= Solution
{c}
Apply the <Abelian group criterion for a product of two prime squares> with the smaller and larger primes in their correct order. For $p<11$, all such <groups> are <Abelian> exactly when $p\nmid11^2-1=120$. Of the primes $2,3,5,7$, only $7$ qualifies. For $p>11$, the criterion is $11\nmid p^2-1$, equivalently $p\not\equiv\pm1\pmod{11}$. Therefore the exact answer is
$$
\boxed{p=7\quad\text{or}\quad p>11\text{ prime with }p\not\equiv1,10\pmod{11}.}
$$
Equivalently, among the stipulated primes $p\ne11$, exclude $2,3,5$ and the residue classes $\pm1$ modulo $11$.
For completeness, the necessity in that criterion supplies actual counterexamples at every excluded prime. If $\ell<q$ and $\ell\mid q^2-1$, the <multiplicative group of a finite field is cyclic>, so choose $\alpha\in\mathbb F_{q^2}^{\times}$ of order $\ell$. Let a generator of $C_{\ell^2}$ act on the additive <group> of $\mathbb F_{q^2}$ by multiplication by $\alpha$. This defines a nontrivial <group action>, hence a nonabelian <semidirect product> of order $\ell^2q^2$. Use $(\ell,q)=(p,11)$ for $p=2,3,5$, and $(\ell,q)=(11,p)$ for $p\equiv\pm1\pmod{11}$ with $p>11$. Thus no excluded prime can be restored to the answer.
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