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.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: