# Maximal subgroups of $p$-groups

The following is exercise 27 from section 6.1 in Dummit and Foote (3rd edition):

Let $$P$$ be a $$p$$-group and let $$\overline{P} = P/frat(P)$$ be elementary abelian of order $$p^r$$. Prove that $$P$$ has exactly $$\dfrac{p^r - 1}{p-1}$$ where $$frat(P)$$ is the Frattini subgroup (intersection of all maximal subgroups of $$P$$).

My attempt:

Any maximal subgroup is contained in $$frat(P)$$ so it suffices by the lattice isomorphism theorem to find maximal subgroups of $$\overline{P}$$. But maximal subgroups of a $$p$$-group are of index $$p$$ (which are in bijective correspondence with subgroups of order $$p$$). Since $$\overline{P}$$ is elementary abelian of order $$p^r$$, $$\overline{P} \cong \prod_{i=1}^r \mathbb{Z}_p$$. Looking at the direct product there are $$r$$ subgroups of order $$p$$. Hence, there are $$r$$ maximal subgroups of $$\overline{P}$$.

But, this does not agree with the conclusion in question. What am I missing?

Thank you,

What you have written is true. What you are missing is: are these $$r$$ subgroups the only ones of order $$p$$? If you believe they are (they are not), try to prove it.
Since $$P/\Phi(P)$$ is elementary abelian, it can be viewed as a vector space over the field $$\mathbb{F}_p$$. Thinking in terms of vector spaces makes the problem a bit easier to handle and understand. The quantities that count the number of subspaces with fixed cardinality of some finite vector space are known as $$q$$-binomial (or Gaussian) coefficients. More explicitly
$$$${n \brack k}_p = \prod_{i=0}^{k-1} \frac{1 - p^{n-i}}{1-p^{i+1}}$$$$ is the number of $$k$$-dimensional subspaces of an $$n$$-dimensional vector space over the field $$\mathbb{F}_p$$. To get your answer, simply put $$n=r$$ and $$k=1$$ if you understand the principle of duality that is in effect in abelian groups. Otherwise, put $$n=r$$ and $$k=r-1$$ to get the same answer.
I suggest you take a little time to try to prove the counting formula. If you don't want to do that, look here (Theorem $$6.3$$).
• Thank you for your response. It was naive of me to think that these were the only subgroups of order $p$.