Solution
= Solution
The <derived series> is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, where the bracket denotes the <commutator subgroup>. The group is <soluble> if
$$
\boxed{G^{(d)}=1\text{ for some }d\geq0.}
$$
Equivalently it has a finite series with abelian factors. The trivial group is included.