See cyclic group, order of a group, Euler's totient function

Let $⟨a⟩$ be a cyclic group of order $n$.
For each $d|n$, the number of elements of $⟨a⟩$ with order $d$ is given by $\phi (d)$.

Proof

Let $⟨a⟩$ be a cyclic group with $|a|=n$.

Let $d|n$.

By There is exactly one subgroup of the order of each divisor of the order of its container cyclic group.,
$⟨a^{\frac{n}{d}}⟩$ is the unique cyclic subgroup of order $d$.

Therefore, the generators of order $d$ are the $(a^{\frac{n}{d}}{)}^m$ such that $⟨(a^{\frac{n}{d}}{)}^m⟩=⟨a^{\frac{n}{d}}⟩$ with $0\le m<d$

By Fundamental Theory of Cyclic Groups#Corollary 3,
$⟨(a^{\frac{n}{d}}{)}^m⟩=⟨a^{\frac{n}{d}}⟩$ if and only if $gcd(m,d)=1$.

By definition of Euler's totient function, $m$ is a totative of $d$, and the number of totatives of $d$ is given by $\phi (d)$.

QED