computation of the order of $\operatorname{GL}(n,\mathbb{F}_{q})$

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$\operatorname{GL}(n,\mathbb{F}_{q})$ is the group of invertible $n\times n$ matrices over the finite field $\mathbb{F}_{q}$ . Here is a proof that $\lvert\operatorname{GL}(n,\mathbb{F}_{q})\rvert=(q^{n}-1)(q^{n}-q)\cdots(q^{n}% -q^{n-1})$ .

Each element $A\in\operatorname{GL}(n,\mathbb{F}_{q})$ is given by a collection of $n$ $\mathbb{F}_{q}$ -linearly independent vectors (http://planetmath.org/LinearIndependence). If one chooses the first column vector of $A$ from $(\mathbb{F}_{q})^{n}$ there are $q^{n}$ choices, but one can’t choose the zero vector since this would make the determinant of $A$ zero. So there are really only $q^{n}-1$ choices. To choose an $i$ -th vector from $(\mathbb{F}_{q})^{n}$ which is linearly independent from $i-1$ already chosen linearly independent vectors $\{V_{1},\dots,V_{i-1}\}$ one must choose a vector not in the span of $\{V_{1},\dots,V_{i-1}\}$ . There are $q^{i-1}$ vectors in this span, so the number of choices is $q^{n}-q^{i-1}$ . Thus the number of linearly independent collections of $n$ vectors in $\mathbb{F}_{q}$ is $(q^{n}-1)(q^{n}-q)\cdots(q^{n}-q^{n-1})$ .

Title	computation of the order of $\operatorname{GL}(n,\mathbb{F}_{q})$
Canonical name	ComputationOfTheOrderOfoperatornameGLnmathbbFq
Date of creation	2013-03-22 13:06:50
Last modified on	2013-03-22 13:06:50
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	15
Author	yark (2760)
Entry type	Proof
Classification	msc 20G15
Related topic	OrderOfTheGeneralLinearGroupOverAFiniteField

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computation of the order of $\operatorname{GL}(n,\mathbb{F}_{q})$