The number of elements in the special linear group over the finite field $\mathbb{Z}/p$

Question

I have $SL_{2}\{\mathbb{Z}/p\}$ for $p$ prime and $\mathbb{Z}$ integers. How do I show that this is a subgroup of $GL_{2}\{\mathbb{Z}/p\}$ and find the number of elements in $SL_{2}\{\mathbb{Z}/p\}$?

Answer 1

Let $\mathbf{F}_p$ denote the field with $\mathbf{Z}/p$. Note that there is a homomorphism

$$ \textrm{det}: \textrm{GL}_n (\mathbf{F}_p) \to \mathbf{F}_p ^* $$ given by taking the determinant of the matrix. Note that the kernel of the homomorphism is exactly $\textrm{SL}_n (\mathbf{F}_p)$. Furthermore, this map is surjective. Therefore, we can use the First Isomorphism Theorem to count the number of elements in the special linear group.

$$|\textrm{SL}_n (\mathbf{F}_p) |= \frac{|\textrm{GL}_n (\mathbf{F}_p)|}{|\mathbf{F}_p ^*|} $$

The formula for the number of elements of $\textrm{GL}_n$ can easily be deduced by first counting it in the $n = 2$ case, which incidentally, is the case that you're interested in.

How can we create a $2 \times 2$ matrix that has determinant nonzero? Well, that's equivalent to being invertible, which is equivalent to the columns being linearly independent. The first column is free; anything nonzero will work, which gives $p^2-1$ choices, since there are $p$ choices for each of the two entries. The next column needs to be chosen outside of the span of the first columm, which has $p$ elements. Therefore, we have $p^2 - p$ choices for the second column.