Circle Group is Group/Proof 1

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The circle group $\struct {K, \times}$ is a group.


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.