# 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 Khan 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 kernel $$SL(n, F)$$. Hence by the 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 $$|GL(n,F)|=(p-1)|SL(n, F)|$$.

• This is rather short but rigorous proof. Thank you. – Henry Choi Sep 1 '20 at 2:05