Circle Group is Group/Proof 1
Jump to navigation
Jump to search
Theorem
The circle group $\struct {K, \times}$ is a group.
Proof
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.
$\blacksquare$
Sources
- 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Subgroups