How many elements are in $\mathrm{SL}_3(\mathbb{F}_q)$?

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In summary, $\mathrm{SL}_3(\mathbb{F}_q)$ is a special linear group of degree 3 over a finite field $\mathbb{F}_q$. It has $q^3(q^3-1)(q^3-q)(q^3-q^2)$ elements, and is an important group in mathematics, with connections to other structures. The elements can also be represented using polynomial equations in the Chevalley basis. Its size is larger than $\mathrm{SL}_2(\mathbb{F}_q)$, but smaller than $\mathrm{SL}_4(\mathbb{F}_q)$.
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Chris L T521
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Here's this week's problem.

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Problem: Compute the cardinality of $\mathrm{SL}_3(\mathbb{F}_q)$, where $\mathbb{F}_q$ is a field of cardinality $q$.

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This week's problem was correctly answered by Deveno. You can find his solution below.

I will choose to solve a somewhat easier problem: the cardinality of \(\displaystyle \ \text{GL}_3(\Bbb F_q)\).

Since these are 3x3 matrices with non-zero determinant, their columns must be linearly independent. We can choose any non-zero vector in \(\displaystyle \Bbb (F_q)^3\) as our first column, there are \(\displaystyle q^3 - 1\) of these. For our second column, we can pick any non-zero vector that is not a scalar multiple of the first. This gives: \(\displaystyle q^3 -1 - (q - 1) = q^3 - q\) choices for the second column, as there are \(\displaystyle q - 1\) non-zero scalars to choose from. Finally, there are \(\displaystyle q^2 - 1\) non-zero linear combinations of the first two chosen vectors, which we must exclude from our choice for the 3rd column, leaving: \(\displaystyle q^3 - 1 - (q^2 - 1) = q^3 - q^2\) choices for column 3. Thus:

\(\displaystyle |\text{GL}_3(\Bbb F_q)| = (q^3-1)(q^3-q)(q^3-q^2)\)

Now we have the short exact sequence:

\(\displaystyle 0 \to \text{SL}_3(\Bbb F_q) \to \text{GL}_3(\Bbb F_q) \xrightarrow{det} (\Bbb F_q)^{\ast} \to 1\)

Which tells us that:

\(\displaystyle |\text{SL}_3(\Bbb F_q)| = \frac{|\text{GL}_3(\Bbb F_q)|}{|(\Bbb F_q)^{\ast}|} = \frac{(q^3-1)(q^3-q)(q^3-q^2)}{q-1} = q^3(q^2+q+1)(q-1)^2(q+1)\)CLT Note: I just left my answer as $q^2(q^3-1)(q^3-q)$, but it doesn't really matter.
 

FAQ: How many elements are in $\mathrm{SL}_3(\mathbb{F}_q)$?

1. What is $\mathrm{SL}_3(\mathbb{F}_q)$?

$\mathrm{SL}_3(\mathbb{F}_q)$ is a special linear group of degree 3 over a finite field $\mathbb{F}_q$. It is a group of invertible 3x3 matrices with entries from $\mathbb{F}_q$ that have a determinant of 1.

2. How many elements are in $\mathrm{SL}_3(\mathbb{F}_q)$?

The number of elements in $\mathrm{SL}_3(\mathbb{F}_q)$ is equal to $q^3(q^3-1)(q^3-q)(q^3-q^2)$, where $q$ is the order of the finite field.

3. What is the significance of the special linear group $\mathrm{SL}_3(\mathbb{F}_q)$?

$\mathrm{SL}_3(\mathbb{F}_q)$ is an important group in mathematics, specifically in the fields of algebra and number theory. It has connections to other mathematical structures, such as Lie algebras and algebraic geometry.

4. Can the elements of $\mathrm{SL}_3(\mathbb{F}_q)$ be represented in a different form?

Yes, the elements of $\mathrm{SL}_3(\mathbb{F}_q)$ can also be represented using polynomial equations over the finite field $\mathbb{F}_q$. This form is known as the Chevalley basis.

5. How does the size of $\mathrm{SL}_3(\mathbb{F}_q)$ compare to other special linear groups?

The size of $\mathrm{SL}_3(\mathbb{F}_q)$ is larger than the size of $\mathrm{SL}_2(\mathbb{F}_q)$, but smaller than the size of $\mathrm{SL}_4(\mathbb{F}_q)$. In general, the size of $\mathrm{SL}_n(\mathbb{F}_q)$ is equal to the number of ways to choose $n$ linearly independent vectors from a vector space of dimension $q^n$.

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