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I am looking for a method to calculate number all subgroups of a finite elementary abelian $p$-group.

Suppose $G$ be an elementary abelian $p$-group of order $p^n$. A proper subgroup $H$ of $G$ is also an elementary abelian $p$-group of order $p^r$ where $r<n$.

We can realize $G$ as $n$ dimensional vector over $\mathbb Z_p$ and number of subgroups of $G$ of order $p^r$ is equal to the number of $r$ dimensional subspaces of the vector space.

Now we have to count the number of linearly independent $r$-tuples of vectors but several tuples may give same subspace.

Now what should I do??

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  • $\begingroup$ you can begin with $\;r=1\;$ and do some rather arithmetic for the $\;1$-dimensional subspaces there...and then some induction, perhaps. $\endgroup$ – DonAntonio Apr 25 '14 at 15:20
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    $\begingroup$ The number of $1$-dimensional subspaces is $(p^n-1)/(p-1)$, the number of $2$-dimensionals is $(p^n-1)(p^n-p)/((p^2-1)(p^2-p))$, etc. That's just doing what you proposed and dividing the number of linearly independent $r$-tuples by the number that give each specific subspace. $\endgroup$ – Derek Holt Apr 25 '14 at 16:47
  • $\begingroup$ @DerekHolt I understand that there are $(p^n-1)(p^n-p)\cdots (p^n-p^{r-1})$ linearly independent $r$-tuples but how do I show that $(p^r-1)(p^r-p)\cdots (p^r-p^{r-1})$ $r$-tuples will give same subspace? $\endgroup$ – user93432 Apr 26 '14 at 6:54
  • $\begingroup$ Because that's the number of linearly independent $r$-tuples in an $r$-dimensional space. (It's the same argument but with $n$ replaced by $r$.) $\endgroup$ – Derek Holt Apr 26 '14 at 10:07
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I am thinking about the Transitive (?) group action of $GL_n(\mathbb F_p)$ on the set of linearly independent $r$-tuples. Number of orbits of this action will give the number of $r$-dimensional subspaces and hence the number of subgroups of order $p^r$. Is my this approach correct??

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