See subgroup, cyclic group

Let $H\le ⟨a⟩$.
If $m$ is the smallest possible integer with $a^m\in H$, then $H=⟨a^m⟩$.

Proof

Assume $H\ne \{e\}$ (trivial case)
Then there exists some nonzero integer $t$ with $a^t\in H$.
Assume $t>0$ without loss of generality.

Let $m>0$ be the smallest possible integer with $a^m\in H$
By closure, $⟨a^m⟩\le H$

Let $b\in H$
then $b=a^k$ for some $k\in \mathbb{Z}$
By Division Algorithm 1, let $k=qm+r$ with $0\le r<m$
Then $a^k=a^{qm+r}=(a^m{)}^q{a}^r$
$\Rightarrow (a^m{)}^{-q}{a}^k=a^r$
NB since $a^m\in H$, and by definition a subgroup must be closed under the operation, so $(a^m{)}^{-q}\in H$.
Furthermore, since $a^k\in H$, by closure, we have $a^r\in H$ which by definition of $m$ implies $r=0$.
Therefore, $b=(a^m{)}^q\in ⟨a^m⟩$
so $H\le ⟨a^m⟩$

Thus, $H=⟨a^m⟩$

QED