Let $G$ be a finite cyclic group with the order $c$, and let $a \in G$ is the generator of the group. For any integer $p, q$, if $\gcd(c, p) = q$, then $<a^p> = <a^q> $.
Proof:
Let $G$ be a finite group with the order $c$. Since $$<a^p> = \{x \in G | x = (a^p)^n, \quad where \quad n \in Z\}$$ and $$<a^q> = \{y \in G | y = (a^q)^m, \quad where \quad m \in Z \}$$ , we are going to show that $<a^p> = <a^q>$.
Let $a_0 \in <a^p>$. Since $\gcd(c, p) = q$, there is some $l \in Z$ such that $lq = p$, so $a_0 = (a^p)^n = a^{(q)(nl)}$.
Since $nl \in Z$, $a_0 \in <a^q>$, so $<a^p> \subseteq <a^q>$.
Similarly, $<a^q> \subseteq <a^p>$, and thence $<a^p> = <a^q>$.
So my question is, is there any flow in this proof ? or any suggestion about any point in the proof ?
Note: $<a> = \{ x \in G | x=a^n$ where $n \in Z \}$