When a rotation has period.

Question

A period of a transformation $f$ in this context is the least natural number $n$ such that $f^{n}=id$, when it exists.

My book says this claim, which I couldn't prove: A rotation have a period if and only if the measure of its rotation angle is a rational multiple of $\pi$.

Can someone help me?

Thanks.

Answer 1

Thanks to Gerry Myerson I was able to find the answer.

$(\Rightarrow)$

Suppose $\theta = \frac{n . \pi}{m}$. Then composing the rotation $m$ times will get to the angle $n . \pi$, which is either congruent to $0 \mod 2\pi$ or congruent to $\pi \mod 2\pi$. Suppose it is congruent to $\pi \mod 2\pi$. Then, by the symmetry of the circle, composing the rotation $m$ times again will get to an angle congruent to $0 \mod 2\pi$. Then we have a non-empty list of natural numbers, which by the Well-Ordering Principle will have least element, then the rotation has a period.

$(\Leftarrow)$ Suppose a rotation has period. Then rotating $m$ times will get to an angle congruent to $0 \mod 2\pi$, which means $m.\theta = 2 . \pi . k$, for some integer $k$, then $\theta = \frac{2. \pi . k}{m}$. Since $2 . k$ is an integer, $\theta$ is rational with $\pi$.

$\square$