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.

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

Let $G$ be finite, and $a \in G$ .

Then $| a^{k} | = \frac{| ⟨ a ⟩ |}{gcd (| ⟨ a ⟩ |, k)}$

If $| a | = n$ ,
Then $| a^{k} | = \frac{n}{gcd (n, k)}$ .

Proof

Let $d = gcd (n, k)$ be an integer. Notably, $d | n$

$(a^{d})^{\frac{n}{d}} = a^{n} = e$
so $| a^{d} | \leq \frac{n}{d}$

Let $0 < i < \frac{n}{d}$
$(a^{d})^{i} = a^{d i}$
nb $d i < d \cdot \frac{n}{d} = n$
and $a^{d i} \neq e$ because that would contradict $| a | = n$
so there is no such $i$ , i.e. $\frac{n}{d}$ is minimal

Therefore:
if $d | n$ , then $| a^{d} | = \frac{n}{d}$

In particular,
$| a^{k} | = | a^{gcd (n, k)} | = \frac{n}{gcd (n, k)}$

QED