Determining the cardinality of $SL_2 (F_3) $ I have been trying to determine the order of $SL_2 (F_3) $. My books says the answer is 24. But the answer that I am getting is 30. 
My method: 
Case 1: Assume that $a_{11} $ is nonzero. Then whatever be the values of the other elements, the value of $a_{22} $ is fixed. Hence, for a nonzero $a_{11} $, there are 2.3.3=18 possibilities. 
Case 2: Assume that $a_{11} $ is 0. Then $a_{22} $ can be anything, and $a_{12} $ and $a_{21} $ can be $\{\pm 1\} $ or $\{\pm 2\} $. There are 4.3=12 such possibilities. 
Hence total = 30. 
Where am I going wrong? Any help would be great. 
 A: Let's first start by finding the cardinality of $GL_2(\mathbb{F}_p)$. The elements can be viewed as pairs of column vectors, which must be linearly independent so the matrix is invertible.
For the first vector, that are $p^2$ vectors but we must remove the zero vector so we have $p^2-1$ choices.
For the second to be linearly independent from the first, it cannot be a multiple of the first, of which there are $p$ such vectors. This means we have $p^2-p$ such options.
Putting this together, $|GL_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)$ so in your case $p=3$ so $|GL_2(\mathbb{F}_3)|=48$.
Now $SL_2(\mathbb{F}_p$ is characterised as the set of matrices with determinant 1, in particular it is the kernel of the determinant map. Now a matrix is invertible if and only if it is nonzero determinant so there are $p-1$ possible choices of determinant for an element $GL_2(\mathbb{F}_p)$.
So $|SL_2(\mathbb{F}_p)|=(p^2-1)(p^2-p)/(p-1)=(p+1)(p^2-p)$. Plugging in $p=3$, we get $|SL_2(\mathbb{F}_3)|=24$.
A: I think your problem is in case 2.
Assuming $a_{11}=0$, you must have $a_{12}a_{21}=-1$, which means that $a_{12}$ can take any non-zero value, and the value of $a_{21}$ is then simply $(-a_{12})^{-1}$. So you have two possibilities, and as you said yourself, $a_{22}$ can take any value. This leaves $2.3=6$ possible matrices.
