Assume in an abelian group $G$ that $\langle b\rangle\cap \langle a\rangle=e$, then the order of $(ab)$ is the lcm of the orders of $a$ and $b$. Essentially, $|ab|=\operatorname{lcm}(|a|,|b|)$.
So far I have the assumption, then suppose $(ab)^k=e$, where $k$ is the smallest integer. So $|ab|=k$. Since $G$ is abelian, $(ab)^k=a^k b^k=e$. Thus, $(a^k)^{-1}=b^k$ and likewise the same goes for, $(b^k)^{-1}=a^k$. Therefore $(b^k)^{-1}\in\langle a\rangle$ and $(a^k)^{-1}\in\langle b\rangle$. So each of $a^k$ and $b^k$ are the identity. Therefore $k$ is a multiple of $n$ and $m$, so we can say $\operatorname{lcm}(m,n)\mid k$. How do I get the last piece I need?
Best Answer
Let $n = \text{lcm}(|a|,|b|)$. So we can write $n = |a|r = |b|s$ for some positive integers $r$ and $s$.
Then $(ab)^{n} = a^{n}b^{n} = (a^{|a|})^{r}(b^{|b|})^{s} = e^{r}e^{s} = e$, so $|ab|$ divides $n$.
On the other hand, your argument above shows that $n$ divides $|ab|$, and so we have equality.