies \left|{\frac 1 z}\right| = 1$

Thus by the Two-step Subgroup Test, $K \le \C^*$.

Thus $K$ is abelian.


 * Next we show that $\left({K, \times}\right)$ is infinite: