= Groups of order p squared q are not simple
{title2=$|G|=p^2q$}
For distinct primes $p,q$, the <Sylow theorems> force a proper nontrivial <normal subgroup> in any group of order $p^2q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a <simple group>.
Back to article page
