Show that $y=x^{k}$ with $gcd(k,n)=1$ is a generator of $G$.

Question

Could someone please verify whether my solution is okay?

Let $G$ be a finite cyclic group with $|G|=n$ and generator $x$. If $y=x^{k}$ and $gcd(k,n)=1$, then show that $y$ is a generator of $G$.

Let $y=x^{k}$ with $gcd(k,n)=1$. Then $|\langle y\rangle|=|y|=\frac{n}{gcd(k,n)}=\frac{n}{1}=n$. Then $G=\langle y\rangle$.

Answer 1

Here's another way:

If we can show that some power of $y$ is $x$, that is, if we can show that for some $a \in \Bbb Z$,

$y^a = x, \tag 1$

then we are done, since then

$x^l = y^{al}, \tag 2$

so that

$\langle x \rangle = \langle y \rangle. \tag 3$

Well, since

$\gcd(n, k) = 1, \tag 4$

we have $a, b \in \Bbb Z$ with

$ak + bn = 1; \tag 5$

then

$x = x^1 = x^{ak + bn} = x^{ak}x^{bn} = x^{ak} = (x^k)^a = y^a, \tag 6$

establishing (1), and we are done!

P.S. to our OP numericalorange:. Your solution looks fine to me!