# Thread: Order of a group

1. ## Order of a group

Let $\mathbb F$ be a ﬁeld. Prove that $SL_2 (\mathbb F)$ be the group of 2 x 2 matrices with coefficient in $\mathbb F$ and determinant equal to 1.
Question : If F has q elements, how many elements are in the group $SL_2 (\mathbb F)$?
I know that $|M_2(\mathbb F)| = q^4$.
$SL_2 (\mathbb F)$ is a subgroup of $GL_2 (\mathbb F)$ the group of units of $M_2(\mathbb F)$.
From this I've tried to solve the problem but I'm not able to find a way to count the elements in any of the groups.

2. Originally Posted by vincisonfire
Let $\mathbb F$ be a ﬁeld. Prove that $SL_2 (\mathbb F)$ be the group of 2 x 2 matrices with coefficient in $\mathbb F$ and determinant equal to 1.
Question : If F has q elements, how many elements are in the group $SL_2 (\mathbb F)$?
Hint: $\text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F}) \simeq \mathbb{F}^{\times}$

3. $|\text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F})| = |\mathbb{F}^{\times}| = q-1$ because any element except 0 has a multiplicative inverse in $\mathbb F$.
In other word, there are $q-1$ equivalent classes (or orbits) of $\text{SL}_2(\mathbb{F})$.
We thus need to find the number of elements in each orbit.
In an equivalent class we have
$A=\left( \begin{array}{cc}q&a\\0&q^{-1} \end{array} \right)$ $A \in \text{SL}_2(\mathbb{F})$
There are $\frac{q}{2}$combinations of $q \text{ and } q^{-1}$.
$2q$combinations of $a$because we can swap $a \text{ and } 0$.
There are therefore $q^2$elements in its equivalent classes.
$|\text{SL}_2(\mathbb{F})| = q^2(q-1) = q^3-q$
Am I right?

4. Originally Posted by vincisonfire
$|\text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F})| = |\mathbb{F}^{\times}| = q-1$ because any element except 0 has a multiplicative inverse in $\mathbb F$.
In other word, there are $q-1$ equivalent classes (or orbits) of $\text{SL}_2(\mathbb{F})$.
We thus need to find the number of elements in each orbit.
In an equivalent class we have
$A=\left( \begin{array}{cc}q&a\\0&q^{-1} \end{array} \right)$ $A \in \text{SL}_2(\mathbb{F})$
There are $\frac{q}{2}$combinations of $q \text{ and } q^{-1}$.
$2q$combinations of $a$because we can swap $a \text{ and } 0$.
There are therefore $q^2$elements in its equivalent classes.
$|\text{SL}_2(\mathbb{F})| = q^2(q-1) = q^3-q$
Am I right?
$| \text{GL}_2(\mathbb{F}) | = (q^2 - 1)(q^2 - q)$

Therefore, $\text{SL}_2(\mathbb{F}) = \frac{\text{GL}_2(\mathbb{F})}{q-1} = \frac{(q^2-1)(q^2 - q)}{q-1} = (q-1)q(q+1)$