Find neutral and inverse element of a group 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 $e_2$?
 A: 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, 
A: 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..
A: 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 $X^2+2$.
