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I want to find the number of elements of $GL(n,\mathbb{Z_p})$, $p$ prime. For $p=2$ I got 6 elements since one of diagonals has to be $0$ and the other $1$. But how should I deal with an arbitrary $n$ ? The expression for the determinant gets complicated.

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marked as duplicate by Derek Holt group-theory Apr 26 '17 at 12:53

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Here is a plan to tackle this question :

  1. The number of elements of $GL(n,\mathbb{Z_p})$ is the number of basis of $\mathbb{Z_p}^n$.
  2. To count the basis of $\mathbb{Z_p}^n$, firstly you chose a non zero vector , so you have $p^n-1$ choices, then to choose the secon vector its must be non zero zero and not colinear with the previous one so you have $p^n-p$ choices. By continuing this reassoning you get : $$|GL(n,\mathbb{Z_p})|=(p^n-1)(p^n-p)\dots(p^n-p^{n-1})$$
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