Determining the proper function for morphisms

$$G_{1}=\lbrace x+y\sqrt{3} \mid x,y \in \mathbb Q,x^2-3y^2=1 \rbrace $$ $$G_{2}=\left\lbrace \begin{pmatrix}x&3y\\ y&x\\\end{pmatrix}\mid x,y \in \mathbb Q,x^2-3y^2=1 \right\rbrace $$ $$(G_{1},\ast) \text{ stable part}$$ $$ a\ast b=a⋅b\text { (real number multiplication)}$$ $$(G_{2},\circ)\text{ stable part}$$ $$A\circ B=A⋅B\text{ (matrix multiplication)} $$ Prove that that G1 and G2 are isomorphic.My question is how to determine the proper function which satisfies the izomophism rule $$ f:G_{1}→G_{2}$$$$f(x\ast y)=f(x)\circ f(y)$$

Solution 1:

Here is an explanation of where those expressions come from.

$ K=\{x+y\sqrt3 : x,y \in \mathbb Q \} $ is a field.

$K$ is also a 2-dimensional vector space over $\mathbb Q$, with basis $\{1, \sqrt3\}$.

Given $a=x+y\sqrt3 \in K$, the map $z \mapsto az$ is a $\mathbb Q$-linear transformation of $K$ whose matrix with respect to the basis $\{1, \sqrt3\}$ is $$ M(a)=\begin{pmatrix}x&3y\\y&x\\\end{pmatrix} $$

The map $M: a\mapsto M(a)$ is an injective ring homomorphism $K \to \mathbb Q^{2\times 2}$.

The map $M$ induces an injective group homomorphism $K^{\times} \to {(\mathbb Q^{2\times 2})}^{\times}=GL(2,\mathbb Q)$.

In this context, $G_1$ is a subgroup of $K^{\times}$ and $G_2=M(G_1)$. So $f=M$ is a group isomorphism.

Note that $G_2 = \ker \det$ and $G_1 = \ker (\det \circ M)$.

Solution 2:

$$\begin{array}{l|rcl} \varphi : & G_1& \longrightarrow & G_2\\ & x+y\sqrt{3} & \longmapsto & \begin{pmatrix} x & 3y \\ y & x \end{pmatrix}\end{array}$$ should work.