# 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.

• What about $\varphi(a +b \sqrt2) = \begin{bmatrix} a & 2b \\ b & a \end{bmatrix}$? Sep 15, 2020 at 13:10
• I was thinking of that, but not sure how to mathematically write a good proof Sep 15, 2020 at 13:12
• Try to prove that $\varphi$ satisfies isomorphism properties. Sep 15, 2020 at 13:13
• I shall try it. Thanks. Sep 15, 2020 at 13:14
• For $G$ and $H$ to be isomorphic means that there exists an isomorphism $\varphi : G \to H$. To prove that there exists an isomorphism $\varphi : G \to H$ there are two steps. First, using your mathematical imagination and experience, write down an appropriate formula for $\varphi : G \to H$; this may require intuition, exploration, experimentaiton, digging around, etc. Second, once you think you have the appropriate formula for $\varphi : G \to H$, prove that it satisfies the definition of an isomorphism. Sep 15, 2020 at 13:23

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$$\psi\left(a+b\sqrt2\right)=\begin{bmatrix}a&2b\\b&a\end{bmatrix}.$$

• I put the bit about a general irreducible polynomial and its companion matrix. Sep 15, 2020 at 14:00

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 $$x^2 + 1,$$

$$M = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$$

$$\{1, \sqrt{2}\}$$ 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 $$\alpha \beta$$.