# Thread: Order of an element and GCD

1. ## 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 $g^k$ is n/d, where d = gcd(n,k).

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

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 $g^m =1$, then r|k. is there any other theorems that i need to know to solve this problems. Thank You!.

2. ## Re: Order of an element and GCD

the trick is to first prove $ = $.

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

$ \subseteq $

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

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

so we also know that

$ \subseteq $

so $ = $. this means that $|g^k| = |g^d|$.

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

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

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

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