Let G be the group $\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} :a,b,c \in \mathbb{R} \}$ and define the action $\pi$ on $\mathbb{R}^2$ by $\pi_g(x) = gx, \forall x \in \mathbb{R}^2$. How many orbits are there?
So what I thought is that $Orb( \begin{pmatrix} 0 \\ 0 \end{pmatrix}) = \{ \begin{pmatrix} 0 \\ 0 \end{pmatrix} \}$ and $Orb( \begin{pmatrix} 0 \\ 1 \end{pmatrix}) = \mathbb{R}^2 \backslash \{\begin{pmatrix} 0 \\ 0 \end{pmatrix} \}$ (since any nonzero vector $\begin{pmatrix} a \\ b \end{pmatrix}$ can be written as $g \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, where $g=\begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix}$ is upper triangular, so there are only the two aforementioned orbits. Is this correct? Thanks in advance!
$G$ is not a group as written, because if $a = 0$ or $c = 0$ then the matrix does not have an inverse.
If you assume $a,c \neq 0$ then you should end up with more than two orbits under the action of $G$ on $\mathbb R^2$.