IMC 1997 · Day 2 · P11

Let $f^n = \underbrace{f \circ f \circ \dots \circ f}_{n \text{ times}}$, $f^0 = id$, $f^{-n} = (f^{-1})^n$ for every natural number $n$. Let $T(x) = \{ f^n(x) : n \in \mathbb{Z} \}$ for every $x \in X$. The sets $T(x)$ for different $x$'s either coinside or do not intersect. Each of them is mapped by $f$ onto itself. It is enough to prove the theorem for every such set. Let $A = T(x)$. If $A$ is finite, then we can think that $A$ is the set of all vertices of a regular $n$ polygon and that $f$ is rotation by $\dfrac{2\pi}{n}$. Such rotation can be obtained as a composition of 2 symmetries mapping the $n$ polygon onto itself (if $n$ is even then there are axes of symmetry making $\dfrac{\pi}{n}$ angle; if $n = 2k+1$ then there are axes making $k \dfrac{2\pi}{n}$ angle). If $A$ is infinite then we can think that $A = \mathbb{Z}$ and $f(m) = m + 1$ for every $m \in \mathbb{Z}$. In this case we define $g_1$ as a symmetry relative to $\dfrac{1}{2}$, $g_2$ as a symmetry relative to 0.