See subgroup, cyclic group

Let $H\le ⟨a⟩$ with $|⟨a⟩|=n$.
For each $k|n$, there exists a unique subgroup of order $k$.

In particular, if $H\le ⟨a⟩$ with $|⟨a⟩|=n$ and $|H|=k$ such that $k|n$,
then $H=⟨a^{\frac{n}{k}}⟩$

Proof

Let $H\le ⟨a⟩$ with $|⟨a⟩|=n$.

Let $k|n$.

Existence:

$⟨a^{\frac{n}{k}}⟩$ has order $k$

Uniqueness:

Suppose $|H|=k$.

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 any cyclic subgroup divides the order of any cyclic group which contains it.,
$|H|\cdot gcd(n,m)=n$

so $gcd(n,m)=\frac{n}{k}$

By Cyclic subgroups are equivalent if one power is expressible as the gcd of the other with the order of the base.,
$H=⟨a^m⟩=⟨a^{gcd(m,n)}⟩$

substituting, $H=⟨a^{\frac{n}{k}}⟩$

QED