Looking for help to understand example of Group I am looking for someone to help me to understand what is going on in the following example, from Hersteins "Topics in Algebra".
It says, 
Let $G$ be the set of all $2*2$ matrices $$\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}$$ where $a,b,c,d$ are integers modulo $p$, $p$ is a prime number such that $ad-bc \neq 0$, with matrix multiplication having its usual definition, then verify that $G$ is a non-abelian finite group.
I think I am just even confused on some definitions. From what I understood previously, to say $a\equiv b \quad mod \quad n$ is to say that $n | a-b$.
So what is it to mean the these entries are integers modulo $p$? Can anyone share some insight about this? Thank you.
 A: $G$ is group because if $x,y \in G$ then $x^{-1} y \in G$ and $G$ is nonabelian because:
$\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix} *\begin{pmatrix} 0 & 1 \\ 1 & 1 \\ \end{pmatrix} \neq \begin{pmatrix} 0 & 1 \\ 1 &  1\\ \end{pmatrix} *\begin{pmatrix} 1 & 0 \\ 1 & 1 \\ \end{pmatrix}$
And at last $G$ is finite because $|G|= (p^2-1)(p^2-p)$
Remarks: addition and multiplication in $G$ is like this:
For example if $p=7$ then $4*5 + 3^2=6+2=1$ ! because $4*5 \equiv 6 \mod {7}$ and $3^2=9 \equiv 2 \mod {7}$ and then $6+2=8 \equiv 1 \mod {7}$ hence:
$\begin{pmatrix} 4 & 3 \\ 0 & 1 \\ \end{pmatrix} *\begin{pmatrix} 5 & 1 \\ 3 & 0 \\ \end{pmatrix}= \begin{pmatrix} 1 & 4 \\ 3 & 0 \\ \end{pmatrix}$
A: The integers come from $0, 1, \dots, p-1$ and addition and multiplication are done $\mod{p}$.
A: Normally, when one talks about the integers mod $p$, what one does is restrict oneself to the representatives of the equivalence classes that lie in: $\{0,1,2,\dots,p-1\}$.  One see the equivalence class of $a$ modulo $p$ denoted variously as: $\overline{a}$, $[a]_p$, or even (via a common abuse of notation) just $a$ itself (one has to then remember that $a$ (mod $p$) is not the same thing as the integer $a$).
Note that for any integer $a$, and any prime $p$, we can write:
$a = qp + r$ for a unique pair of integers $q,r$ with $0 \leq r < p$.
Since $a - r = qp$, ir follows that $a \equiv r$ (mod $p$). This justifies only considering the representatives of the equivalence classes  in my first paragraph, since these account for all of them.
So, for your purposes, it suffices to consider $a,b,c,d \in \{0,1,2,\dots,p-1\}$, writing $\overline{a},\overline{b},\overline{c},\overline{d}$ for the actual matrix entries (addition and multiplication are "different" for the integers modulo $p$, then in the "ordinary" integers- for example, $\overline{p-1} + \overline{1} = \overline{0}$, which certainly never happens for any positive prime integer).
The important thing to show, in this example, is that we have closure, to which end you might consider the properties of determinants (which still work in the integers modulo $p$, since these form a field-something you might not yet have covered at this point in Herstein).
