# 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.

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Presumably, that's what you are asked to do. – Alex B. Apr 21 '11 at 11:22
Let $q=3$, and take, say, $n=4$. The first row of the matrix can be anything but the $0$-vector, $q^4-1$ possibilities. For any one of these, the second row is anything but a multiple of the first row, so there are $q^4-q$ possibilities. For any specific choice of first two rows, the third row is anything but linear combinations of the first two rows. The number of linear combinations $au+bv$ of linearly independent $u$, $v$ is just the number of choices for the pair $(a,b)$, namely $q^2$. So for every choice of first two rows, there are $q^4-q^2$ choices of third row. Continue. – André Nicolas Apr 21 '11 at 12:42
Continuing...user6312's observation and the multiplication principle of counting will get you only the answer to 1). The easiest way to do 2) is to use 1) and a little group theory, if you know some. The determinant function $\det \colon \mathrm{GL}_n (\mathbb{F}_3) \rightarrow \mathbb{F}_3^{\times}$ is a group homomorphism whose kernel is $\mathrm{SL}_n (\mathbb{F}_3)$. Now use the fact that all cosets of a subgroup of a finite group have the same cardinality. – Barry Smith Apr 21 '11 at 13:06
Jenna? Are you there? Hello, Jenna? – Gerry Myerson Apr 26 '11 at 9:56
@André: make it a full-fledged answer! ;) – Raskolnikov Oct 19 '11 at 8:23

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 $$|SL_n(F_q)|=\frac{1}{q-1}|GL_n(F_q)|.$$

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Can you tell what is your field? $F_{3}$ consists only 3 memebers it must be e.g. $\{0,1,2\}$ in mod $3$ for $n=2$ , $GL_{n}(F_{3})$ has $48$ members.

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For $n=2$ it is 48 instead. – Soarer Sep 19 '11 at 6:59
yes you are right. – Vahid Sep 19 '11 at 7:15