 Solution

If $H\le G$, induction gives $H^{(i)}\le G^{(i)}$, because commutators of elements of a subgroup are also commutators in the larger group. Thus termination of the derived series of $G$ forces termination for $H$.

For a normal subgroup $N$, the quotient map sends $[x,y]$ to the commutator of their images. Surjectivity then givesConsequently subgroups and quotient groups of a soluble group are soluble. The assertion about a quotient uses a normal subgroup; it is not a quotient by an arbitrary subgroup.