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Let $G$ be a group and let $a,b\in G$, with $ord(a)=n$ and $a=b^k$. I can see how $\langle a\rangle=\langle b\rangle$ implies that $k$ and $n$ are relatively prime since $\langle b^k\rangle = \langle b\rangle$ and $b$ generates $b^k$ if and only if $gcd(k,ord(b))=1$. Since $\langle a\rangle=\langle b\rangle$, we have $ord(a)=ord(b)$, thus $gcd(k,n)=1$. For the converse, clearly $\langle a\rangle\subseteq \langle b\rangle$ because $a$ is a power of $b$. Where I am stuck is proving that $\langle b\rangle\subseteq \langle a\rangle$.

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