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

\begin{equation} {n \brack k}_p = \prod_{i=0}^{k-1} \frac{1 - p^{n-i}}{1-p^{i+1}} \end{equation} 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 $6.3$).

  • $\begingroup$ Thank you for your response. It was naive of me to think that these were the only subgroups of order $p$. $\endgroup$
    – Mike
    Jul 16 '19 at 3:17

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