Order of General Linear Group over Galois Field

Theorem
Let $F$ be a Galois field with $p$ elements.

Then the order of the general linear group $\operatorname {GL} \left({n, F}\right)$ is:


 * $\displaystyle \prod_{j \mathop = 1}^n \left({p^n - p^{j-1} }\right)$

Proof
Let $F$ be a Galois field with $p$ elements: $\left\vert{F}\right\vert = p$.

Let $A = \left[{a_{ij} }\right]_{n, n}$ be a matrix such that $\left|{A}\right| \ne 0$ and $a_{ij} \in F$.

How many such matrices can be constructed?

In order to avoid a zero determinant, the top row of the matrix, $\left\{{a_{1j}}\right\}_{j \mathop = 1, \dotsc, n}$ 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 $F$, minus the one sequence $0, 0, \dotsc, 0$.

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 $p^{\left({j - 1}\right)}$ sequences which are linear combinations of previous rows.

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


 * $\displaystyle \prod_{j \mathop = 1}^n \left({p^n - p^{j-1} }\right)$