Let $G = \left\{ \begin{bmatrix} \alpha & \beta \\ 2\beta & \alpha \end{bmatrix} \;\Bigg| \; \alpha,\beta \in \mathbb Z_{3} (\alpha,\beta) \neq (0,0) \right\}$
Let $\mathbb F = \mathbb Z_{3}[x]/\langle x^{2}+1 \rangle$
Show that $\phi : G \to \mathbb F\backslash\{0\}$ defined by
$\phi\left(\begin{bmatrix} \alpha & \beta \\ 2\beta & \alpha \end{bmatrix}\right) = \alpha+\beta x$
defines a group isomorphism between $G$ and the multiplication group of field $\mathbb F$
So I have to show that $\phi$ is bijective, and that $\forall g_{1},g_{1}^{'} \in G : \phi(g_{1} *_{1} g_{1}^{'}) = \phi(g_{1}) *_{2} \phi(g_{1}^{'})$, correct?
Injective: I can see why this is true since for $\phi(g_{1}) = \alpha + \beta x$ and $\phi(g_{2}) = \gamma + \delta x$. Then $\alpha + \beta x = \gamma + \delta x$ only if $\alpha = \gamma$ and $\beta = \delta$, but how can I prove it?
Similiar for surjective. I think since it can be divided by a polynomial with degree 2 (ie. $x^{2}+1)$), it can also be divided by $\alpha + \beta x$, but I'm not sure on this and how to write this formally.
Also for the last condition I get: Let $g_{1}=\begin{bmatrix} \alpha & \beta \\ 2\beta & \alpha \end{bmatrix}$ and let $g_{2}=\begin{bmatrix} \gamma & \delta \\ 2\delta & \gamma \end{bmatrix}$. $\phi(g_{1}g_{2}) = \begin{bmatrix} \alpha & \beta \\ 2\beta & \alpha \end{bmatrix}\begin{bmatrix} \gamma & \delta \\ 2\delta & \gamma \end{bmatrix} = \begin{bmatrix} \alpha \gamma + 2\delta\beta & \alpha\delta+\beta\gamma \\ 2\beta\gamma+2\delta\alpha & 2\beta\delta+\alpha\gamma \end{bmatrix} \to \alpha\gamma + 2\delta\beta + (\alpha\delta+\beta\gamma)x$ $\phi(g_{1})\phi(g_{2}) = (\alpha+\beta x)(\gamma+\delta x) = \alpha\gamma + \alpha\delta x + \beta x \gamma + \beta\delta x^{2}$
But $\alpha\gamma + 2\delta\beta + (\alpha\delta+\beta\gamma)x \neq \alpha\gamma + \alpha\delta x + \beta x \gamma + \beta\delta x^{2}$, correct? I feel like I'm making mistakes here.
Hints:
$$\Bbb F=\{a+bw\;;\;a,b\in\Bbb F_3\;,\;\;w^2=-1\}\;\;\text{and operations modulo}\;\;3$$
This field is a vector space of dimension two voer $\;\Bbb F_3\;$, with basis $\;\{1,w\}\;$ , and from here the solution to your first question (injectivity + surjectivity)
As for the product: use that $\;w^2=-1\;$, though in your post you call $\;x=w\;$ , which can be a little confusing because the variable of the polynomials...