Order of a subgroup of 2x2 matrices Take $H$ to be the subgroup of $G$, a group of 2x2 integers mod $p$
matrices ($p$ is a prime number) with nonzero determinants and the
operation of matrix multiplication,
such that for $h \in H, det(h) = 1.$ I want to figure out the
order of this subgroup. However, I'm having issues making inferences
about this subgroup. I understand that it might be good to do this
proof combinatorially, i.e. given
$$h = \left(\begin{array}{cc} a&b \\ c&d\end{array}\right),$$
we know that $det(h) = ad-bc = 1.$ We know that there are at
most $p$ options for $a,b,c,$ and $d$ given integers modulo
$p$, but I'm having issues figuring out the number of
equivalence class combinations that I can assign to 
$a,b,c,$ and $d$ such that $ad-bc = 1.$ Any suggestions would be appreciated.
 A: First let us count the non-singular matrices. There are $p^2-1$ choices for the first row. Whatever the first row was, we need to avoid the $p$ multiples of that row, so we have $p^2-p$ choices for the second row, for a total of $(p^2-1)(p^2-p)$. 
Whatever the determinant is, we can make it $1$  by dividing (modulo $p$) the second row by one of $1$, $2$, and so on up to $p-1$.
So the number of matrices with determinant $1$ is $\frac{(p^2-1)(p^2-p)}{p-1}$. 
A: Let $F$ be a finite field with $p$ elements and $GL_n(F)$ denote the group of all $n \times n$ nonsingular matrices over $F$, and  $SL_n(F)$ denote the subgroup of $GL_n(F)$ consisting of matrices with determinant $1$.
 The determinant clearly induces a homomorphism from $GL_n(F)$ onto the multiplicative group $F^*$, which has $p —1$ elements. The kernel of the homomorphism is $SL_n(F)$, and the cosets with respect to this kernel are the elements of $GL_n(F)$ which have the same determinant. Since all cosets of a group must have the same order, it follows that the order of $SL_n(F)$, is: $\frac{|GL_n(F)|}{p-1}$.
Now we want to find: $|GL_n(F)|$. A matrix $A$ is nonsingular if and only if its rows  are linearly independent vectors in $F^n$. Therefore, the first row $A_1$ can be any nonzero vector in $F^n$, so there are $p^n — 1$ possibilities. Once the first row is chosen, the second row, $A_2$, can be any vector which is not a multiple of the first row, that is, $A_2\neq cA_1$, where $c\in F$, leaving $p^n —p$ choices for $A_2$. In general, the row $A_i$, can be any vector which cannot be written in the form:
$$c_1A_1 + c_2A_2 + ... + c_{i-1}A_{i-1}$$ 
where $c_j \in F$ for $j=1,2,...,i-1$. Hence, there are $p^n-p^{i-1}$ possibilities for $A_i$. By multiplying these together we see that: $|GL_n(F)|=\Pi_{k=0}^{n-1}(p^n-p^k)$.
