Circle Group
Theorem
Let $\mathbb S$ be the set of all complex numbers of unit modulus:
- $\mathbb S = \left\{{z \in \C: \left|{z}\right| = 1}\right\}$.
Then $\left({\mathbb S, \cdot}\right)$ is an infinite abelian group under the operation of complex multiplication.
This is called the circle group or the complex unit circle (group).
Proof
We note that $\mathbb S \ne \varnothing$ as the identity element $1 + 0 i \in \mathbb S$.
Since all $z \in \mathbb S$ 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 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 $x,y \in \mathbb S$ then $x\cdot y^{-1} \in \mathbb S$.
Let $x, y \in \mathbb S$ 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:
| \(\displaystyle \) | \(\displaystyle \exp \left({i t}\right) \exp \left({-i t}\right)\) | \(=\) | \(\displaystyle \exp \left({ i \left({t - t}\right) }\right)\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \exp \left({0}\right)\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle 1\) | \(\displaystyle \) |
So $y^{-1} = \exp \left({-i t}\right)$. We note that this lies in $\mathbb S$.
Furthermore, we have:
| \(\displaystyle \) | \(\displaystyle xy\) | \(=\) | \(\displaystyle \exp \left({i s}\right) \exp(-2 \pi i t)\) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \exp \left({ i \left({s - t}\right) }\right)\) | \(\displaystyle \) |
We conclude that $xy \in \mathbb S$.
By the Two-Step Subgroup Test, $\mathbb S$ is a subgroup of $\C$ under complex multiplication.
$\blacksquare$
Sources
- Seth Warner: Modern Algebra (1965): $\S 7$: Exercise $7.2$
- Ian D. Macdonald: The Theory of Groups (1968): $\S 1$: Example $1.08$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 31$