Prove That if |a|=m and |b|=n and ⟨a⟩∩⟨b⟩={e} then, gcd(m, n)=1

Question

$G$ is a group and $a,b,\in G$.

To summarize the question, if the cyclic group generated by $a$ and $b$ only has the identity element in common, then the orders of $a$ and $b$ are relatively prime.

I'm not sure how to proceed. What should I start with?

Answer 1

This is false. Take, for instance, the group $S_3$, and then take $a=(1\ \ 2)$, and $b=(1\ \ 3)$. Then $\langle a\rangle\cap\langle b\rangle=\{e\}$. However, but both $a$ and $b$ have order $2$.