Order of $x^k$ in cyclic subgroup of order $n$ generated by $x$

Question

$ord(x)=n$. Prove that $ord(x^k)= \frac nd$ where $d=gcd(k,n)$

This is what I have so far: Let $k=ad$ and $n=bd$.

We need to find the smallest $p$ such that $(x^k)^p = 1$.

$(x^{ad})^b = x^{abd} = x^{an} = (x^n)^a = 1$. We now need to prove that $b=\frac nd$ is the smallest $p$ that satisfies the equation $(x^k)^p = 1$. How do I do this?

Answer 1

Assume that $o(x)=n$. We have $(x^k)^{\frac{n}{(n,k)}}=1$ and so $o(x^k)\,|\,\frac{n}{(n,k)}$. Since $(x^k)^{o(x^k)}=1$, we can conclude that $n=o(x)\,|\,k{o(x^k)}$. Hence $\frac{n}{(n,k)}\,|\,o(x^k)$.

Why $n=o(x)\,|\,k{o(x^k)} \rightarrow \frac{n}{(n,k)}\,|\,o(x^k)$?

There exists $q\in \mathbb Z$ such that $n=qk{o(x^k)}$. Now, we divide by $(n,k)$. Thus we have $\frac{n}{(n,k)}=\frac{k}{(n,k)}q{o(x^k)}$. and so $\frac{n}{(n,k)}\,|\,\frac{k}{(n,k)}o(x^k)$. Therefore, $\frac{n}{(n,k)}\,|\,o(x^k)$ as, $(\frac{n}{(n,k)},\frac{k}{(n,k)})=1$.