Problem from Herstein (Group Theory) This is the problem from Topics in Algebra by I. N. Herstein.
Part of Example No. 2.2.9: 
Let $G$ be the set of all $2 \times 2$ matrices 
$ \left( {\begin{array}{cc}
   a & b \\
   c & d \\
  \end{array} } \right) $
where $a, b, c, d$ are integers modulo $p$, ($p$ a prime number), such that $ ad-bc \neq 0$. Under matrix multiplication, prove that $G$ is a non-abelian finite group for any general $p$. All the multiplication and additions of the entries are those with modulo $p$.
Till now, I am solving it for small values of $p$, writing down elements explicitly as suggested in the example itself. I would like to know proof for any general $p$.  
 A: For general $p$ prime, the group in question is $GL_{2}(\mathbb{Z}/p)$, the general linear group of degree $2$ over the field $\mathbb{Z}/p$, the ring of integers modulo $p$. For a matrix $A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, $ad-bc = \text{det}(A)$. For two matrices $A,B \in G$, we have that $\text{det}(AB)=\text{det}(A)\text{det}(B) \neq 0$, because the unit group of $\mathbb{Z}/p$ consists of all non-zero elements, i.e $\text{det}:G \rightarrow \mathbb{Z}/p^{\times}$ is a group homorphism. So $G$ is closed under matrix multiplication. The normal formula for the inverse of a $2$x$2$ matrix holds, the identity element is $I_{2}$, and associativity holds because it holds in $\mathbb{Z}/p$. So $G=GL_{2}(\mathbb{Z}/p)$ is a group.
$G$ is nonabelian for any $p$ prime as $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, but $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$ 
The finiteness of $G$ is clear because the total number of possible $2$x$2$ matrices over $\mathbb{Z}/p$ is $p^{4}$, and $G$ is a subset of this. $\square$
A: For an elementary method that simply involve multiplying out everything...
To check for closure, simply multiplying out everything.
Associative can once again be checked by multiplying out everything.
The identity is $\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}$
The inverse is $(ad-bc)^{p-2}\begin{pmatrix}d & -b\\-c & a\end{pmatrix}$ (get this from normal rule for 2x2 matrix inverse, and the inverse of determinant is just from FLT).
To prove that it is nonabelian, consider $\begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}$ and $\begin{pmatrix}1 & 0\\1 & 1\end{pmatrix}$ which is guaranteed to work for any $p$.
Finiteness is from the fact that each component is finite.
A: lemma: if $gcd(a,n)=1$ then $a^{-1}$ exist in $\mathbb Z_n$.
Let $A\in G$ then $A^{-1}=(det(A))^{-1}Adj(A)$, In that case, we need to check whether $det(A)^{-1}$ exist in $\mathbb Z_p$.
Since $p\nmid det(A)$ then $gcd(p,det(A))=1$. 
Since $det(AB)=det(A)det(B)$ which shows that if $A,B\in G$ then $AB\in G$.
We are done.
Finiteness is obvious as  number of the all possible matrices is $p^4$ so this number is less than $p^4$.
I left you to show that it is nonabelian.
