 Solution

The commutator subgroup is $G^{\prime }=⟨xy{x}^{-1}{y}^{-1}:x,y\in G⟩$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G^{\prime }\le H$, then $gh{g}^{-1}{h}^{-1}\in H$ for every $g\in G,h\in H$, so $H$ is normal. Every commutator becomes trivial in $G/H$, making that quotient Abelian. Conversely, the quotient homomorphism to an Abelian group kills each commutator, and hence kills $G^{\prime }$. ThereforeSet $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)}{)}^{\prime }$. The group is soluble exactly when this derived series reaches the identity after finitely many steps.

If a nontrivial simple group is soluble, its normal subgroup $G^{\prime }$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G^{\prime }=1$ and $G$ is Abelian. Every nonidentity element then generates $G$, since its cyclic subgroup is normal. An infinite cyclic group is not simple, and a finite cyclic group is simple exactly when its order is prime. This proves soluble simple group has prime order.