Let $G$ be a group of order $p^2q^2$ , $p$ does not divide $q^2-1$ , $q$ does not divide $p^2-1$ , then is $G$ abelian?

Question

Let $G$ be a group of order $p^2q^2$ ,where $p,q$ are primes , $p$ does not divide $q^2-1$ ,and $q$ does not divide $p^2-1$ , then is $G$ abelian ?

Answer 1

Let G be such a group. Consider n the cardinal of the set S of Sylow subgroup H of G of order $p^2$, we know that $n$ divides $q^2$. Suppose that $n=q$ or $n=q^2$, since $n-1=0$ mod $p$, we deduce that $p$ divides $q-1$ or $p$ divides $q^2-1$ this implies that $p$ divides $q^2-1$ since $q-1$ divides $q^2-1$. This is a contradiction with the hypothesis so $n=1$.

The same argument shows that the number of Sylow $q^2$ groups is 1.

Let H be the Sylow $p^2$ subgroup and L be the Sylow $q^2$ subgroup. H and L are normal. Let $u$ in H and $v$ in L, we have $uvu^{-1}v^{-1}=u(vu^{-1}v^{-1})=(uvu^{-1})v^{-1}\in L\cap H=1$.

This implies that $uv=vu$. Since a group of order $p^2$ is commutative, we deduce that G is commutative.