= Solution
{c}
Write $|G|=pq$ with $p<q$. The <Sylow theorems> give $n_q\mid p$ and $n_q\equiv1\pmod q$. Since $p<q$, necessarily $n_q=1$. Its unique <Sylow subgroup> $N$ is a <normal subgroup> and has prime order $q$, so
$$
\boxed{N\cong C_q\trianglelefteq G.}
$$
A <Sylow subgroup> $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. Thus
$$
\boxed{G\cong C_q\rtimes C_p.}
$$
The <semidirect product> action is a <group homomorphism> $C_p\to\operatorname{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the <Chinese remainder theorem>; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a <metacyclic group> and hence a <solvable group>.
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