Atlas and transition maps for a Lie group Consider the group $G$ consisting of real $2\times 2$ matrices of the form
$\begin{pmatrix}
x & 0\\
y & z
\end{pmatrix}$ of nonzero determinant with multiplication as the operation. I believe that this is a Lie group because as a space it is the intersection of $GL(2,\mathbb{R})$ and the variety $V(w)$, where $w$ is the upper right matrix entry (I seem to recall a theorem saying that such intersections give Lie groups). Furthermore, if we define $Z(G)$ to be the center of $G$, consisting of matrices of the form
$\begin{pmatrix}
\lambda & 0 \\
0 & \lambda
\end{pmatrix}$, we have that $Z(G)$ is a normal subgroup of $G$ which is closed and thus a Lie subgroup. This makes the group $G' = G/Z(G)$ a (2-dimensional) Lie group.
My question is twofold. First, is $G$ in fact a Lie group? Second, can someone give an example of an atlas for $G'$ and the corresponding transition maps between charts?
 A: A quotient of a Lie group by a closed subgroup is always a Lie group.
Can you describe the quotient group as an abstract group? When you do that, you will have charts for it...
A: I did not realised that the structure of this group would be as simple as it turned out to be, mostly because I had simply assumed that working with this group would be beyond me given that I've never worked with Lie groups before. After looking again, I believe I've proved that $G'$ is a Lie group and found an atlas for it.
The group $G'$ is isomorphic to the group of real $2\times 2$ matrices with nonzero determinant of the form
$\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix}$ as the equivalence class of
$\begin{pmatrix}
x & 0 \\
y & z
\end{pmatrix}$ contains precisely $1$ matrix of such form, specifically
$\begin{pmatrix}
\frac{x}{z} & 0 \\
\frac{y}{z} & 1
\end{pmatrix}$ (note that $z\neq 0$ because $xz \neq 0$).
$G'$ can be made into a topological space by equipping it with the metric defined as $d\left(\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix},\begin{pmatrix}
x' & 0 \\
y' & 1
\end{pmatrix}\right) = \sqrt{(x-x')^2 + (y-y')^2}$. An atlas for $G'$ is given by the charts $\alpha: G'_+ \rightarrow \mathbb{R}^2$ and $\beta: G'_- \rightarrow \mathbb{R}^2$, where $G'_+$ is the (open) set of matrices with positive determinant and $G'_-$ the (open) set with negative determinant, defined by $\alpha \left(\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix}\right) = \begin{pmatrix}
x \\
y
\end{pmatrix}$ and $\beta \left(\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix}\right) = \begin{pmatrix}
x \\
y
\end{pmatrix}$. These charts are non-intersecting, and this makes $G'$ a smooth manifold.
In order to show that $G'$ is a Lie group we need only show that the map $\phi: G'\times G' \rightarrow G'$ defined as $\phi \left(\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix},\begin{pmatrix}
x' & 0 \\
y' & 1
\end{pmatrix},\right) = {\begin{pmatrix}
x & 0 \\
y & 1
\end{pmatrix}}^{-1} \begin{pmatrix}
x' & 0 \\
y' & 1
\end{pmatrix} = \begin{pmatrix}
\frac{x'}{x} & 0 \\
\frac{xy'-yx'}{x}  & 1
\end{pmatrix}$ is smooth. This follows easily from the fact that rational functions on $\mathbb{R}^2$ are smooth whereever their denominator is not $0$ and that $x\neq 0$ in $G'$.
