Let $G$ be a $p$-group, i.e. $|G|=p^n$. Call $\Phi(G)$ the Frattini group of $G$.

Then we have that $G/\Phi(G)\simeq(C_p)^d$ ($d$ copies of the cyclic group of order $p$, i.e. $\overbrace{C_p\times\cdots\times C_p}^{d-times}$), for some $d\in\mathbb N$. And till here it's all right.

Then my teacher said that the numbers of maximal subgroup of $G$ is $$ \frac{p^d-1}{p-1}. $$ What I can't understand is:

  1. Why the numbers of maximal subgroups of a $p$-groups are of the form $$ \frac{p^m-1}{p-1}=1+p+p^2+\cdots+p^{m-1} $$ for some $n\in\mathbb N$.
  2. Why $m=d$, hence in which way $d$ is related to the numbers of maximal subgroup of $G$.

Any help would be appreciated so much. Thank you all.


Let $G=C_p\times C_p\times...\times C_p=(C_p)^d$ Since $C_p$ is a field, we can think $G$ as a vector space over $C_p$ with dimension $d$.

Notice that any $d-1$ dimensional subspace of a vector space can be uniquely defined by a orthogonal complement of a $1$ dimensional subspace. Thus, it is enough to find number of the $1$ dimensional vector space.

We have $p^d-1$ nontrivial elements and $\langle v\rangle=\langle cv\rangle$ where $c\in\{1,2,..,p-1\}$ which means we have $$\dfrac{p^d-1}{p-1}$$ one dimensional vector space, so we are done.

If $G$ is any $p$ group then there is one to coresspondence between maximal subgroups of $G$ and maximal subgroups of $G/\Phi(G)$ as $\Phi(G)\leq M$ for any $M$ which conclude the result.

Note: Above argument shows also that number of the subgroups of index $p$ is equal to number of the subgroups of order $p$ in elementary abelian groups.

  • $\begingroup$ Thank you! I noticed too the corrispondence between maximal subgroup of $G$ and maximal subgroups of $C_p^d$... but the maximal subgroups of $C_p^d$ are of the form $C_p^r\times1\times C_p^s$ where $r+s=d-1$ (obviously $r,s\in\mathbb Z_{\ge0}$), but this yelds to $d-1$ as the number of maximal subgroups of $G$. Where is my mistake? $\endgroup$ – Joe May 23 '14 at 18:26
  • $\begingroup$ @Joe: A subgroup of $H\times K$ need not be in the form $M\times N$ where $M\leq H$ and $N\leq K$ and to see this notice that $Z_p\times Z_p$ has $p+1$ maximal subgroup, (take $p$=2 and examine $Z_2\times Z_2$ you will find $3$ maximal subgroup not $2$) $\endgroup$ – mesel May 23 '14 at 18:37
  • $\begingroup$ @Joe: I guess you do not like my answers :) $\endgroup$ – mesel May 24 '14 at 8:19
  • $\begingroup$ You're right. His answer was more direct... however I must acknowledge that yours is smarter. Hence, sorry spin, seriously, but I have to change. $\endgroup$ – Joe May 24 '14 at 11:14
  • $\begingroup$ @mesel Sorry sir, i have a question. how we can conclude that number of the subgroups of index $p$ is equal to number of the subgroups of order $p$ in elementary abelian groups? $\endgroup$ – Little girl Jun 17 '18 at 14:12

Let $V = (C_p)^n$, which is a $n$-dimensional vector space over $C_p$. Then the number of $d$-tuples $(x_1, \ldots, x_d)$ of linearly independent vectors is $$f(n,d) = (p^n - 1)(p^n - p) \cdots (p^n -p^{d-1})$$

Now if $K$ is the number of $d$-dimensional subspaces, we have $K \cdot f(d,d) = f(n,d)$.


$$K = \frac{p^n -1}{p^d -1} \cdot \frac{p^{n-1}-1}{p^{d-1} -1} \cdots \frac{p^{n-d+1} -1}{p -1}$$

So when $d = n-1$, you get $K = \frac{p^n - 1}{p-1}$.

Since $\Phi(G)$ is contained in every maximal subgroup of $G$, you have a bijection between the maximal subgroups of $G$ and $G/\Phi(G)$. Combining this with the above result you have what you need.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.