I have the following problem: Consider a finite group G with elements $a,b\in G$ such that $ab=ba$ and $\langle a\rangle\cap\langle b\rangle=\langle e \rangle$. Prove that $|ab|=lcm(|a|,|b|)$. I found a similar problem in my textbook that assumes $(|a|,|b|)=1$ and asks to prove that $|ab|=|a||b|$. My problem seems to be the generalization of this problem, but I am stuck in the details.
I let $|a|=n$ and $|b|=m$ and assumed that some $(ab)^k=e$. I need to show that $k=lcm(|a|,|b|)$. I was able to show that $m|nk$ and $n|mk$, but I don't know how to use the fact that the cyclic subgroups generated by $a$ and $b$ are distinct, other than identity, and how that gets me to $k=lcm(m,n)$.
Any help would be appreciated.
Well, if $(ab)^k = e$, then $a^k = b^{-k} \in \langle b \rangle$. Hence $a^k = e$ and $n$ divides $k$. Similarly $m$ divides $k$ and therefore $lcm(n, m)$ divides $k$.