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I know this question has been asked many times, but I did not found a particular answer to my question. Assume $G$ is a group, $o(g) = n$ and $m \in \mathbb{Z}$. Then $$o(g^m) = \frac{n}{\gcd(m,n)}$$

I mean obviously $$(g^m)^{n/\gcd(m,n)} = (g^n)^{m/\gcd(m,n)} = 1$$ so $o(g^m)| n/\gcd(m,n)$. But how would I prove the equality?

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    What happens if you raise it to a smaller power? – 2017-01-24
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    Page$\#95$ of https://archive.org/details/NumberTheory_862 – 2017-01-24
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    @TobiasKildetoft Thanks. I think I got it now: Say $k < n/\gcd(m,n)$ and $o(g^m) = k$. Then $k\gcd(m,n) < n$ or with Bézout $kxm + kxn < n$. Then $g^{k\gcd(m,n)} = g^{kxm}g^{kxn} = 1$ and therefore $o(g) \leq k\gcd(m,n) < n$ contradicting the minimality of $n$. – 2017-01-24
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    The cyclic group $\{1,g,g^2,\ldots,g^{n-1}\}$ generated by $g$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z},+)$. Modulo $n$ : the order of $1$ is $n$, the order of $2$ is $n / gcd(n,2)$, the order of $m$ is $n/gcd(n,m)$ ... This is really the definition of the $gcd$ and the order of an element – 2017-01-24

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Let be $o(g^m)=r$ then $(g^m)^r=g^{mr}=e$ and $n|mr$, Let be $mr=tn; t\in \mathbb Z$ and $d=gcd(m,n)$ and $m=ud, n=vd ; u,v \in \mathbb Z$ and $gcd(u,v)=1$ then from $mr=tn$ we have $udr=tvd\Rightarrow ur=tv$ because $gcd(u,v)=1$ we have $u|t,v|r \Rightarrow \frac{n}{d}|r $.

On other hand $(g^m)^{\frac{n}{d}}=g^{\frac{nm}{d}}=g^{\frac{nud}{d}}=g^{un}=(g^n)^u=e^u=e$ and we have that $o(g^m)=r$ then $r|\frac{n}{d}$

then we have $r=\frac{n}{d}$ or $o(g^m)=\frac{n}{gcd(m,n)}$

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The order of $g^m$ is the smallest $r$ such that $g^mr=1$, i.e. the smallest $r$ such that $rm$ is a multiple of $n$. By definition, this is the l.c.M. of $m$ and $n$, and it is a well-known formula that $$\operatorname{lcm}(m,n)\times\gcd(m,n)=m\times n,\;\text{whence}\enspace\operatorname{lcm}(m,n)=\frac{m}{\gcd(m,n)}\times n, \;\text{so that}\quad r=\frac{m}{\gcd(m,n)}.$$