Show that a group of order \(p^2 q^2\) where \(q\) does not divide \(p^2-1\) and \(p\) does not divide \(q^2-1\) is abelian.
The hypothesis \(p\neq q\) is needed: for \(p=q\) both divisibility conditions hold, and a group of order \(p^4\) need not be abelian, as the dihedral group of order \(16\) shows.
Solution
Let \(p\neq q\) be primes and \(|G|=p^2q^2\), and let \(n_p\), \(n_q\) be the numbers of Sylow \(p\)- and \(q\)-subgroups.
1. The Sylow \(p\)-subgroup \(P\) is normal.
By Sylow’s theorems, \(n_p\mid q^2\) and \(n_p\equiv1\pmod p\), so \(n_p\in\{1,q,q^2\}\). If \(n_p=q\) or \(n_p=q^2\), then \(p\mid q-1\) or \(p\mid q^2-1\); in both cases \(p\mid q^2-1\), contrary to hypothesis. So \(n_p=1\).
2. The Sylow \(q\)-subgroup \(Q\) is normal.
Exchange \(p\) and \(q\) in step 1, using \(q\nmid p^2-1\).
3. \(G\cong P\times Q\).
\(P\cap Q=1\) since \(\gcd(p^2,q^2)=1\), and \(|PQ|=|P||Q|=|G|\), so \(G=PQ\). For \(x\in P\) and \(y\in Q\), normality of \(Q\) gives \(xyx^{-1}y^{-1}\in Q\) and normality of \(P\) gives \(x(yx^{-1}y^{-1})\in P\), so \(xyx^{-1}y^{-1}\in P\cap Q=1\). Thus \(P\) and \(Q\) commute elementwise and \((x,y)\mapsto xy\) is an isomorphism \(P\times Q\to G\).
4. Q.E.D.
Groups of order \(p^2\) and \(q^2\) are abelian, so \(P\) and \(Q\) are abelian, and by step 3 so is \(G\cong P\times Q\).