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Let $a$ and $b\neq a^{-1}$ be two elements in an abelian group with order $m$ and $n$ respectively. Is it true that the order of the element $ab$ will be $\text{lcm}(m,n)$?

If not, then please give a counter example.

Watson
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This is not true. In the group $G = \{z \in \Bbb C \mid z^9=1\}$ of $9$-th roots of unity, $a=\zeta_9=e^{2\pi i /9}\;$ has order $9$, $b=a^2$ has order $9$ but $ab=a^3=e^{2\pi i /3}$ has order $3$.

However, there exists an element in your group whose order is $\text{lcm}(m,n)$.

Moreover, if two elements $x,y$ in a group $G$ commute, and if their order $m$ and $n$ are coprime, then $xy$ has order $mn$. In general, if $xy=xy$, then the order of $xy$ divides $\text{lcm}(m,n)$.

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Watson
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    Good. +1 for you. I was about to answer that the conjecture was true. Perhaps one can say instead that it is true provided $b \notin \langle a\rangle\setminus {e}$? – MPW Aug 25 '16 at 12:31
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    @MPW : But if I consider $x=(\zeta_9,\zeta_9^2) \in G^2$ and $y=(\zeta_9^2,\zeta_9) \in G^2$, both have order $9$, $y$ is not a power of $x$ (if $y=x^k$ then $k \equiv 2 \pmod 9, 2k \equiv 1 \pmod 9$), and $xy$ has order $3$. Am I missing something? – Watson Aug 25 '16 at 12:38
  • Nope. You're completely correct. +1 again. I'm being lazy and musing without putting any effort into it. – MPW Aug 25 '16 at 15:08