Show that no group of order \(p^2q^2\) is simple for primes \(p<q\).
Solution
If \(p=q\), then \(G\) is a \(p\)-group and has nontrivial center, so it is not simple. Assume \(p<q\).
1. The number \(n_q\) of Sylow \(q\)-subgroups is either \(1\) or \(p^2\).
Sylow gives \[ n_q\mid p^2, \qquad n_q\equiv1\pmod q. \] Thus \(n_q\in\{1,p,p^2\}\). Since \(1<p<q\), the possibility \(n_q=p\) is impossible. Hence \(n_q=1\) or \(p^2\).
2. If \(n_q=p^2\), then \((p,q)=(2,3)\).
The congruence \(p^2\equiv1\pmod q\) gives \[ q\mid(p-1)(p+1). \] Because \(q>p\), one cannot have \(q\mid p-1\), so \(q\mid p+1\). Since \(0<p+1<2q\), this forces \(q=p+1\). The only consecutive primes are \(2\) and \(3\).
3. Unless \(|G|=36\), the Sylow \(q\)-subgroup is normal, so \(G\) is not simple.
If \(n_q=1\), the unique Sylow \(q\)-subgroup is a nontrivial proper normal subgroup. By 2, the only remaining case is \(p=2\), \(q=3\).
4. Suppose \(|G|=36\) and no Sylow \(3\)-subgroup is normal. Then \(n_3=4\), and conjugation on the four Sylow \(3\)-subgroups gives\[ \varphi:G\longrightarrow S_4 \]with kernel \(K\) of order \(3\) or \(9\).
The action is transitive. The stabilizer of a Sylow \(3\)-subgroup \(Q\) is \(N_G(Q)\), of order \(36/4=9\), so \(K\subseteq N_G(Q)\) and \(|K|\mid9\). Since \(G/K\) embeds in \(S_4\), \(K\) cannot be trivial because \(36\nmid24\). Hence \(|K|=3\) or \(9\).
5. If \(|K|=9\), then \(K\) itself is a normal Sylow \(3\)-subgroup. Thus in the remaining case \(|K|=3\) and \(G/K\cong A_4\).
When \(|K|=3\), the image has order \(12\). A subgroup of order \(12\) in \(S_4\) has index \(2\), hence is \(A_4\).
6. In the remaining case, \(K\subseteq Z(G)\).
Conjugation on \(K\cong C_3\) gives a homomorphism \[ G\longrightarrow\Aut(K)\cong C_2. \] If it were nontrivial, its kernel \(H\) would be normal of order \(18\). Sylow’s theorem in \(H\) gives a unique Sylow \(3\)-subgroup of order \(9\), hence a characteristic subgroup of \(H\) and therefore a normal Sylow \(3\)-subgroup of \(G\), contradiction. Thus conjugation on \(K\) is trivial.
7. The remaining order-\(36\) case has a normal Sylow \(2\)-subgroup.
Let \(V_4\trianglelefteq A_4\) be the Klein four subgroup and put \[ N=\varphi^{-1}(V_4). \] Then \(N\trianglelefteq G\) and \(|N|=12\). Let \(P\) be a Sylow \(2\)-subgroup of \(N\), so \(|P|=4\). Since \(K\cap P=1\), \(K\subseteq Z(G)\), and \(|KP|=12\), we have \(N=KP\).
For \(k\in K\) and \(p'\in P\), centrality of \(K\) gives \[ (kp')P(kp')^{-1}=p'P(p')^{-1}=P. \] Hence \(P\trianglelefteq N\). It is therefore the unique Sylow \(2\)-subgroup of \(N\), hence characteristic in \(N\). Since \(N\trianglelefteq G\), we get \(P\trianglelefteq G\).
Thus every group of order \(p^2q^2\) has a nontrivial proper normal subgroup, so no such group is simple.