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.

• 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
• @BarrySmith So this then follows from the first isomorphism theorem? – Anthony Peter Nov 2 '16 at 22:08
• If by "this", you mean the answer to task 2, then yes. – Barry Smith Nov 3 '16 at 0:36

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

• Excellently explained. This was of great help. – Cauchy Jul 1 '17 at 23:39
• @andre well done. So it means Special linear groups arr not normal subgroups of General linear group, right? – Prince Thomas Oct 28 '17 at 14:26
• Great and simple solution! +1 for that :) Thanks a lot for that :) – ZFR Apr 17 '18 at 15:10
• @PrinceThomas The special linear group \emph{is} a normal subgroup of the general linear group. The special linear group is the kernel of the determinant, which is a homomorphism from the general linear group to the underlying field of units. – frito_mosquito Aug 5 '18 at 14:07

Determinant function is a surjective homomorphism from $GL(n, F)$ to $F^*$ with kernal $SL(n, F)$. Hence by fundamental isomorphism theorem $\frac{GL(n,F)}{SL(n,F)}$ is isomorphic to $F^*$, the multiplicative group of nonzero elements of F.

Thus if $F$ is finite with $p$ elements then $|GL(n,F)|=(p-1)|SL(n, F)|$