Solution: Let us assume that $o(a) = m$, $o(b) = n$ and $o(ab) = k$. Then, \[ a^m = e = b^n = (ab)^k. \] Since $a$ and $b$ commutes, we have \begin{align*} (ab)^k = a^k b^k & \implies a^k b^k = e \\ & \implies a^k = b^{-k} \\ & \implies a^{nk} = b^{-nk} = e \\ & \implies m \mid nk \\ & \implies m \mid k \quad (\text{ since } \gcd (m,n) = 1). \end{align*}

Similarly, \begin{align*} b^k = a^{-k} & \implies b^{mk} = a^{-mk} = e \\ & \implies n \mid mk \\ & \implies n \mid k . \end{align*} Since $\gcd (m,n) = 1$ we must have $mn \mid k$.

We also have \begin{align*} (ab)^ {mn} = a^{mn} b^{mn} = e \implies k \mid mn. \end{align*} Since $m,n$ and $k$ are positive integers, we have $mn = k$ and therefore, we proved that \[ o(ab) = o(a) o(b). \]