Order of an element and GCD

I have the following problem which i can only solve half of it.

__Problem:__ Let G = <g> be a cyclic group of order n. Prove that the order of $\displaystyle g^k$ is n/d, where d = gcd(n,k).

__Solution: __Since G has order n, order of g is n. i.e. $\displaystyle g^n$=1. Also, d|k, giving da = k for some integer a. Hence, $\displaystyle ({g^k})^{n/d} = g^{(nk)/d} = g^{(ndb)/d} = {(g^n)}^b = 1$.

However, i am unable to show that n/d is the order. The only theorem about order that i know is that: If r is the order of g and $\displaystyle g^m =1$, then r|k. is there any other theorems that i need to know to solve this problems. Thank You!.

Re: Order of an element and GCD

the trick is to first prove $\displaystyle <g^k> = <g^d>$.

now k = db, so $\displaystyle g^k = g^{db} = (g^d)^b$ therefore:

$\displaystyle <g^k> \subseteq <g^d>$

writing d = ns + kt (which we can by the euclidean algorithm), we have:

$\displaystyle g^d = g^{ns+kt} = (g^{ns})(g^{kt}) = (g^n)^s(g^k)^t = (g^k)^t$

so we also know that

$\displaystyle <g^d> \subseteq <g^k>$

so $\displaystyle <g^k> = <g^d>$. this means that $\displaystyle |g^k| = |g^d|$.

now obviously $\displaystyle (g^d)^{n/d} = e$. suppose that $\displaystyle (g^d)^u = e$

for some 0 < u < n/d. then 0 < du < n, but $\displaystyle g^{du} = (g^d)^u = e$

contradicting the fact that the order of g is n. thus there can be no such u,

so $\displaystyle |g^k| = |g^d| = n/d$.