# Circle Group is Infinite Abelian Group/Proof 2

## Theorem

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

## Proof

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 $\theta \in \hointr 0 {2 \pi}$, 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:

 $\displaystyle \map \exp {i t} \map \exp {-i t}$ $=$ $\displaystyle \map \exp {i \paren {t - t} }$ $\displaystyle$ $=$ $\displaystyle \map \exp 0$ $\displaystyle$ $=$ $\displaystyle 1$

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

We note that this lies in $K$.

Furthermore, we have:

 $\displaystyle x y$ $=$ $\displaystyle \map \exp {i s} \map \exp {-2 \pi i t}$ $\displaystyle$ $=$ $\displaystyle \map \exp {i \paren {s - t} }$

We conclude that $x y \in K$.

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

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.

$\blacksquare$