Exercise on Cyclic Group Let $G$ be a finite cyclic group of order $n$. Let $a$ be a generator. Let $r$ be an integer $\neq 0$, and relatively prime to n.
Show that $a^r$ is also a generator of $G$.
Proof: $G=\langle a\rangle$ such that $a^n=1$
$|a^r|=\frac{|a|}{\gcd(r,n)}=|a|$
Since $|a^r|=|a|$m then $G=\langle a^r\rangle $.
Is my proof correct ?

Another way:
$$g.c.d(r,n)=1\implies \;\exists\,x,y,\in\Bbb Z\;\;s.t.\;\; xr+yn=1$$
and from here
$$a=a^1=a^{rx+yn}=\left(a^r\right)^x\left(a^n\right)^y=\left(a^r\right)^x\implies \langle a\rangle \subset \langle a^r\rangle$$
