Circle Group is Group/Proof 3

Proof
Taking the group axioms in turn:

So $\struct {\mathbb S, \cdot}$ is closed.

Complex Multiplication is Associative.

From Complex Multiplication Identity is One we have that the identity element of $K$ is $1 + 0 i$.

We have that:
 * $\cmod z = 1 \implies \dfrac 1 {\cmod z} = \cmod {\dfrac 1 z} = 1$

But:
 * $z \times \dfrac 1 z = 1 + 0 i$

So the inverse of $z$ is $\dfrac 1 z$.

All the criteria are satisfied, and the result follows.