I am trying to prove the following result.
Let $G$ be a group containing elements $x$ and $y$ such that the orders of $x$, $y$, and $xy$ are pairwise relatively prime; prove that $G$ is not solvable.
If $G$ is solvable, it has an abelian and normal subgroup $H$. If $x,y\in H$, then $o(xy)=o(x)o(y)$, a contradiction. How should I reach a contradiction if either one of $x$ and $y$ is not in $H$?
Also, I observe that $\langle x\rangle\langle y\rangle$ is not a group, but I don't know how to leverage that.
You can do induction on the derived length of $G$. Let $n$ be minimal such that there is counterexample $G$ which is solvable of derived length $n$. Then there is an abelian normal subgroup $N$ of $G$ such that the derived length of $G/N$ is less than $n$, and so the result is true in $G/N$.
You need to consider the cases when $0,1,2$ or $3$ elements of the elements $x,y,xy$ are in $N$ separately (but $2$ cannot occur, and $1$ is inconsistent with the hypothesis).