Powers of a rotation: when are they dense?

Write $\theta$ for the angle modulo $2\pi$. Note that $\alpha/\pi\in\mathbb{Q}$ iff $\alpha/2\pi\in\mathbb{Q}$.

Rational case. If $\alpha/2\pi=p/q$, then $R_{q\alpha}=R_{2\pi p}=I$, so $\{R_{k\alpha}\}$ has at most $q$ elements: finite, hence closed and not dense (as $SO(2)$ is infinite).

Irrational case. Suppose $\alpha/2\pi\notin\mathbb{Q}$. Then the $R_{k\alpha}$, $k\in\mathbb{Z}$, are pairwise distinct: $R_{j\alpha}=R_{k\alpha}$ would give $(j-k)\alpha\in2\pi\mathbb{Z}$.

Fix $N\ge1$ and cut the circle into $N$ arcs of length $2\pi/N$. Among the $N+1$ distinct points $R_{0},R_\alpha,\dots,R_{N\alpha}$ two lie in the same arc, say $R_{i\alpha}$ and $R_{j\alpha}$ with $i<j$. Then $R_{(j-i)\alpha}=R_\delta$ for some angle $\delta$ with $0<|\delta|<2\pi/N$ (nonzero by distinctness, and we may represent $\delta$ in $(-\pi,\pi]$; its absolute value is the angular distance between two points of the same arc).

The powers $R_{m\delta}=R_{m(j-i)\alpha}$ belong to our set, and the angles $m\delta$, $m\in\mathbb{Z}$, form an arithmetic progression of step $|\delta|<2\pi/N$ going around the circle. So every point of the circle is within $2\pi/N$ (even $|\delta|/2$) of some $R_{m\delta}$.

Since $N$ was arbitrary, the set is dense.

Remark. This is Kronecker's theorem in dimension one: the orbit $\{k\alpha\bmod2\pi\}$ of an irrational rotation is dense (and in fact equidistributed). Lesson. Pigeonhole produces arbitrarily small elements in an infinite subgroup of a compact group.