Circle Group is Group/Proof 3
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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: $\S 31$