Circle Group is Infinite Abelian Group

Theorem
Let $$K = \left\{{z \in \C: \left|{z}\right| = 1}\right\}$$.

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 $$z \in K \implies z^{-1} \in K$$:


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

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

Thus $$K$$ is abelian.


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