I have $GL_2({\Bbb{Z}}_2)$ in which ${\Bbb{Z}}_2$ consists of the integers $\{0,1\}$.
We observe that $|M_2({\Bbb{Z}}_2)|=2^4$.
Now let's define Y to be the set of all non-zero elements of ${\Bbb{Z}}_2\times {\Bbb{Z}}_2$
$Y = [y_1,y_2,y_3]$
Where:
$$ y1= \begin{pmatrix} 1\\ 0\\ \end{pmatrix} $$
$$ y2 = \begin{pmatrix} 0\\ 1\\ \end{pmatrix} $$
$$ y3= \begin{pmatrix} 1\\ 1\\ \end{pmatrix} $$
Observe that $GL_2({\Bbb{Z}}_2)$ acts on Y via the function $GL_2({\Bbb{Z}}_2)\times Y\to Y$ given by $(C,y)\mapsto C_y$ where $C_y$ denotes the products of the matrices.
Consider $\gamma:GL_2({\Bbb{Z}}_2)\to S_3$ corresponding to this action.
Here, $Y\mapsto\{1,2,3\}$ by $y\mapsto i$, we identify $S_y$ with the symmetric group $S_3$
To start, I found the elements of $GL_2({\Bbb{Z}}_2)$ which contains $6$ elements which are $2\times 2$ matrices with determinants of non-zero.
I am then told to write down for each $C\in GL_2({\Bbb{Z}}_2)$, the element of $\gamma(C)$ of $S_3$ in which I have to write down as cycle notation.
How am I supposed to do this?
And then next, I am to explain why $\gamma$ is an isomorphism from the group $GL_2({\Bbb{Z}}_2)$ onto the group $S_3$.
My explanation for this is that there exists two elements of the group in which it shows non-abelian and we know that $S_3$ is non-abelian, therefore isomorphism.
I feel as is if my explanation isn't enough, could someone also elaborate? Or am I right?
Thank you!