Let \(p\ne q\) be primes. Show that a group \(G\) of order \(p^2q^2\) is abelian if \[ q\nmid(p^2-1)\qquad\text{and}\qquad p\nmid(q^2-1). \]
Solution
Let \(n_p\) be the number of Sylow \(p\)-subgroups. Sylow’s theorem gives \[ n_p\mid q^2, \qquad n_p\equiv1\pmod p. \] Thus \(n_p\in\{1,q,q^2\}\). If \(n_p=q\), then \(p\mid(q-1)\), hence \(p\mid(q^2-1)\), contrary to hypothesis. If \(n_p=q^2\), then again \(p\mid(q^2-1)\), contradiction. Therefore \[ n_p=1. \] So the Sylow \(p\)-subgroup \(P\) is normal.
Similarly, \[ n_q\mid p^2, \qquad n_q\equiv1\pmod q. \] The possibilities \(n_q=p\) and \(n_q=p^2\) would imply respectively \(q\mid(p-1)\) or \(q\mid(p^2-1)\), both impossible under the stated hypothesis. Hence \[ n_q=1, \] so the Sylow \(q\)-subgroup \(Q\) is normal.
Since \(P,Q\normal G\) and \(|P|,|Q|\) are coprime, \[ [P,Q]\le P\cap Q=1. \] Hence \(P\) and \(Q\) commute elementwise. Also \(|PQ|=|P||Q|=|G|\), so \[ G=P\times Q. \] Every group of order \(p^2\) or \(q^2\) is abelian, so both \(P\) and \(Q\) are abelian. Therefore \(G\) is abelian.