Circle Group is Group/Proof 3

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The circle group $\struct {K, \times}$ is a group.


Taking the group axioms in turn:

Group Axiom $\text G 0$: Closure

\(\ds z, w\) \(\in\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds \cmod z\) \(=\) \(\ds 1 = \cmod w\)
\(\ds \leadsto \ \ \) \(\ds \cmod {z w}\) \(=\) \(\ds \cmod z \cmod w\)
\(\ds \leadsto \ \ \) \(\ds z w\) \(\in\) \(\ds K\)

So $\struct {K, \times}$ is closed.


Group Axiom $\text G 1$: Associativity

Complex Multiplication is Associative.


Group Axiom $\text G 2$: Existence of Identity Element

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


Group Axiom $\text G 3$: Existence of Inverse Element

We have that:

$\cmod z = 1 \implies \dfrac 1 {\cmod z} = \cmod {\dfrac 1 z} = 1$


$z \times \dfrac 1 z = 1 + 0 i$

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


All the group axioms are satisfied, and the result follows.