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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.

  • 7
    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.2011-04-21
  • 5
    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.2011-04-21
  • 0
    @BarrySmith So this then follows from the first isomorphism theorem?2016-11-02
  • 0
    If by "this", you mean the answer to task 2, then yes.2016-11-03

2 Answers 2