$M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?
Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix}
x_{1} && x_{2}\\
x_{3} && x_{4}\\
\end{pmatrix}$
so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$. 
The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.
Any help would be much appreciated.

The group $M$, which I am going to rename $G$, acts transitively on $\mathbb{R}^2 \setminus \{(0, 0)\}$ which is clearly homeomorphic to $\mathbb{R} \times S^1$. The stabilizer of the vector 
$$
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
is the set of matrices of the form
$$
\begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix}
$$
which is clearly homeomorphic to $\mathbb{R}$. 
If you've seen group actions, that's a complete answer. 
EDIT: I translated this into a language not using group actions in my previous answer, but made quite a mistake (the equations I wrote always admit a trivial solution).
