# Thread: Order of a group

1. ## Order of a group

Let $\displaystyle \mathbb F$ be a ﬁeld. Prove that $\displaystyle SL_2 (\mathbb F)$ be the group of 2 x 2 matrices with coefficient in $\displaystyle \mathbb F$ and determinant equal to 1.
Question : If F has q elements, how many elements are in the group $\displaystyle SL_2 (\mathbb F)$?
I know that $\displaystyle |M_2(\mathbb F)| = q^4$.
$\displaystyle SL_2 (\mathbb F)$ is a subgroup of $\displaystyle GL_2 (\mathbb F)$ the group of units of $\displaystyle M_2(\mathbb F)$.
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 $\displaystyle \mathbb F$ be a ﬁeld. Prove that $\displaystyle SL_2 (\mathbb F)$ be the group of 2 x 2 matrices with coefficient in $\displaystyle \mathbb F$ and determinant equal to 1.
Question : If F has q elements, how many elements are in the group $\displaystyle SL_2 (\mathbb F)$?
Hint: $\displaystyle \text{GL}_2(\mathbb{F})/\text{SL}_2(\mathbb{F}) \simeq \mathbb{F}^{\times}$

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

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

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