Circle Group is Group/Proof 3

Theorem

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

Proof

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.

$\Box$

Group Axiom $\text G 1$: Associativity

Complex Multiplication is Associative.

$\Box$

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

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

$\Box$

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

We have that:

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

But:

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

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

$\Box$

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

$\blacksquare$

Sources

• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Examples of Group Structure: $\S 31$