Circle Group is Group

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

Proof 1

First we note that $K \subseteq \C$.

So to show that $K$ is a group it is sufficient to show that $K$ is a subgroup of the multiplicative group of complex numbers $\struct {\C_{\ne 0}, \times}$.

From Complex Multiplication Identity is One, the identity element $1 + 0 i$ is in $K$.

Thus $K \ne \O$.

We now show that $z, w \in K \implies z w \in K$:

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

Next we see that $z \in K \implies z^{-1} \in K$:

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

Thus by the Two-Step Subgroup Test:

$K \le \C_{\ne 0}$

Thus $K$ is a group.


Proof 2

We note that $K \ne \O$ as the identity element $1 + 0 i \in K$.

Since all $z \in K$ have modulus $1$, they have, for some $\theta \in \hointr 0 {2 \pi}$, the polar form:

$z = \map \exp {i \theta} = \cos \theta + i \sin \theta$

Conversely, if a complex number has such a polar form, it has modulus $1$.

Observe the following property of the complex exponential function:

$\forall a, b \in \C: \map \exp {a + b} = \map \exp a \map \exp b$

We must show that if $x, y \in K$ then $x \cdot y^{-1} \in K$.

Let $x, y \in K$ be arbitrary.

Choose suitable $s, t \in \hointr 0 {2 \pi}$ such that:

$x = \map \exp {i s}$
$y = \map \exp {i t}$

We compute:

\(\ds \map \exp {i t} \map \exp {-i t}\) \(=\) \(\ds \map \exp {i \paren {t - t} }\)
\(\ds \) \(=\) \(\ds \map \exp 0\)
\(\ds \) \(=\) \(\ds 1\)

So $y^{-1} = \map \exp {-i t}$.

We note that this lies in $K$.

Furthermore, we have:

\(\ds x y\) \(=\) \(\ds \map \exp {i s} \map \exp {-2 \pi i t}\)
\(\ds \) \(=\) \(\ds \map \exp {i \paren {s - t} }\)

We conclude that $x y \in K$.

By the Two-Step Subgroup Test, $K$ is a subgroup of $\C$ under complex multiplication.


Proof 3

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.


Proof 4

Consider the complex modulus function $\cmod {\, \cdot \,}: \C \to \R, z \mapsto \cmod z$.

By Complex Modulus is Norm, we have that $\cmod z \ge 0$ for all $z \in \C$, and:

$\cmod z = 0 \iff z = 0$

Let $\C_{\ne 0} := \C \setminus \set 0$ denote the complex numbers without zero.

From Group of Units of Field and Complex Numbers form Field, we have that $\struct {\C_{\ne 0}, \times}$ is a group.

By the above observation, the modulus has a restriction to $\C_{\ne 0}$:

$\cmod {\, \cdot \,}: \C_{\ne 0} \to \R_{>0}, z \mapsto \cmod z$

From $\cmod 1 = 1$ and Modulus of Product, it follows that $\phi$ is in fact a group homomorphism:

$\phi: \struct {C_{\ne 0}, \times} \to \struct {\R_{>0}, \times}, z \mapsto \cmod z$

Now $K$ is by definition the kernel of $\phi$.

Hence, by Kernel of Group Homomorphism is Subgroup, $\struct {K, \times}$ is a subgroup of $\struct {\C_{\ne 0}, \times}$.


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