Let be a prime, let be a finite -group, and let be elementary abelian of rank . Prove that has exactly maximal subgroups.
We begin with a lemma.
Lemma: If is a finite elementary abelian -group of rank , then contains exactly distinct maximal subgroups. Proof: By Sylow’s Theorem, the maximal subgroups of are precisely those of index . In this previous exercise, we showed that the number of subgroups of index is equal to the number of subgroups of order . Now every nonidentity element of generates an order -subgroup, and each such subgroup is generated by elements. Thus the desired number is .
Recall from the Lattice Isomorphism Theorem that the maximal subgroups of correspond bijectively to the maximal subgroups of which contain ; by definition, this is all the maximal subgroups of . Thus the number of maximal subgroups of is the same as the number of maximal subgroups in , which is by the lemma.