= Solution
\b[Yes: every finite group of rotations about the origin in the plane is cyclic.] A <planar rotation> is determined by an angle modulo $2\pi$, and composition adds those angles. If $G$ consists only of the identity, it is a <cyclic group>. Otherwise, finiteness allows us to choose the smallest positive representative angle $\alpha\in(0,2\pi)$ occurring in $G$.
Take any <planar rotation> in $G$ with representative angle $\theta\in[0,2\pi)$. Write $\theta=q\alpha+r$, where $q$ is an integer and $0\leq r<\alpha$. The composition $R_\theta R_\alpha^{-q}$ belongs to $G$ and has angle $r$. Minimality of $\alpha$ forces $r=0$, so every element of $G$ is a power of $R_\alpha$.
Similarly, write $2\pi=m\alpha+r$ with $0\leq r<\alpha$. If $r>0$, the <planar rotation> $R_\alpha^{-m}$ has representative angle $r$, a contradiction. Thus $\alpha=2\pi/m$ and
$$
\boxed{G=\langle R_{2\pi/m}\rangle.}
$$
This proves <finite groups of planar rotations are cyclic> by an elementary angle argument.
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