Theorem Suppose $X$ is a set. Then a mapping $f:X\to X$ of period $n$ is a bijection.

Proof. If $n=1$ , the claim is trivial; $f$ is the identity mapping. Suppose $n=2,3,\ldots$ . Then for any $x\in X$ , we have $x=f\big(f^{n-1}(x)\big)$ , so $f$ is an surjection. To see that $f$ is a injection, suppose $f(x)=f(y)$ for some $x,y$ in $X$ . Since $f^n$ is the identity, it follows that $x=y$ .