Linear Algebra problem, old Berkeley exam. I came across this problem:

Let $G$ be the group of $2 \times 2$ matrices with determinant $1$ over the four-element field $\bf F$. Let $S$ be the set of lines through the origin in $\bf F^2$. Show that $G$ acts faithfully on $S$. (The action is faithful if the only element of $G$ wich fixes every element of $S$ is the identity.)

I solved it (not really sure about the correctness of the proof) without using any abstract algebra.The only "official" solution i found to this problem uses some abstract algebra and is more elegant.
Nonetheless I tried my best and would like to have some feeback (is my proof correct?) on my proof. I'm really thankful for your help. Here is my solution: 

Solution: Suppose $G$ acts not faithfully on $S$.Then $\exists g \neq id \in G : gs=s ,   \ \forall s\in S$. We saw the multip. table of $\bf F$ in the lecture. I want to identify only $2$ elements of $S$. Those will suffice for our proof. Each line in $ \bf F$ is determinated by it's slope wich can be $0,1,a,b,$ or $\infty$. We consider the lines  $S_1$ and $S_a$ with slopes $1$ and $a$ respectivly.

Define $g$ as $g:=\begin{pmatrix} g_1 & g_2  \\g_3 & g_4   \end{pmatrix}$ with $g_{1,2,3,4} \in \bf F$. It's easy to verify that
$$
S_a=\left \{(0,0),(1,a)\right \} \text{ and } S_1=\left \{ (0,0),(1,1),(a,a),(b,b) \right \}.
$$
Now the idea is to apply $g$ on $(1,1)$ and $(1,a)$ and try to find $g_{1,2,3,4}$ wich fit with the conditions on $g$. Applying $g$ on $(1,1)$ we get the condition:
$$g_1+g_2=1.\tag{1}$$
While applying $g$ on $(1,a)$ we get:
\begin{align}
g_1+g_2a&=1,\tag{2}\\
g_3+g_4a&=a.\tag{3}
\end{align}
Combining $(1)$ and $(2)$, we get $g_2a=g_2$, hence we have two possibilities. Either $a=1$ (wich cannot be, checking the mult. table) or $g_2=0$. But we know another important thing about $g_{1,2,3,4}$, namely $g_1g_4-g_2g_3=1$ since the determinant of $g$ is $1$. Knowing that $g_2=0$ this equation becomes $g_1g_4=1$. Now plugging $g_2=0$ in $(1)$ we get $g_1=1$ hence from the last equation $g_4=1$. Now,finally, considering $(3)$ and the fact that $g_4=1$ we get $g_3+a=a \Rightarrow  g_3=0$ hence $g=id$, contradiction!

Hope the works and the proof is not too confusing.
Besides the feedback (wich is actually what i need), alternative solutions are of course also welcome!
 A: The problem with your proof is that you assumed that 
$$g (1,a) = (1,a)$$
and $$g(1,1) = (1,1)$$
But the matrices act on lines, so when you have, for example, the line $s=\{ (1,1) ,(0,0) , (a,a) , (b,b) \}$ , then $gs=s$ only means that $\{ g(1,1) ,g(0,0) , g(a,a) , g(b,b) \} = \{ (1,1) ,(0,0) , (a,a) , (b,b) \}$. i.e. $g(1,1)$ could be any one of the four elements of $s$. That will make everything a bit more complicated and maybe you'd have to consider several cases, but if you're willing to solve even more linear equations, something similar to your attempt could work to solve the problem. 
A: Apart from Bursko651's critique, I also want to point out that $S_a\ne\left \{(0,0),(1,a)\right \}$. The line should contain four elements: $(0,0),\,(1,a),\,(a,b),\,(b,1)$.
Also, you said that your proof hasn't used any abstract algebra. Yet there is no way to prove the statement without using abstract algebra. Suppose I claim that the problem statement is false, and a "counterexample" is given by the multiplicative group $G$ generated by $M=aI$ for some $a\ne0,1$. Note that $M$ does fix every line in $S$. So, unless one can explain why $\det(M)=a^2\ne1$, one cannot refute this "counterexample". Yet to show that $a^2\ne1$, we must know something about the underlying field. (Edit: As Marc van Leeuwen comments, you know the multiplication table of $\mathbf{F}$ and you have used it in your proof. So, yes, one can prove the problem statement without using abstract algebra directly.)
At any rate, suppose $g\in G$ fixes every line in $S$ and $\det(g)=1$. Then $g\pmatrix{1\\ 0}=\pmatrix{x\\ 0}$ and $g\pmatrix{0\\ 1}=\pmatrix{0\\ y}$ for some $x,y\ne0$. That is, $g=\pmatrix{x&0\\ 0&y}$ with $xy=\det(g)=1$. Now, the field $\mathbf{F}$ of four elements is unique up to isomorphism. Its four elements are $\{0,1,a,b\}$ with $a^2=b,\ b^2=a$ and $ab=1$. It follows that $(x,y)=(1,1),(a,b)$ or $(b,a)$ (and this is where we use abstract algebra). However, the latter two cases are impossible, or else $g$ will not fix the line spanned by $(1,1)^\top$. Hence $g$ must be equal to $I$. QED
A: I don't know which "abstract algebra" solution was given, but here is one that can easily be made concrete, thus shedding the "abstract" part. If a $n\times n$ matrix fixes every line through the origin, this means that every nonzero vector is an eigenvector. If two eigenvectors $v,w$ are such that $v+w$ is also an eigenvector, or$~0$, then $v,w$ must be eigenvectors for the same eigenvalue, which is proved by a direct calculation. So all nonzero vectors are eigenvectors can only happen if there is only one eigenvalue, and our matrix is a multiple of the identity. This part is "abstract" only because I mentioned nothing concrete.
In $GL(2,F)$ one has $\det(\lambda I_2)=\lambda^2$, and in $\Bbb F_4$ the equation $\lambda^2=1$ has $\lambda=1$ as unique solution. You can prove this "abstractly" by using $X^2-1-(X-1)^2$ in characteristic$~2$, or you can just try all $\lambda$.
To make the first part really concrete, just take two vectors that must be eigenvectors, $v=(1,0)$ and $w=(0,1)$ seems the best choice; then so are $(x,0)$ and $(0,y)$ for all nonzero $x,y$, and if every $(x,y)$ is to be an eigenvector, conclude that $v,w$ are eigenvectors for the same eigenvalue$~\lambda$.
In summary, being abstract is just being lazy, when the details don't matter.
