> [!NOTE] Theorem > Let $G$ be a [[Group|group]]. Let $g\in G$ with finite [[Order of Group Element|order]]. Then for any $m\in \mathbb{Z},$ $\text{ord}(g^{m})= \frac{\text{ord}(g)}{\gcd(m, \text{ord}(g))}$where $\gcd(m,n)$ denotes the [[Greatest Common Divisor (GCD)|GCD]] of $m$ and $n$ and $g^{n}$ denotes the $n$th [[Integer Power of Group Element|power]] of $g.$ **Proof**: Let $d=\gcd(m,n).$ Then there exists $m',n'\in \mathbb{Z}$ such that $m=dm'$ and $n=dn'.$ Then $\begin{align} (g^{m})^{n'} &= (g^{dm'})^{n'} \\ & = (g^{dn'})^{m'} \\ &= (g^{n})^{m'} \\ &= e^{m'} \\ &= e \end{align}$Now BWOC, suppose order of $g^{m}$ is $k<n'.$ By [[Bézout's Lemma]], there exists $x,y\in\mathbb{Z}$ so that $mx+ny=d.$ Thus $\begin{align} g^{dk} &= g^{(mx+ny)k} \\ &= g^{mxk}g^{nyk} \\ &=e \end{align}$But $dk<dn'=n$ contradicting the fact that $g$ has order $n.$ Therefore $|g^{m}|=n'= \frac{n}{\gcd(n,m)} .$ # Applications See [[Subgroups of Cyclic Group]] and [[Number of Primitive Roots Modulo Prime]].