= Solution
If $H\leq G$, induction gives $H^{(i)}\leq G^{(i)}$, 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 gives
$$
(G/N)^{(i)}=G^{(i)}N/N.
$$
Consequently \b[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.
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