# Circle Group is Group/Proof 1

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## 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