# computation of the order of $\mathrm{GL}\beta \x81\u2018(n,{\mathrm{\pi \x9d\x94\xbd}}_{q})$

$\mathrm{GL}\beta \x81\u2018(n,{\mathrm{\pi \x9d\x94\xbd}}_{q})$ is the group of invertible^{} $n\Gamma \x97n$ matrices
over the finite field^{} ${\mathrm{\pi \x9d\x94\xbd}}_{q}$.
Here is a proof that
$|\mathrm{GL}\beta \x81\u2018(n,{\mathrm{\pi \x9d\x94\xbd}}_{q})|=({q}^{n}-1)\beta \x81\u2019({q}^{n}-q)\beta \x81\u2019\mathrm{\beta \x8b\u2015}\beta \x81\u2019({q}^{n}-{q}^{n-1})$.

Each element $A\beta \x88\x88\mathrm{GL}\beta \x81\u2018(n,{\mathrm{\pi \x9d\x94\xbd}}_{q})$ is given by a collection^{} of $n$
${\mathrm{\pi \x9d\x94\xbd}}_{q}$-linearly independent vectors (http://planetmath.org/LinearIndependence).
If one chooses the first column vector^{} of $A$ from ${({\mathrm{\pi \x9d\x94\xbd}}_{q})}^{n}$
there are ${q}^{n}$ choices, but one canβt 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 ${({\mathrm{\pi \x9d\x94\xbd}}_{q})}^{n}$
which is linearly independent from $i-1$ already chosen
linearly independent vectors $\{{V}_{1},\mathrm{\beta \x80\xa6},{V}_{i-1}\}$
one must choose a vector not in
the span of $\{{V}_{1},\mathrm{\beta \x80\xa6},{V}_{i-1}\}$.
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 ${\mathrm{\pi \x9d\x94\xbd}}_{q}$
is $({q}^{n}-1)\beta \x81\u2019({q}^{n}-q)\beta \x81\u2019\mathrm{\beta \x8b\u2015}\beta \x81\u2019({q}^{n}-{q}^{n-1})$.

Title | computation of the order of $\mathrm{GL}\beta \x81\u2018(n,{\mathrm{\pi \x9d\x94\xbd}}_{q})$ |
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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 |