IMC 2005 · Day 1 · P6

Write $d = \gcd(m, n)$. It is easy to see that $\langle G(m), G(n) \rangle = G(d)$; hence, it will suffice to check commutativity for any two elements in $G(m) \cup G(n)$, and so for any two generators $a^m$ and $b^n$. Consider their commutator $z = a^{-m} b^{-n} a^m b^n$; then the relations $$z = (a^{-m} b a^m)^{-n} b^n = a^{-m} (b^{-n} a b^n)^m$$ show that $z \in G(m) \cap G(n)$. But then $z$ is in the center of $G(d)$. Now, from the relation $a^m b^n = b^n a^m z$, it easily follows by induction that $$a^{ml} b^{nl} = b^{nl} a^{ml} z^{l^2}.$$ Setting $l = m/d$ and $l = n/d$ we obtain $z^{(m/d)^2} = z^{(n/d)^2} = e$, but this implies that $z = e$ as well.