Euclidean plane/Finite subgroup/Proper/Cyclic/Fact/Proof

Proof

Every element in ${}G$ is, due to fact, a rotation of the plane with a determined angle ${}\theta$. We consider the surjective group homomorphism

$\mathbb {R} \longrightarrow \operatorname {SO} _{2}\!,\theta \longmapsto D(\theta )={\begin{pmatrix}\operatorname {cos} \,\theta &-\operatorname {sin} \,\theta \\\operatorname {sin} \,\theta &\operatorname {cos} \,\theta \end{pmatrix}},$

which assigns to an angle its corresponding rotation. Let ${}H\subseteq \mathbb {R}$ be the preimage of ${}G$ under this mapping; that is, ${}H$ consists in all rotation angles of those rotations that belong to ${}G$. The group ${}H$ is generated by a representing system for the elements of ${}G$ and by ${}2\pi$. In particular, ${}H$ is a finitely generated subgroup of ${}\mathbb {R}$. Since every group element of ${}G$ has a finite order, every ${}\theta \in H$ has the form ${}\theta =2\pi q$ with a rational number ${}q\in \mathbb {Q}$. This means that ${}H$ id a finitely generated subgroup of ${}2\pi \mathbb {Q} \subseteq \mathbb {R}$. Hence, ${}H$ is isomorphic to a finitely generated subgroup of the rational numbers. According to exercise, ${}H$ is cyclic, say ${}H=\mathbb {Z} \alpha$, with a uniquely determined angle ${}\alpha \in [0,2\pi )$. Then the group ${}G$, which is the image group of ${}H$, is cyclic as well.

To fact