Matrices with entries from $\mathbb{Z}_2$ Let $G$ be the set of all $\ 2\times 2$ matrices $\begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$ where $a,b,c,d$ are integers modulo $2$, such that $ad-bc\neq 0$. Using matrix multiplication as the operation in $G$, prove that $G$ is a group of order $6$.
Proof: We know that $\mathbb{Z}_2=\{[0],[1]\}$ where $[0], \ [1]$ are equivalence classes.
Let $A=\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix}\in G$ and $B=\begin{pmatrix}
a_2 & b_2 \\
c_2 & d_2
\end{pmatrix}\in G$ then $$AB=\begin{pmatrix}
a_1a_2+b_1c_2 & a_1b_2+b_1d_2 \\
a_2c_1+d_1c_2 & b_2c_1+d_1d_2
\end{pmatrix}$$ since $a_i,b_i,c_i,d_i\in \mathbb{Z}_2$ then entries of $AB$ also belong to $\mathbb{Z}_2$. Also $\text{det}(AB)=\text{det}(A)\text{det}(B)\neq 0$ Right?
So matrix multiplication is binary operation. Associativity holds for any matrices and it is true also in this case. The identity matrix $E\in G$
But when i was checking inverse matrix I stucked. If $A=\begin{pmatrix}
a_1 & b_1 \\
c_1 & d_1
\end{pmatrix}\in G$ then $$A^{-1}=\dfrac{1}{ad-bc}\begin{pmatrix}
d & -b \\
-c & a
\end{pmatrix}$$
Why this matrix lies in $G$? The first entry is $\dfrac{d}{ad-bc}$ and it need to be an integers. 
Can anyone please explain this moment to me, please?
 A: We are told that $a,b,c$ and $d$ are integers modulo $2$. What does that mean? It means that they're not elements of $\Bbb Z$, but of the group (actually a field) $\Bbb Z/2\Bbb Z=\{[0],[1]\}$ (also called $\Bbb Z_2$ for short). The operation $+$ in that group is defined by $[0]+[0]=[1]+[1]=[0]$ and $[0]+[1]=[1]+[0]=[1]$.
Thus, when we compute $ad-bc$, we obtain, not an ordinary integer, but either $[0]$ or $[1]$. Since we're excluding the case $ad-bc=[0]$, then we have $ad-bc=[1]$.
Now, we ask, what is meant by $\frac{1}{ad-bc}$? If this expression is to be meaningful, its value must be the unique solution to the equation $(ad-bc)x=[1]$; i.e., we're looking for a multiplicative inverse. Since we have $[1]\cdot[1]=[1]$, we can "divide", and obtain $\frac{[1]}{[1]}=[1]$. A less confusing notation might be: $[1]^{-1}=[1]$.
The omission of brackets is common, and unfortunately confusing. :/
This interpretation is necessary in order for the problem to make sense. If we allowed for a matrix whose determinant is $2$ (ignoring for the moment that $2$ is not, itself, an element of $\Bbb Z_2$), then we would not end up with a group at all.
A: If $ad-bc\ne0$ in $\mathbf F_2$, then necessarily $ad-bc=1$. Furthermore, we have a group because $G$ is actually  defined by the condition $\det A=1$, and the determinant is multiplicative (over any field), i.e. $\det (AB)=\det A\cdot\det B$.
