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.

  • $\begingroup$ The value of $a_{11}$ doesn't fix $a_{22}$. (Also, how would you get an element of order $5$ in $SL_2(\mathbb{F}_3)$? Consider the Jordan normal form of an arbitrary element.) $\endgroup$ – anomaly Jul 8 '15 at 15:49
  • $\begingroup$ Expanding on @anomaly's comment, knowing $a_{11}$ only fixes $a_{22}$ if one of the other two entries is zero; you can see this from the relation $a_{11}a_{22}-a_{21}a_{12}=1$. $\endgroup$ – Matt B Jul 8 '15 at 15:58
  • $\begingroup$ @anomaly- I did not mean to say that knowing $a_{11} $ fixes $a_{22} $. I'm saying that knowing three of the elements fixes the fourth one. $\endgroup$ – user67803 Jul 8 '15 at 16:43

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.

  • $\begingroup$ Brilliant! Thanks $\endgroup$ – user67803 Jul 8 '15 at 16:47

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 $|SL_2(\mathbb{F}_3)|=24$.

  • $\begingroup$ Thanks for the elaborate answer, Matt! $\endgroup$ – user67803 Jul 8 '15 at 16:48
  • 1
    $\begingroup$ You're quite welcome. This method also extends to $GL_n(\mathbb{F}_p)$ where the size becomes $(p^n-1)(p^n-p)(p^n-p^2)...(p^n-p^{n-1})$ which is a nice exercise! $\endgroup$ – Matt B Jul 8 '15 at 18:11

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