Partitions by a subgroup of $GL_2(\mathbb R)$ of another subgroup of $GL_2(\mathbb R)$

Question

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.

• 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.

• 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 $G$.

Answer 1

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 $x > 0$.

Answer 2

• This question was REALLY DIFFICULT TO UNDERSTAND because this section comes before product groups and quotient groups, each of which has a definition of product set

• $$AB=\{ab | a \in A, b\in B\}$$

• Now that I understand the question, the solution is much much less difficult than the question.

Right coset: For some fixed $g \in G$ with $g=\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}$ for some $a > 0, b \in \mathbb R$, the right cosets have the form $$Hg = \{\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix}| u >0 \}\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix} = \{\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}| u >0 \} = \{\begin{bmatrix} ua & ub\\ 0 & 1 \end{bmatrix} | u > 0 \}$$

Looking at any $Hg$ a subset of $G$, which can be represented by a point in the right half plane, $Hg$ is therefore a graph parametrized by $u>0$ s.t. $x(u)=ua,y(u)=ub$, commonly known as $y=\frac{b}{a}x$, which is allowed because $a>0$ and $x=x(u)>0$ because $u,a>0$. $\therefore, Hg$ is the line passing through the origin with slope $\frac b a$ without the origin and everything below it.

• My mistake was allowing my $g$ to depend on $h \in H$. This would no longer be a fixed $g$. $g$ is instead $g=g_x$, and '$Hg_x$' is nonsensical! This is a little like allowing $\delta$ to depend on $x$ in $\lim_{x \to x_0} f(x)$. $\delta$ can depend on $\varepsilon$ (as usual), and $\delta$ can depen on $x_0$ (non-uniform continuity), but $\delta$ cannot depend on $x$ because $x$ depends on $\delta$!

Left cosets: With a second viewing, I am correct but for the wrong reasons. Let any $g \in G$ be fixed and with again as with the right cosets, $g=\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}$ for some $a > 0, b \in \mathbb R$. Consider now a $g' \in G$ also fixed but with this specific form $g'=\begin{bmatrix} 1 & b'\\ 0 & 1 \end{bmatrix}$ for some $a'=1 > 0, b' \in \mathbb R$. As it turns out $gH$ and $g'H$ have the graph.

$$g'H = \begin{bmatrix} 1 & b\\ 0 & 1 \end{bmatrix}\{\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix} \mid u > 0\} = \{\begin{bmatrix} 1 & b\\ 0 & 1 \end{bmatrix}\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix} \mid u > 0\} = \{\begin{bmatrix} u & b\\ 0 & 1 \end{bmatrix} \mid u > 0\}$$

$$gH = \begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}\{\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix} \mid u > 0\} = \{\begin{bmatrix} a & b\\ 0 & 1 \end{bmatrix}\begin{bmatrix} u & 0\\ 0 & 1 \end{bmatrix} \mid u > 0\} = \{\begin{bmatrix} ua & b\\ 0 & 1 \end{bmatrix} \mid u > 0\}$$

In either case $y=y(u)=b$ and $x=x(u) \in (0,\infty)$. I got lucky that the particular choice of $a' = 1$ did not affect this outcome, and that I did not need to try to attempt to let $g'$ depend on $h \in H$.

• Another mistake is that I did not consider $$\bigcup_{b \in \mathbb R, a>0} \{\begin{bmatrix} ua & ub\\ 0 & 1 \end{bmatrix} | u > 0 \}$$ or $$\bigcup_{b \in \mathbb R, a>0} \{\begin{bmatrix} ua & b\\ 0 & 1 \end{bmatrix} | u > 0 \}$$

to be equal to $G$ (which they both are) just because the forms are different.