See subgroup, cyclic group

Let $H\le ⟨a⟩$.
If $|⟨a⟩|=n$, then the order of any subgroup divides $n$.

In particular, for $H=⟨a^m⟩\le ⟨a⟩$, with $|⟨a⟩|=n$,
$|H|\cdot gcd(n,m)=n$

Proof

Let $H\le ⟨a⟩$ with $|⟨a⟩|=n$.

By Every subgroup of a cyclic group is itself cyclic.,
Let $m$ be the smallest possible integer with $a^m\in H$. Then $H=⟨a^m⟩$.

By The order of a group element to some power is equal to the order of the generator, divided by the gcd of the order of the generator with the power of the element.,
$|H|=\frac{n}{gcd(n,m)}$

then $|H|\cdot gcd(n,m)=n$

so $|H|$ divides $n$.

QED

Corollary 1

For a group $⟨a⟩$,
$|a^m|=\frac{|a|}{gcd(|a|,m)}$