= Solubility of groups of order p squared q
{title2=$|G|=p^2q,\quad p,q\text{ distinct primes}\implies G\text{ soluble}$}
The assertion follows using only <Sylow theorems>. Write the order instead as $pq^2$ 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.
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