 Solubility of groups of order p squared q

The assertion follows using only Sylow theorems. Write the order instead as $p{q}^2$ to separate the prime of multiplicity one. If $p<q$, the number of Sylow $q$-subgroups divides $p$ and is one modulo $q$, so one such subgroup is normal. If $p>q$ and the Sylow $p$-subgroup is not normal, its number must be $q^2$, forcing $p\mid (q-1)(q+1)$. Since $p>q$, this forces $p=q+1$ and hence $p=3,q=2$. In that order-twelve exception, four order-three subgroups account for eight nonidentity elements. The four remaining elements contain every order-four Sylow subgroup, so that subgroup is unique and normal. In every case a normal Sylow subgroup and its quotient have prime or prime-square order, hence are abelian groups. They provide a series with abelian factors and prove solubility. Orders with the squared exponent lowered to one or zero follow by the same simpler counting.