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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.

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 $G$.

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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 $x > 0$.

Andreas Caranti
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  • Thank you Andreas Caranti, but I don't think I even understand the question. Did I do something wrong in my solution? In your solution, you divide the right half of $\mathbb R^2$ plane $x>0$ by $y=\frac{b}{a}x$? –  Oct 13 '18 at 00:23
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    @JackBauer, I don't want to impose, but quite simply yours is no solution. It is a (correct) calculation from which you draw a conclusion that isn't actually there.

    What I did is what was requested, that is, I computed the elements of a fixed, but arbitrary, right coset, and then characterised them as the elements of a ray (a half-line, or whatever one wants to call it).

     – Andreas Caranti Oct 13 '18 at 10:09
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    @JackBauer, so for the left cosets the partition is given by the horizontal rays, whereas for the right coset the partition is given by the rays departing from the origin (which of course is not in the group, but still). – Andreas Caranti Oct 13 '18 at 10:15
  • Oh I think I see it. I'm being selective in my choice of $g$ for $gH$ or $Hg$ when it should be any $g \in G$? No imposing at all. I knew I wasn't understanding the question somehow. You are answering my question :) –  Oct 14 '18 at 00:38
  • For right cosets: I realized my mistake. The matrix I chose depends on $x$ for the particular matrix in $H$, but the matrix I chose shouldn't on the particular matrix in $H$! It's like choosing a $\delta$ that depends on $x$ in $\lim_{x \to x_0} f(x)$: $\delta$ can depend on $\varepsilon$ (as it usually does) and can depend on $x_0$ (for non-uniform continuity), but $\delta$ depending on $x$ is meaningless or circular. Do I understand correctly? –  Oct 16 '18 at 05:36
  • For left cosets: Did I get it right even if for the wrong reasons? I posted an answer if it helps. –  Oct 16 '18 at 06:24
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    $\newcommand{\Set}[1]{\left{ #1 \right}}$$\newcommand{\R}{\mathbb{R}}$@JackBauer, your left cosets look ok to me. Here is how I would write it, in the notation of my answer. $\begin{bmatrix} a & b\ 0 & 1 \end{bmatrix} H = \Set{ \begin{bmatrix} u a & b\ 0 & 1 \end{bmatrix} : u \in \R, u > 0 } = \Set{ \begin{bmatrix} x & b\ 0 & 1 \end{bmatrix} : x \in \R, x > 0 }$, so this is the horizontal ray given by the intersection of the line $y = b$ with the right open half-plane. – Andreas Caranti Oct 16 '18 at 07:50
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  • 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.