# Circle Group is Infinite Abelian Group

## Contents

## Theorem

The circle group $\struct {K, \times}$ is an uncountably infinite abelian group under the operation of complex multiplication.

## 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 $\left({\C_{\ne 0}, \times}\right)$.

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

Thus $K \ne \varnothing$.

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

\(\displaystyle z, w\) | \(\in\) | \(\displaystyle K\) | |||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left\vert{z}\right\vert\) | \(=\) | \(\displaystyle 1 = \left\vert{w}\right\vert\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle \left\vert{z w}\right\vert\) | \(=\) | \(\displaystyle \left\vert{z}\right\vert \left\vert{w}\right\vert\) | ||||||||||

\(\displaystyle \implies \ \ \) | \(\displaystyle z w\) | \(\in\) | \(\displaystyle K\) |

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

- $\left\vert{z}\right\vert = 1 \implies \left\vert{\dfrac 1 z}\right\vert = 1$

Thus by the Two-Step Subgroup Test:

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

Thus $K$ is a group.

From Complex Multiplication is Commutative it also follows from Subgroup of Abelian Group is Abelian that $K$ is an abelian group.

Finally we have that the Circle Group is Uncountably Infinite.

$\blacksquare$

## Proof 2

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

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

- $z = \exp \left({i \theta}\right) = \cos \left({\theta}\right) + i \sin \left({\theta}\right)$

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

Observe the following property of the complex exponential function:

- $\forall a, b \in \C: \exp \left({a + b}\right) = \exp \left({a}\right) \exp \left({b}\right)$

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 $s, t \in \left[{0 .. 2 \pi}\right)$ such that:

- $x = \exp \left({i s}\right)$
- $y = \exp \left({i t}\right)$

We compute:

\(\displaystyle \exp \left({i t}\right) \exp \left({-i t}\right)\) | \(=\) | \(\displaystyle \exp \left({ i \left({t - t}\right) }\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \exp \left({0}\right)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 1\) |

So $y^{-1} = \exp \left({-i t}\right)$. We note that this lies in $K$.

Furthermore, we have:

\(\displaystyle xy\) | \(=\) | \(\displaystyle \exp \left({i s}\right) \exp(-2 \pi i t)\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \exp \left({ i \left({s - t}\right) }\right)\) |

We conclude that $xy \in K$.

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

That the operation $\times$ on $K$ is commutative follows from Complex Multiplication is Commutative and Restriction of Commutative Operation is Commutative.

That is, $\times$ is commutative on $K$ because it is already commutative on $\C$.

Finally we have that the Circle Group is Uncountably Infinite.

$\blacksquare$

## Proof 3

Taking the group axioms in turn:

### G0: Closure

\(\displaystyle z, w\) | \(\in\) | \(\displaystyle K\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \cmod z\) | \(=\) | \(\displaystyle 1 = \cmod w\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \cmod {z w}\) | \(=\) | \(\displaystyle \cmod z \cmod w\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle z w\) | \(\in\) | \(\displaystyle K\) |

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

$\Box$

### G1: Associativity

Complex Multiplication is Associative.

$\Box$

### G2: Identity

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

$\Box$

### G3: Inverses

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 $\dfrac 1 z$.

$\Box$

### C: Commutative

We have that Complex Multiplication is Commutative.

We also have from Restriction of Commutative Operation is Commutative that $\times$ is likewise commutative on $K$.

$\Box$

### Uncountably Infinite

Circle Group is Uncountably Infinite.

$\Box$

All the criteria are satisfied, and the result follows.

$\blacksquare$

## Proof 4

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

By Complex Modulus is Norm, we have that $\left\vert{z}\right\vert \ge 0$ for all $z \in \C$, and:

- $\left\vert{z}\right\vert = 0 \iff z = 0$

From Group of Units of Field and Complex Numbers form Field, we have $\C^\times = \C \setminus \left\{{0}\right\}$.

By above observation, the modulus has a restriction to $\C^\times$:

- $\left\vert{\cdot}\right\vert: \C^\times \to \R_{>0}, z \mapsto \left\vert{z}\right\vert$

From $\left\vert{1}\right\vert = 1$ and Modulus of Product, it follows that it is in fact a group homomorphism:

- $\phi: \left({C^\times, \cdot}\right) \to \left({\R_{>0}, \cdot}\right), z \mapsto \left\vert{z}\right\vert$

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

Hence, by Kernel of Group Homomorphism is Subgroup, $\left({K, \times}\right)$ is a subgroup of $\left({\C^\times, \cdot}\right)$.

By Subgroup of Abelian Group is Abelian and Complex Multiplication is Commutative, it is also abelian.

Finally, note that $\left({K, \times}\right)$ is uncountable from Circle Group is Uncountably Infinite.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $7.2$ - 1968: Ian D. Macdonald:
*The Theory of Groups*... (previous) ... (next): $\S 1$: Some examples of groups: Example $1.08$