An exercise from Artin's Algebra:
Let G and H be the following subgroups of $GL_2(\mathbb R)$:
$$G = \{ \begin{bmatrix}
x & y\\
0 & 1
\end{bmatrix} \} , H= \{\begin{bmatrix}
x & 0\\
0 & 1
\end{bmatrix}\}$$with x and y real and x > O.
An element of G can be represented by a point in the right half plane. Make sketches showing the partitions of the half plane into left cosets and into
right cosets of H.
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This same exercise was asked on reddit 7 years ago. Here is the link: Abstract Algebra confusion. One comment looks to be agreement with my answer for the left cosets.
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This question Left Cosets and Right Cosets. looks to be somehow related. It looks like a generalization.
I noticed that
$$\begin{bmatrix}
x & b\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
x & 0\\
0 & 1
\end{bmatrix}\begin{bmatrix}
1 & \frac{b}{x}\\
0 & 1
\end{bmatrix}, \begin{bmatrix}
x & b\\
0 & 1
\end{bmatrix} = \begin{bmatrix}
1 & b\\
0 & 1
\end{bmatrix}\begin{bmatrix}
x & 0\\
0 & 1
\end{bmatrix}, \ \text{and} \ \begin{bmatrix}
1 & b\\
0 & 1
\end{bmatrix}, \begin{bmatrix}
1 & \frac{b}{x}\\
0 & 1
\end{bmatrix} \in G$$
Hence:
The left cosets are horizontal rays $y=b, x>0$. For each left coset, $x>0$ is divided into everything above $y=b, x>0$ and everything below $y=b, x>0$.
The right cosets are right-halves of hyperbolas $y=\frac{b}{x}, x>0$. For each right coset, $x>0$ is divided into everything above $y=b, x>0$ and everything below $y=b, x>0$.
Am I correct? After seeing Andreas Caranti's answer, I don't think I understand the question. I think these are left and right cosets because they are disjoint because the rays and hyperbolas do not intersect and their union is all of $G$.
Best Answer
We have $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\R}{\mathbb{R}}$ $$ H = \Set{ \begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix} : u \in \R, u > 0 }. $$ Then we have for fixed $a, b \in \R$, $a > 0$, $$ H \begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix} = \Set{ \begin{bmatrix} u a & u b\\ 0 & 1 \end{bmatrix} : u \in \R, u > 0 } = \Set{ \begin{bmatrix} x & y \\ 0 & 1 \end{bmatrix} : x, y \in R, x > 0, x^{-1} y = a^{-1} b }, $$ so you see this is the part of the line $y = a^{-1} b x$ for $x > 0$.