Prove that the order of the cyclic subgroup $\langle g^k\rangle $ is $n/{\operatorname{gcd}(n,k)}$ [duplicate]
Solution 1:
From where you are:
$n \vert kl$
Let $m = \gcd(n,k)$. Then (since $\gcd (\frac{n}{m}, \frac{k}{m}) = 1)\mid l \Rightarrow$
$$\frac{n}{m} \leq l $$
But $(g^k)^{\frac{n}{m}} = g^{(n\frac{k}{m})}= e^{\frac{k}{m}}=e \Rightarrow$
$$l \leq \frac{n}{m}$$
We conclude that:
$$l = \frac{n}{m}$$
Solution 2:
Note that $kl$ must be the least common multiple of $k$ and $n$.
Then make use of: $$nk=\gcd(n,k)\times\text{lcm}(n,k)$$