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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.

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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.

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  • $\begingroup$ 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
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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 $K \cdot f(d,d) = f(n,d)$.

Hence

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

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.

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