= Solution
The <commutator subgroup> is $G'=\langle xyx^{-1}y^{-1}:x,y\in G\rangle$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G'\leq H$, then $ghg^{-1}h^{-1}\in H$ 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 $G'$. Therefore
$$
\boxed{G'\leq H\iff H\triangleleft G\text{ and }G/H\text{ is Abelian}.}
$$
Set $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)})'$. 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 $G'$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G'=1$ 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>.
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