[proofplan] We apply the Sylow theorems to constrain the number $n_p$ of Sylow $p$-subgroups of $G$. By [Sylow's Third Theorem](/theorems/3251), $n_p \equiv 1 \pmod{p}$ and $n_p \mid q$. Since $q < p$, the only divisor of $q$ that is congruent to $1$ modulo $p$ is $1$ itself, forcing $n_p = 1$. A unique Sylow $p$-subgroup is normal, so $G$ is not simple (the Sylow $p$-subgroup is a proper nontrivial normal subgroup). [/proofplan] [step:Determine the divisibility and congruence constraints on $n_p$] Let $n_p$ denote the number of Sylow $p$-subgroups of $G$. Since $|G| = p^2 q$ and $p \nmid q$ (as $p$ and $q$ are distinct primes), we have the factorisation $|G| = p^2 \cdot q$ with $p \nmid q$, so the Sylow $p$-subgroups have order $p^2$ and the complementary factor is $m = q$. By [Sylow's Third Theorem](/theorems/3251) applied to $G$ with the prime $p$, the number $n_p$ satisfies two conditions: (i) $n_p \equiv 1 \pmod{p}$, and (ii) $n_p \mid q$. [guided] We want to show that $G$ has a unique Sylow $p$-subgroup. The tool for counting Sylow subgroups is [Sylow's Third Theorem](/theorems/3251), which requires a finite group $G$ and a prime $p$ dividing $|G|$. We verify: $G$ is finite with $|G| = p^2 q$, and $p$ divides $p^2 q$ since $p$ is prime and $p \mid p^2$. Writing $|G| = p^a \cdot m$ with $p \nmid m$, we identify $a = 2$ and $m = q$. The exponent is $a = 2$ because $p \nmid q$ (since $p$ and $q$ are distinct primes), so $p^2$ is the largest power of $p$ dividing $|G|$. Sylow's Third Theorem then gives two constraints on $n_p$, the number of Sylow $p$-subgroups: (i) $n_p \equiv 1 \pmod{p}$, and (ii) $n_p \mid m = q$. Why are these constraints useful? Because $q$ is prime, its only positive divisors are $1$ and $q$, which severely limits the possibilities for $n_p$. The congruence condition will eliminate all but one candidate. [/guided] [/step] [step:Show that $n_p = 1$ by eliminating $n_p = q$] Since $q$ is prime, the positive divisors of $q$ are $1$ and $q$. By constraint (ii), $n_p \in \{1, q\}$. We rule out $n_p = q$. If $n_p = q$, then by constraint (i) we would need $q \equiv 1 \pmod{p}$, i.e., $p \mid (q - 1)$. But $q < p$ by hypothesis, so $q - 1 < p$, and hence $0 < q - 1 < p$. Since $p$ cannot divide a positive integer strictly less than $p$, we have $p \nmid (q - 1)$, contradicting $q \equiv 1 \pmod{p}$. Therefore $n_p = 1$. [guided] Since $q$ is prime, the divisor condition $n_p \mid q$ restricts $n_p$ to exactly two candidates: $n_p = 1$ or $n_p = q$. We want to show $n_p = 1$, so we proceed by eliminating the other possibility: assume for contradiction that $n_p = q$. If $n_p = q$, then the congruence condition $n_p \equiv 1 \pmod{p}$ forces $q \equiv 1 \pmod{p}$. This means $p \mid (q - 1)$, so $q - 1 \ge p$ (since the smallest positive multiple of $p$ is $p$ itself). But we are given $p > q$, which gives $q - 1 < q < p$, so $q - 1 < p$. This is a contradiction: $q - 1$ cannot simultaneously be at least $p$ and strictly less than $p$. This is the step where the hypothesis $p > q$ is consumed: without it, the argument fails. If instead $q > p$, the congruence $q \equiv 1 \pmod{p}$ need not lead to a contradiction — for instance, $p = 2$ and $q = 3$ gives $q \equiv 1 \pmod{2}$, which is consistent, and groups of order $12 = 2^2 \cdot 3$ can indeed have multiple Sylow $2$-subgroups. Having ruled out $n_p = q$, the only remaining possibility from the divisor condition is $n_p = 1$. We conclude $n_p = 1$: there is exactly one Sylow $p$-subgroup of $G$. [/guided] [/step] [step:Conclude that the unique Sylow $p$-subgroup is normal and $G$ is not simple] Let $P$ denote the unique Sylow $p$-subgroup of $G$. By [Sylow's Second Theorem](/theorems/3250), all Sylow $p$-subgroups of $G$ are conjugate. Since $P$ is the only Sylow $p$-subgroup, for every $g \in G$ the conjugate $gPg^{-1}$ is also a Sylow $p$-subgroup and hence $gPg^{-1} = P$. Therefore $P \trianglelefteq G$. Since $|P| = p^2 \ge 4$ (as $p \ge 2$) and $|G| = p^2 q > p^2 = |P|$, the subgroup $P$ is a proper nontrivial normal subgroup of $G$. Therefore $G$ is not simple. [guided] We have established $n_p = 1$, so there is a unique Sylow $p$-subgroup $P \le G$ with $|P| = p^2$. We must show $P \trianglelefteq G$, i.e., that $P$ is closed under conjugation by every element of $G$. Why does uniqueness imply normality? The key is that conjugation is an automorphism of $G$: for any $g \in G$, the map $x \mapsto gxg^{-1}$ carries subgroups to subgroups of the same order. In particular, $gPg^{-1}$ is a subgroup of $G$ satisfying $|gPg^{-1}| = |P| = p^2$, so $gPg^{-1}$ is itself a Sylow $p$-subgroup of $G$. By [Sylow's Second Theorem](/theorems/3250), any two Sylow $p$-subgroups are conjugate: if $Q$ is any Sylow $p$-subgroup, there exists $h \in G$ with $Q = hPh^{-1}$. Since $P$ is the only Sylow $p$-subgroup, $gPg^{-1}$ must equal $P$ for every $g \in G$. This is precisely the definition of $P$ being normal in $G$, so $P \trianglelefteq G$. To confirm that this normal subgroup witnesses non-simplicity, we check that $P$ is both nontrivial and proper. It is nontrivial because $|P| = p^2 \ge 4$ (as $p \ge 2$). It is proper because $|P| = p^2 < p^2 q = |G|$, which holds since $q \ge 2$. A group is simple if and only if it has no proper nontrivial normal subgroups; since $P$ is one such subgroup, $G$ is not simple. This completes the proof. [/guided] [/step]