So, one of the homework problems I am working on is
If $a$ and $b$ are elements of a group that commute and $\langle a\rangle\cap \langle b\rangle = \{1\}$, what is the order of $ab$ if the order of $a$ is $m$ and the order of $b$ is $n$? Prove your assertion. Show by example that your assertion is false in general, in the case that $a$ and $b$ do not commute.
What I'm thinking is that the order if $ab$ would be $mn$, because, $\langle ab\rangle$ would contain $a^1 b^1$, $a^1b^2,\ldots ,a^1b^n$, $a^2b^1$, $a^2b^2,\ldots,a^m b^n$. Because, for every $a$, there would be $n$ $b$'s, and there are $m$ $a$'s, so there should be $nm$ in $\langle ab\rangle$, right?
However, I don't think this is exactly correct, because what I have shown there has nothing to do with $a$ and $b$ being able to commute.
I would like some help, am I going in the right direction? What might I need to do in order to show that it is false in general when $a$ and $b$ do not commute?