Let \(G\) be a group of order \(p^2q\), where \(p\) and \(q\) are primes. Show that \(G\) has a normal Sylow subgroup.
Solution
If \(p=q\), then \(G\) itself is the unique Sylow \(p\)-subgroup, so assume \(p\ne q\).
1. If \(p>q\), the Sylow \(p\)-subgroup is normal.
Sylow gives \[ n_p\mid q, \qquad n_p\equiv1\pmod p. \] Thus \(n_p\in\{1,q\}\). Since \(q<p\), the value \(q\) cannot be congruent to \(1\) modulo \(p\), so \(n_p=1\).
2. Suppose \(p<q\). Then either the Sylow \(q\)-subgroup is normal or \((p,q)=(2,3)\).
Sylow gives \[ n_q\mid p^2, \qquad n_q\equiv1\pmod q. \] Hence \(n_q\in\{1,p,p^2\}\). Since \(1<p<q\), the value \(p\) is impossible. If \(n_q=p^2\), then \[ q\mid p^2-1=(p-1)(p+1). \] Because \(q>p\), this forces \(q\mid p+1\), hence \(q=p+1\). The only consecutive primes are \(2\) and \(3\).
3. In the exceptional order-\(12\) case, a Sylow subgroup is still normal.
If the Sylow \(3\)-subgroup is unique, we are done. Otherwise \(n_3=4\). The four order-\(3\) subgroups intersect only in the identity, so they contribute \(4(3-1)=8\) nonidentity elements. Exactly three nonidentity elements remain.
Every Sylow \(2\)-subgroup has order \(4\), and its three nonidentity elements cannot lie in an order-\(3\) subgroup. Hence every Sylow \(2\)-subgroup consists of the identity together with exactly those same three remaining elements. Therefore the Sylow \(2\)-subgroup is unique and normal.
Thus every group of order \(p^2q\) has a normal Sylow subgroup.