The question: Let $G$ be a group and $a$, $b \in G$. Let $ord(a) = n $. Suppose that $a = b^{k}$. Show that $\langle a\rangle = \langle b\rangle $ if and only if $k$ and $n$ are relatively prime.
$\Leftarrow$ Let $k$ and $n$ be relatively prime. Let $x \in \langle a \rangle $. Then, $x = a^{q}$. So $a^{q} = (b^{k})^{q} = b^{kq} \in \langle b\rangle $. So $x \in \langle b\rangle$ which means $ \langle a \rangle \subset \langle b\rangle$.
Let $x \in \langle b\rangle$. Then $x = b^{q}$. Since k and n are relatively prime, this means there exists integers, $j$ and $m$, such that $mk + jn = 1$. This is where I am stuck in this portion of the proof. Can anyone give some hints on how to finish the rest of this proof? Thank you very much.