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While proving some facts about matrix group operations on finite fields, I stumbled across the following question:

What is the order of the group of invertible $n\times n$ matrices over a finite field of prime order $p$?

The answer seems to be $$\left|\mathrm{GL}_n(\mathbb F_p)\right|=\prod_{k=1}^n\left(p^n-p^{k-1}\right)\text,$$ but I have not yet seen a satisfying proof: I took a look at this one, but found it to be somewhat sloppy. Is there a nice formal proof of this fact?

Ahaan S. Rungta
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user201344
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    Maybe you could elaborate on what you found sloppy about it? It seems fine to me. – Tobias Kildetoft Dec 15 '14 at 13:37
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    This is almost certainly a duplicate, although I can't find any dupe. That particular proof (counting matrices with linearly independent rows) is already very simple. The only way it could be made more formal is by shoring up the argument with induction (although it's reasonably OK the way it currently is.) – rschwieb Dec 15 '14 at 13:42
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    Judging by the question title alone you may not believe that it is a duplicate, but André's answer covers this case as well. Anyway, I was happy enough with it to use my AA-powervote. – Jyrki Lahtonen Dec 15 '14 at 13:49
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    @JyrkiLahtonen I gave that question a new title to better match the content (and the more general answer than what was originally asked for). – Tobias Kildetoft Dec 15 '14 at 14:02
  • A good idea, @Tobias. – Jyrki Lahtonen Dec 15 '14 at 14:03

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To count invertible $n\times n$ matrices is the same as counting bases (the columns of such a matrix form a basis; in other words: A matrix is invertible if and only if the image of the standard basis is a basis) or in fact just to count linearly independent families of $n$ vectors. Given $k\ge0$ linearly independent vectors $v_1,\ldots,v_k\in V=\mathbb F_q^n$, among the $q^n$ elements of $V$ there are $q^k$ that are in the space spanned by the first $k$ vectors, hence there are only $q^n-q^k$ possible choices for $v_{k+1}$ to produce a larger familiy of linearly independent vectors. So if $a_k$ is the number of linearly independent families of $k$ vectors in $\mathbb F_q^n$, we have $$ a_0=1,\qquad a_{k+1}=(q^n-q^k)a_k$$ and conclude $$ a_k=\prod_{i=0}^{k-1}(q^n-q^i)$$ and specifically $$ |\operatorname{GL}_n(\mathbb F_q)|=a_n=\prod_{i=0}^{n-1}(q^n-q^i)=\prod_{k=1}^{n}(q^n-q^{k-1}).$$

Hagen von Eitzen
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