Let \(G\) be a group of order \(p^2q\) for \(p, q\) prime. Show that \(G\) has a nontrivial normal subgroup.
Solution
Let \(|G|=p^2q\), and let \(n_p,n_q\) be the numbers of Sylow \(p\)- and \(q\)-subgroups.
1. If \(p=q\), then \(G\) has a proper nontrivial normal subgroup.
\(|G|=p^3\), so \(Z(G)\neq1\) by the class equation. If \(Z(G)\neq G\), it is the required subgroup; if \(G\) is abelian, every subgroup of order \(p\) (which exists by Cauchy’s theorem) is normal.
2. If \(p>q\), then the Sylow \(p\)-subgroup is normal.
\(n_p\mid q\) and \(n_p\equiv1\pmod p\); since \(1<q<p\), \(q\not\equiv1\pmod p\), so \(n_p=1\).
3. If \(p<q\), then a Sylow \(p\)- or Sylow \(q\)-subgroup is normal.
\(n_q\mid p^2\) and \(n_q\equiv1\pmod q\), and \(p\not\equiv1\pmod q\) since \(1<p<q\), so \(n_q\in\{1,p^2\}\). If \(n_q=1\), the Sylow \(q\)-subgroup is normal. If \(n_q=p^2\), distinct Sylow \(q\)-subgroups (of prime order \(q\)) meet trivially, so they contain \(p^2(q-1)\) elements of order \(q\). The remaining \(p^2q-p^2(q-1)=p^2\) elements contain every Sylow \(p\)-subgroup, each of order \(p^2\), so there is exactly one Sylow \(p\)-subgroup, and it is normal.