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Given a group $(\mathbb{R^2\setminus (0,0),\cdot})$ where $\cdot$ is defined as $$(x,y)\cdot(u,v)=(xu-2yv,xv+yu)$$ Find neutral and inverse element.

$$(e_1,e_2)\cdot (x,y)=(x,y)\cdot (e_1,e_2)=(x,y), \forall(x,y)\in \mathbb{R^2}$$ $$(e_1,e_2)\cdot (x,y)=(e_1x-2e_2y,e_1y+e_2x)=(x,y)$$ $$e_1x-2e_2y=x$$ $$e_1y+e_2x=y$$

How to find $e_1$ and $e_2$?

user300045
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    Are you sure that it should be a group? The pair $(0,0)$ has no inverse since $(x,y) \dot (0,0) = (0,0)$. In fact, this structure is isomorphic to the multiplicative monoid of $\mathbb{R}[X]/(X^2+2)$, which fails to be a group due to 0. – lisyarus Sep 08 '15 at 22:55
  • Ok, I have forgotten: $(\mathbb{R^2}\setminus (0,0),\cdot)$. Element $(0,0)$ is excluded. Could you explain this without isomorphism notation? – user300045 Sep 08 '15 at 23:02
  • I'll give an answer then. – lisyarus Sep 08 '15 at 23:05
  • This is just a pair of (non-degenerate) simultaneous equations! (Remember that $e_1,e_2$ are just real numbers.) In particular, if a solution exists it is unique. If you need to you can just solve these as with any other pair of simultaneous equations, but you should be able to spot a solution. – Mathmo123 Sep 08 '15 at 23:12

3 Answers3

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Let $G$ denote this magma (i.e. the set $\mathbb R^2 \backslash \{(0,0)\}$ with this operation). The map $(u,v) \cdot - : G \to G$ is simply defined using a linear map! $$ \begin{bmatrix} x \\ y \end{bmatrix} \mapsto \begin{bmatrix} xu - 2yv \\ xv + yu \end{bmatrix} = \begin{bmatrix} u & -2v \\ v & u \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$ So now that you look at it this way...

First : which matrix corresponds as the identity map on $\mathbb R^2 \backslash \{(0,0)\}$?

Second : Given $(u,v)$ and the above map (which is linear), what is its inverse? Does it have the same form, e.g. the form $(u',v') \cdot -$ ?

Hope that helps,

Patrick Da Silva
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Hint: Compare with the construction of $\mathbf C$ as the set $\mathbf R^2$ with addition componentwise, and multiplication that looks odd at first sight..

Bernard
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Since $(e_1, e_2) \cdot (x,y) = (x,y)$ should work for any $x,y \in \mathbb{R}$, we can pick some usefull values for them, like $0$ or $1$.

Let $(x,y) = (1,0)$. Then $(e_1, e_2) \cdot (1,0) = (e_1, 0)$ should be equal to $(1,0)$. So, $e_1 = 1$.

Let $(x,y) = (0,1)$. Then $(e_1, e_2) \cdot (0,1) = (e_2, 0)$ should be equal to $(0,1)$. So, $e_2 = 0$.

Now we can check that the pair $(1,0)$ works as an identity. In fact, it does: $(1, 0) \cdot (x,y) = (x\cdot 1 - 2 \cdot y \cdot 0, y \cdot 1 + x \cdot 0) = (x, y)$. Multiplication on the right can be checked similarily.

To make the whole construction more clear, you can notice that it resembles complex numbers multiplication, but with a "$2$" inside the multiplication formula. This is a consequence of the fact that this formula is exactly the multiplication in factor ring $\mathbb{R}[X]/(X^2 + 2)$ (while ordinary complex numbers are defined as $\mathbb{R}[X]/(X^2 + 1)$), that is, a factorization of the ring of polynomials over $\mathbb{R}$ by the ideal generated by $X^2+2$.

lisyarus
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