 Abelian group criterion for a product of two prime squares

Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ 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 $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.