## Count the maximal subgroups of a finite p-group

Let $p$ be a prime, let $P$ be a finite $p$-group, and let $P/\Phi(P)$ be elementary abelian of rank $r$. Prove that $P$ has exactly $(p^r-1)/(p-1)$ maximal subgroups.

We begin with a lemma.

Lemma: If $G$ is a finite elementary abelian $p$-group of rank $r$, then $G$ contains exactly $(p^r-1)/(p-1)$ distinct maximal subgroups. Proof: By Sylow’s Theorem, the maximal subgroups of $G$ are precisely those of index $p$. In this previous exercise, we showed that the number of subgroups of index $p$ is equal to the number of subgroups of order $p$. Now every nonidentity element of $G$ generates an order $p$-subgroup, and each such subgroup is generated by $p-1$ elements. Thus the desired number is $(p^r-1)/(p-1)$. $\square$

Recall from the Lattice Isomorphism Theorem that the maximal subgroups of $P/\Phi(P)$ correspond bijectively to the maximal subgroups of $P$ which contain $\Phi(P)$; by definition, this is all the maximal subgroups of $P$. Thus the number of maximal subgroups of $p$ is the same as the number of maximal subgroups in $P/\Phi(P)$, which is $(p^r-1)/(p-1)$ by the lemma.