How to prove that two groups $G$ and $H$ are isomorphic? Let $G = \big\{a + b\sqrt2 \mid a,b \in\mathbb{Q}\big\}$.
Let $H = \bigg\{\begin{bmatrix} a & 2b \\ b & a \end{bmatrix} \biggm |a,b \in\mathbb{Q}\bigg\} $
And denote $0_{2\times 2} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
I have proven that both $G$ and $H$ are abelian/commutative because after some computations we have $G1 + G2 = G2 + G1$ and $H1 + H2 = H2+ H1$.
Now I have to show that $G$ and $H$ are isomorphic. I know that an isomorphism from $G1$ to $G2$ is a bijective homomorphism. We call $G1$ and $G2$ isomorphic, and write $G1 \cong G2$ if an isomorphism from $G1$ to $G2$ exists.
I am struggling how to construct such a proof.
Thanks in advance.
 A: Since$$\left(a+b\sqrt2\right)\left(c+d\sqrt2\right)=\color{red}{ac+2bd}+(\color{blue}{ad+bc})\sqrt2$$and since$$\begin{bmatrix}a&2b\\b&a\end{bmatrix}.\begin{bmatrix}c&2d\\d&c\end{bmatrix}=\begin{bmatrix}\color{red}{ac+2bd}&2(\color{blue}{ad+bc})\\\color{blue}{ad+bc}&\color{red}{ac+2bd}\end{bmatrix},$$simply take$$\psi\left(a+b\sqrt2\right)=\begin{bmatrix}a&2b\\b&a\end{bmatrix}.$$
A: If you begin with the rationals and some monic polynomial $f(x)$ of degree $n$ is irreducible, then we get two fields. One is
$$   \mathbb Q [x] / (f(x))  $$
The other: take the companion matrix (or its transpose)  $M.$ By Cayley-Hamilton, $f(M) = 0$ as matrices.  We get a ring from all matrices of the form
$$ a_0 I + a_1 M + a_2 M^2 +  \cdots + a_{n-1} M^{n-1}   $$
Plus, any polynomial expression (arbitrary degree) in $M$ is equal to such an expression. This ring of matrices is also a field. They are isomorphic as fields.
Your polynomial is  $x^2 - 2$ and its companion matrix is
$$
M = \left(
\begin{array}{cc}
 0 & 2 \\
1 & 0 \\
\end{array}
\right)
$$
This is exactly the construction that gives the complex numbers (rational coefficients), polynomial $x^2 + 1,$
$$
M = \left(
\begin{array}{cc}
 0 & -1 \\
1 & 0 \\
\end{array}
\right)
$$
A: $\{1, \sqrt{2}\}$ is a basis for the space $\{a + \sqrt{2} b \mid a,b \in \mathbb Q\}$.
We can represent multiplication by $a + \sqrt{2} b$ as a matrix by noting how it acts on the basis vectors.

*

*$(a + \sqrt{2} b) \cdot 1 = a + \sqrt{2} b$

*$(a + \sqrt{2} b) \cdot \sqrt{2} = 2 b + \sqrt{2} a$
so $1$ gets mapped to $(a,b)$ and $\sqrt{2}$ gets mapped to $(2b,a)$.
So we can tabulate this into a matrix $$M = \left(
\begin{array}{cc}
a & 2b \\
b & a \\
\end{array}
\right)$$
and indeed

*

*$M(1,0) = (a,b)$

*$M(0,1) = (2b,a)$
Secondly if $M$ represents $\alpha$ and $N$ represents $\beta$ it can be seen that the matrix product $MN$ represents $\alpha \beta$.
