A group is <solvable group>[solvable] when its <derived series>
$$
G^{(0)}=G,\qquad G^{(r+1)}=[G^{(r)},G^{(r)}]
$$
eventually reaches the trivial group; equivalently, it has a subnormal series with <abelian group>[abelian] quotients.

The <symmetric group> $S_4$ is solvable because
$$
\{1\}\triangleleft V_4\triangleleft A_4\triangleleft S_4
$$
has abelian quotients $V_4$, $C_3$, and $C_2$, where $V_4$ is the <Klein four-group>. Thus
$$
\boxed{S_4\text{ is solvable}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 18G / c / i / Solution

{title2=$G^{(n)}$}
{wiki}

The derived series is defined by
$$
G^{(0)}=G,\qquad G^{(n+1)}=[G^{(n)},G^{(n)}].
$$
A group is <solvable group>[solvable] exactly when this series reaches the trivial group after finitely many steps.

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