Let \(p,q\) be primes and let \(|G|=p^2q\). Show that \(G\) has a nontrivial proper normal subgroup.
Solution
If \(p=q\), then \(G\) is a finite \(p\)-group. Its center is nontrivial, so \(G\) has a nontrivial normal subgroup. Thus assume \(p\ne q\).
If \(p>q\), Sylow’s theorem gives \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Since \(q<p\), the only possibility is \[ n_p=1. \] Hence the Sylow \(p\)-subgroup is normal.
Now suppose \(p<q\). Sylow gives \[ n_q\mid p^2, \qquad n_q\equiv1\pmod q. \] Thus \(n_q\in\{1,p,p^2\}\). The value \(p\) is impossible because \(p<q\). If \(n_q=p^2\), then \[ q\mid p^2-1=(p-1)(p+1). \] Since \(q>p\), this forces \(q\mid p+1\), hence \[ q=p+1. \] The only consecutive primes are then \[ p=2,\qquad q=3. \] So except for groups of order \(12\), one has \(n_q=1\) and hence a normal Sylow \(q\)-subgroup.
It remains to treat \(|G|=12\). If \(n_3=1\), the Sylow \(3\)-subgroup is normal. Otherwise \(n_3=4\), and the four Sylow \(3\)-subgroups contain \[ 4(3-1)=8 \] distinct nonidentity elements. Hence exactly four elements remain, including the identity. Any Sylow \(2\)-subgroup has order \(4\) and cannot contain an element of order \(3\), so it must consist exactly of these four remaining elements. Thus the Sylow \(2\)-subgroup is unique and normal.
Therefore every group of order \(p^2q\) has a nontrivial proper normal subgroup.