= Solution
If a finite <group> is not simple, choose a maximal proper normal <subgroup> and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.
For $S_4$ one <composition series> is
$$
S_4\triangleright A_4\triangleright V_4
\triangleright\langle(12)(34)\rangle\triangleright1.
$$
The successive quotients have orders $2,3,2,2$, hence are simple. As <subgroups> of $S_4$, the <groups> $S_4,A_4,V_4$, and $1$ are normal. The selected order-two <subgroup> is not normal, since conjugation permutes the three double transpositions.
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