Number of elements of each order in a cyclic group

Statement

Let $G$ be a cyclic group of finite order $n$, and let $d$ be a positive divisor of $n$. Then $G$ has exactly $\varphi(d)$ elements of order $d$, where $\varphi$ is Euler's totient function.

Remarks

Here $\varphi(d)$ is the number of integers $k$ with $1 \leq k \leq d$ and $\gcd(k, d) = 1$. The elements of order $d$ are the generators of the unique subgroup of order $d$. Counting all elements by their orders gives the identity $\sum_{d \mid n} \varphi(d) = n$. In particular, $G$ has exactly $\varphi(n)$ generators.

Explanations

L1 human

In $\mathbb{Z}/6\mathbb{Z}$, the element $0$ has order $1$ and the element $3$ has order $2$. The elements $2$ and $4$ have order $3$, and $1$ and $5$ have order $6$. These counts are $1, 1, 2, 2$, which are $\varphi(1), \varphi(2), \varphi(3), \varphi(6)$. The totient function counts the elements of each order.