= Solution
Factor the <group order> as $588=2^2\cdot3\cdot7^2$. The <Sylow theorems> say that the number of <Sylow subgroups> for the prime seven divides $12$ and is congruent to one modulo seven. Among $1,2,3,4,6,12$, only one qualifies. Thus there is a unique <normal subgroup> $N$ of order $49$.
Every <group> of prime-square order is <Abelian>. Indeed its <center of a group> is nontrivial, and a cyclic quotient by the <center of a group> forces the whole <group> to be abelian. Consequently $N$ is either $C_{49}$ or $C_7\times C_7$, and in either case it has two <composition factors>, both $C_7$.
It remains to consider $Q=H/N$, of order twelve. Its number of <Sylow subgroups> for the prime three is either one or four:
* If it is one, the normal order-three <subgroup> has factor $C_3$, and the remaining quotient has order four. An order-four <group> is abelian and has two factors $C_2$.
* If it is four, these order-three <subgroups> have disjoint nonidentity elements and account for eight elements. Any <Sylow subgroup> for the prime two, of order four, must consist of the four remaining elements, since its nonidentity elements have power-of-two orders. It is therefore unique and normal. Its two factors are $C_2$, and the quotient has order three, giving $C_3$.
Lift the <composition series> of $Q$ to $H$ using the <subgroup correspondence theorem>, and join it to a <composition series> inside $N$. This gives
$$
\boxed{C_2,\ C_2,\ C_3,\ C_7,\ C_7}
$$
as the multiset of <composition factors>. The <Jordan–Hölder theorem> makes this independent of the chosen series. In particular every <group> of order $588$ is a <solvable group>, with composition length five; that is a statement about composition length, not necessarily its <derived length>.
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