Order of General Linear Group over Galois Field

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Theorem

Let $\GF$ be a Galois field with $p$ elements.

Then the order of the general linear group $\GL {n, \GF}$ is:

$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$

Proof

Let $\GF$ be a Galois field with $p$ elements: $\card \GF = p$.

Let $A = \sqbrk {a_{i j} }_{n, n}$ be a matrix such that $\size A \ne 0$ and $a_{i j} \in \GF$.

How many such matrices can be constructed?

In order to avoid a zero determinant, the top row of the matrix, $\sequence {a_{1 j} }_{j \mathop = 1, \dotsc, n}$ must have at least one non-zero element.

Therefore there are $p^n - 1$ possibilities for the top row:

the $p^n$ possible sequences of $n$ values from $\GF$, minus the one sequence $\paren {0, 0, \dotsc, 0}$.

The only restriction on the second row is that it not be a multiple of the first.

Therefore, there are the $p^n$ possible sequences again, minus the $p$ sequences which are multiples of the first row.

Thus, continuing in this fashion, the $j^{th}$ row can be any of the $p^n$ possible sequences, minus the $p^{\paren {j - 1} }$ sequences which are linear combinations of previous rows.

The number of possible matrices satisfying the conditions of $A$, then, is:

$\ds \prod_{j \mathop = 1}^n \paren {p^n - p^{j - 1} }$

$\blacksquare$