Exercise 9.1.11

Proof. Let $G$ a fixed cyclic group of order $n$, by example $G={𝕌}_n$. If $A_d$ is the set of elements of order $d$ in $G$, then $G$ is the disjoint union of the $A_n$, so $|G|={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{d=0}^n|A_d|$.

By the proof of Exercise 10, $|A_d|=\varphi (d)$ if $d\mid n$, and $|A_d|=0$ if $d\nmid n$, so

Note: as an alternative proof, we can take the degrees in the formula $x^n-1={\prod <!--\; FUNCTION\; APPLICATION\; -->}_{d\mid n}{\Phi }_d(n)$. □