See greatest common divisor, cyclic group, order of a group

Let $G$ be finite, and $a\in G$.
$|a|=n$.
Then $⟨a^k⟩=⟨a^{gcd(n,k)}⟩$.

Proof

Let $d=gcd(n,k)$.

Let $k=dr$ for some integer $r$.
$a^k=a^{dr}=(a^d{)}^r$
then $a^k\in ⟨a^d⟩$
so $⟨a^k⟩\subseteq ⟨a^d⟩$

Greatest Common Divisor is expressible as a linear combination of its arguments,
so there exist integers $s,t$ with $d=ns+kt$
$a^d=a^{ns+kt}=(a^n{)}^s(a^k{)}^t$
since $|a|=n$, $a^n=e$
so $a^d=(a^k{)}^t$
then $a^d\in ⟨a^k⟩$
so $⟨a^d⟩\subseteq ⟨a^k⟩$

Therefore, $⟨a^k⟩=⟨a^d⟩=⟨a^{gcd(n,k)}⟩$.

QED