= Abelian group criterion for a product of two prime squares
{title2=$p<q:\quad(\forall |G|=p^2q^2,\ G\text{ abelian})\iff p\nmid q^2-1$}
Let $p<q$ be distinct <prime numbers>. Every <group> of <group order> $p^2q^2$ is <Abelian> if and only if $p$ does not divide $q^2-1$.
For sufficiency, the <Sylow theorems> make the number of <Sylow p-subgroups> one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is <normal>. The number of <Sylow subgroups> for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other <Sylow subgroup> is normal too. Their intersection is trivial, and their <group commutators> lie in that intersection, so the <group> is their <direct product of groups>. A <group> of prime-square order is abelian: its nontrivial <center of a group> is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.
For necessity, suppose $p\mid q^2-1$. The <multiplicative group of a finite field is cyclic>, so $\mathbb F_{q^2}^{\times}$ contains $\alpha$ of order $p$. Let the <cyclic group> $C_{p^2}$ act on the additive <group> of $\mathbb F_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial <group action>, since $\alpha^{p^2}=1$. The resulting <semidirect product> has order $p^2q^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian <group> of this order exists exactly when $p\mid q^2-1$.
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