Order of General Linear Group over Galois Field


 * In progress

Theorem
If $$F \ $$ is a field with $$ p $$ elements, the order of $\text{GL}_n(F) \ $ is:


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

Proof
Let $$A=[a_{ij}]_{n,n} \ $$ be a matrix such that $$|A|\neq 0 \ $$ and $$a_{ij}\in F \ $$, where $$F \ $$ is a field of finitely many elements: $$|F|=p \ $$.

How many such matrices can be constructed? In order to avoid a zero determinant, the top row of the matrix, $$\left\{{a_1j}\right\}_{j=1,\dots,n} \ $$ 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 $$0,0, \dots, 0 \ $$.

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.

Continuing in this fashion, then, the $$j^{th} \ $$ row could be any of the $$p^n \ $$ possible sequences, minus the $$p^{(j-1)} \ $$ sequences which are multiples of one of the previous rows.

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


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