Let $G$ be a finite group where for all $a, b \in G$ with $\operatorname{gcd}(o(a), o(b)) = 1$, $o(ab) = o(a)o(b)$. Is $G$ nilpotent?
My try: Let's work by induction on $\lvert G \rvert$. Since every proper subgroup of $G$ is nilpotent it is solvable. Every $p$-complement is nilpotent, thus if $\lvert G \rvert$ has more than $3$ prime divisors it is nilpotent. Since $p$-group is nilpotent, it suffices to show when $\lvert G \rvert$ has $2$ prime divisors. Assume $\operatorname{gcd}(o(a), o(b)) = 1, a \ne e, b \ne e$. If $\langle a, b\rangle < G$, $\langle a, b\rangle$ is nilpotent thus $a$ and $b$ commute. So if $G$ is never generated by two nonidentity elements of coprime order, it is nilpotent. Thus we can assume that $G = \langle a, b\rangle$. Then $o(a)$ and $o(b)$ must be prime powers and the primes are exactly the two prime divisors of $\lvert G \rvert$. How can I proceed? Is there more elementary proof?