Show that a group of order \(p^2 q^2\) has a normal Sylow subgroup.
Solution
If \(p=q\), then \(G\) itself is the unique Sylow \(p\)-subgroup, so assume \(p\neq q\) and, after interchanging the primes, \(p<q\).
1. The number \(n_q\) of Sylow \(q\)-subgroups is either \(1\) or \(p^2\).
Sylow’s theorems give \[ n_q\mid p^2, \qquad n_q\equiv1\pmod q. \] Thus \(n_q\in\{1,p,p^2\}\). Since \(1<p<q\), the possibility \(n_q=p\) is incompatible with \(n_q\equiv1\pmod q\). 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 \(3\), so \(p=2\) and \(q=3\).
3. Unless \(|G|=36\), \(G\) has a normal Sylow \(q\)-subgroup.
If \(n_q=1\), the Sylow \(q\)-subgroup is normal. By 2, the only way \(n_q\neq1\) can occur is \(p=2\), \(q=3\), which gives \(|G|=36\).
4. Now suppose \(|G|=36\). If the Sylow \(3\)-subgroup is not normal, then \(n_3=4\) and conjugation on the four Sylow \(3\)-subgroups gives a homomorphism\[ \varphi:G\longrightarrow S_4 \]whose kernel \(K\) has order \(3\) or \(9\).
Sylow gives \(n_3=1\) or \(4\). Assume \(n_3=4\). The action is transitive, and the stabilizer of a Sylow \(3\)-subgroup \(Q\) is \(N_G(Q)\), of order \(36/4=9\). Thus \(K\subseteq N_G(Q)\), so \(|K|\mid9\). Since \(G/K\) embeds in \(S_4\), \(K\) cannot be trivial because \(36\nmid24\). Hence \(|K|=3\) or \(9\).
5. If \(|K|=9\), then \(K\) is a normal Sylow \(3\)-subgroup. Hence, when no Sylow \(3\)-subgroup is normal, \(|K|=3\) and \(G/K\cong A_4\).
The first assertion is immediate from \(K\trianglelefteq G\). In the remaining case \(|K|=3\), the image has order \(36/3=12\). A subgroup of order \(12\) in \(S_4\) has index \(2\) and is therefore the unique index-\(2\) subgroup \(A_4\).
6. Under the assumptions of 5, \(K\subseteq Z(G)\).
5. Therefore the conjugation action on \(K\) is trivial, so \(K\subseteq Z(G)\).
7. In the remaining order-\(36\) case, \(G\) has a normal Sylow \(2\)-subgroup.
Let \(V_4\trianglelefteq A_4\) be its unique Sylow \(2\)-subgroup and put \[ N=\varphi^{-1}(V_4). \] Then \(N\trianglelefteq G\) and \(|N|=|K||V_4|=12\). Let \(P\) be a Sylow \(2\)-subgroup of \(N\); then \(|P|=4\), so \(P\) is also a Sylow \(2\)-subgroup of \(G\). Since \(K\cap P=1\) and \(K\subseteq Z(G)\) by 6, \[ |KP|=|K||P|=12=|N|, \] so \(N=KP\). Every element of \(N\) is \(kp'\) with \(k\in K\), \(p'\in P\), and centrality of \(K\) gives \[ (kp')P(kp')^{-1}=p'Pp'^{-1}=P. \] Thus \(P\trianglelefteq N\). It is therefore the unique Sylow \(2\)-subgroup of \(N\), hence characteristic in \(N\). Since \(N\trianglelefteq G\), we get \(P\trianglelefteq G\).
8. Therefore every group of order \(p^2q^2\) has a normal Sylow subgroup.
Combine 3 with the order-\(36\) analysis in <1>4–<1>7.