Let $G=GL(2,GF(3))$ be the group of all matrices $ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $

Question

Let $G=GL(2,GF(3))$ be the group of all matrices $ \begin{pmatrix} a & b \\ c & d \end{pmatrix} $ whose determinants are not zero. and whose entries $a,b,c,d$ are taken from the finite field $GF(3)=\{0,1,2\}$. and $H = \{\begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G : ad - bc = 1\}$ What is $o(H)$?

First of all, I've computed $o(G) = 48$, The columns have to be linearly independent. so first column is $3\times 3 = 9$ options, and $(0,0)$ can't be too, so total $8$ options. Second column, same thing but we need to take into consideration that there could be a scalar multiplcation with the first column, so we need to substract two scalars $1,2$ and we stay with $6$ options. Total is $o(G) = 48$. I've computed this value so I can use lagrange later.

Now, I need to find the matrices that their determinant gives $1$. Is there a way to find them efficiently without counting one after one? I think there're 24 options in total, but I'm not sure how to prove it without counting one by one.

I thought about taking

$ad = 0 \rightarrow bc = 2$, total of $5\times 2 = 10$ options

$ad = 1, \rightarrow bc = 0$, total of $5\times 2 = 10$ options

$ad = 2 \rightarrow bc = 1$, total of $2\times 2 = 4$ options.

total of $10+10+4 = 24$ options.

Then $o(H) | o(G) = 24 | 48 = 2$, and lagrange works.

Answer 1

You method works. Here's an efficient way, though:

Hint: Note that $H$ is a subgroup, and that the coset $$ \pmatrix{2&0\\0&1} H $$ consists of all elements whose determinant is $2$.

Answer 2

Note that you have a surjective homomorphism \begin{align*} \operatorname{GL}_2(\Bbb F_3)&\to\Bbb F_3^\times\\ M&\mapsto\det M \end{align*} which has kernel precisely $H = \operatorname{SL}_2(\Bbb F_3) = \{M\in\operatorname{GL}_2(\Bbb F_3)\mid\det M = 1\}$. Thus, $$ \operatorname{GL}_2(\Bbb F_3)/\operatorname{SL}_2(\Bbb F_3)\cong\Bbb F_3^\times, $$ and hence $2 = \#\Bbb F_3^\times = \#(\operatorname{GL}_2(\Bbb F_3)/\operatorname{SL}_2(\Bbb F_3)) = \#\operatorname{GL}_2(\Bbb F_3)/\#\operatorname{SL}_2(\Bbb F_3)$.