Let \(p\) and \(q\) be distinct prime numbers.
Let \(G\) be a group with order \(p^2q\). We show that \(G\) is not simple.
Then, by Sylow Theorem (3), we know that \(n_p \mid q\) and \(n_p \equiv 1 \mod p\).
This gives us that that \(n_p = 1\) or \(n_p = q\). However, \(q < p\) and thus \(q \not\equiv 1 \mod p\). Thus, \(n_p = 1\).
This gives us that \(G\) is not simple.
Using Sylow Theorem (3) again gives us that \(n_q \in \{1, p, p^2\}\). If \(n_q = 1\), then we are done. As \(p < q\), we again get that \(n_q \neq p\). Now, we consider the case \(n_q = p^2\).
Thus, there are \(p^2\) Sylow-\(q\) subgroups of \(G\). Now, note that each such Sylow-\(q\) subgroup has order \(q\), a prime and thus, the intersection of two distinct Sylow-\(q\) subgroups must be trivial, id est, \((1)\).
Thus, the union of these Sylow-\(q\) subgroups contains \(d = p^2(q-1)\) elements, after removing the identity.
All of these elements have order \(q\). Now, we are left with \(d' = p^2q - d = p^2\) elements. Now, we know that \(n_p \ge 1\) and none of the \(d\) elements counted above can be a part of a Sylow-\(p\) subgroup. Thus, the remaining \(d'\) must form the unique Sylow-\(p\) subgroup and we get that \(n_p = 1\).
Thus, \(G\) is not simple!