Circle Group is Infinite Abelian Group/Proof 2

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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 $\theta \in \left[{0 .. 2 \pi}\right)$, 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.

$\blacksquare$