Let $\mathbb{F}_3$ be a field with three elements and let $V = \mathbb{F}_3^2$. Let $\alpha,\beta,\gamma$ and $\delta$ be the four one-dimensional subspaces spanned by $\begin{bmatrix}1\\0 \end{bmatrix},\begin{bmatrix}0\\1 \end{bmatrix},\begin{bmatrix}1\\1 \end{bmatrix}$ and $\begin{bmatrix}1\\-1 \end{bmatrix}$, respectively. Let $\operatorname{GL}_3 (\mathbb{F_3})$ act on $\{\alpha,\beta,\gamma,\delta\}$ by matrix multiplication.
1) Prove that the kernel of the homomorphism $\phi : G \to S_4$ corresponding to this action is $\{\pm I_2\}$.
2) Prove that $\phi$ is surjective.
Attempt: So I think that I showed what the kernel is correctly. Here is my argument: Suppose that $A\in \ker (\phi)$. Let $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$. Then applying $A$ to each of the 4 subspaces and insuring that these subspaces be fixed, we see that
$$ \begin{align*} A \cdot \alpha = \alpha \implies \left\{\begin{bmatrix}a\\c \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-a\\-c \end{bmatrix}\right\} &= \left\{\begin{bmatrix}1\\0 \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-1\\0 \end{bmatrix}\right\}\\ A \cdot \beta = \beta \implies \left\{\begin{bmatrix}b\\d \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-b\\-d \end{bmatrix}\right\} &= \left\{\begin{bmatrix}0\\1 \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}0\\-1 \end{bmatrix}\right\}\\ A \cdot \gamma = \gamma \implies \left\{\begin{bmatrix}a+b\\c+d \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-a-b\\-c-d \end{bmatrix}\right\} &= \left\{\begin{bmatrix}1\\1 \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-1\\-1 \end{bmatrix}\right\}\\ A \cdot \delta = \delta \implies \left\{\begin{bmatrix}a-b\\c-d \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-a+b\\-c+d \end{bmatrix}\right\} &= \left\{\begin{bmatrix}1\\-1 \end{bmatrix}, \begin{bmatrix}0\\0 \end{bmatrix}, \begin{bmatrix}-1\\1 \end{bmatrix}\right\}\\ \end{align*} $$
The first condition implies that $c=0$, $a=\pm 1$. The second condition implies that $b=0, d=\pm 1$. The third condition implies that $a=1,d=1$ or $a=-1,d=-1$, and this is also what the fourth condition implies. So we see that the only values that satisfy all four cases are $a=1,d=1,c=0,b=0$ or $a=-1,d=-1,c=0,b=0$. Hence, $\ker (\phi) = \{\pm I_2\}$. Is this argument sufficient? Also, I'm having a hard time showing that $\phi$ is surjective, and this is kind of where I get stuck.
Let me label $\alpha$, $\beta$, $\gamma$, and $\delta$ with numbers $1$, $2$, $3$, and $4$. Observe that the transposition $(1\;2)$ is given by the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}\in G$. You can get $(2\;3)$ by the matrix $\begin{pmatrix}-1&1\\0&1\end{pmatrix}\in G$. Also, $(3\;4)$ is given by $\begin{pmatrix}1&0\\0&-1\end{pmatrix}\in G$. Since $S_4$ is generated $(1\;2)$, $(2\;3)$, and $(3\;4)$, we are done. However, Dietrich Burde's hint seems to be the simplest way.