Find all cosets of $\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}$ in $\operatorname{GL}_2(\mathbb{R})$ I am trying to tackle this question:

Find all left and right cosets of
$$ H = \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} $$
in $G = \operatorname{GL}_2(\mathbb{R})$, where $a, c \in \mathbb{R}^*$ and $b \in \mathbb{R}$.

Supposedly, the solution is to take
$$ g_1 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in G $$
and
$$ h_1 = \begin{pmatrix} 1 & -\frac{b}{a} \\ 0 & 1 \end{pmatrix} \in H $$
to get
$$ g_2 = g_1 h_1 = \begin{pmatrix} a & 0 \\ c & \frac{ad-bc}{a} \end{pmatrix}, $$
and then take
$$ h_2 = \begin{pmatrix} \frac{1}{a} & 0 \\ 0 & \frac{a}{ad-bc} \end{pmatrix} \in H $$
to yield
$$ g_3 = g_2 h_2 = \begin{pmatrix} 1 & 0 \\ \frac{c}{a} & 1 \end{pmatrix} . $$
Apparently $g_3$ for all $c \in \mathbb{R}$ is a coset. It's also claimed
$$ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$
is the only other coset.

Two questions:

*

*How are you supposed to go about choosing matrices like $g_1$, $h_1$, and $h_2$? It almost seems like they were pulled out of thin air. How would you approach a question like this (still in $\operatorname{GL}_2(\mathbb{R})$) that has a lot of cosets, e.g. 4 or more, without being bogged down by the sheer number of possibilities to compute?

*How do we know $g_3$ is a complete coset and can't be "broken down" further?

 A: I would approach this question using group actions. The group $GL_2(\Bbb{R})$ acts on the set $X=\Bbb{P}^1(\Bbb{R})$ of 1-dimensional subspaces of the plane $\Bbb{R}^2$. That is, lines through the origin. Basic linear algebra says that this action is transitive. In other words, if $L_1$ and $L_2$ are two such lines, there exists a matrix $g\in GL_2(\Bbb{R})$ such that $g(L_1)=L_2$.
What's more, the subgroup is the stabilizer of the special line
$$
L_0=\left\{\binom x0\mid x\in\Bbb{R}\right\}.
$$
That is, $g(L_0)=L_0$ if and only if $g\in H$.
Basic properties of group actions then imply that the cosets of $H$ are in a bijective correspondence with the lines from $\Bbb{P}^1(\Bbb{R})$. If $g(L_0)=L'$ then
$$
gH=\left\{g'\in GL_2(\Bbb{R})\mid g'(L_0)=L'\right\}.
$$
So it suffices to enumerate the lines in $\Bbb{P}^1(\Bbb{R})$ and find matching coset representatives.
Well, from school years we know that a line through the origin is uniquely determined by its slope, and then we have the vertical line $L_{\infty}=\{\binom 0y\mid y\in\Bbb{R}\}$
of infinite slope. The line with slope $k$ is
$L_k=\{\binom x{kx}\mid x\in\Bbb{R}\}.$ To get all the cosets of $H$ you need to find a matrix $g_k$ such that $g_k(L_0)=L_k$, one for each $k$. And then one matrix $g_\infty$
such that $g(L_0)=L_\infty$.
