# Order of general- and special linear groups over finite fields.

Let $\mathbb{F}_3$ be the field with three elements. Let $n\geq 1$. How many elements do the following groups have?

1. $\text{GL}_n(\mathbb{F}_3)$
2. $\text{SL}_n(\mathbb{F}_3)$

Here GL is the general linear group, the group of invertible n×n matrices, and SL is the special linear group, the group of n×n matrices with determinant 1.

First question: We solve the problem for “the” finite field $$F_q$$ with $$q$$ elements. The first row $$u_1$$ of the matrix can be anything but the $$0$$-vector, so there are $$q^n-1$$ possibilities for the first row.
For any one of these possibilities, the second row $$u_2$$ can be anything but a multiple of the first row, giving $$q^n-q$$ possibilities.

For any choice $$u_1, u_2$$ of the first two rows, the third row can be anything but a linear combination of $$u_1$$ and $$u_2$$. The number of linear combinations $$a_1u_1+a_2u_2$$ is just the number of choices for the pair $$(a_1,a_2)$$, and there are $$q^2$$ of these. It follows that for every $$u_1$$ and $$u_2$$, there are $$q^n-q^2$$ possibilities for the third row.

For any allowed choice $$u_1$$, $$u_2$$, $$u_3$$, the fourth row can be anything except a linear combination $$a_1u_1+a_2u_2+a_3u_3$$ of the first three rows. Thus for every allowed $$u_1, u_2, u_3$$ there are $$q^3$$ forbidden fourth rows, and therefore $$q^n-q^3$$ allowed fourth rows.

Continue. The number of non-singular matrices is
$$(q^n-1)(q^n-q)(q^n-q^2)\cdots (q^n-q^{n-1}).$$

Second question: We first deal with the case $$q=3$$ of the question. If we multiply the first row by $$2$$, any matrix with determinant $$1$$ is mapped to a matrix with determinant $$2$$, and any matrix with determinant $$2$$ is mapped to a matrix with determinant $$1$$.

Thus we have produced a bijection between matrices with determinant $$1$$ and matrices with determinant $$2$$. It follows that $$SL_n(F_3)$$ has half as many elements as $$GL_n(F_3)$$.

The same idea works for any finite field $$F_q$$ with $$q$$ elements. Multiplying the first row of a matrix with determinant $$1$$ by the non-zero field element $$a$$ produces a matrix with determinant $$a$$, and all matrices with determinant $$a$$ can be produced in this way. It follows that
$$|SL_n(F_q)|=\frac{1}{q-1}|GL_n(F_q)|.$$