Finding the Orbit and Stabiliser of a specific line in $\mathbb{R^2}$

Question

Let $G=GL(2,\mathbb{R})$, that is the group of all invertible 2 by 2 real matrices.

Let $G$ act on the points of $\mathbb{R}^2$ by matrix multiplication. Consider the action of $G$ on the straight lines in $\mathbb{R}^2$ through the origin.

If $L$ is the line $y=2x$, that is, $L=\{(x,2x) :x \in \mathbb{R}\}$, find $G(L)$ and $G_L$.

Here $G(L)$ is the orbit, and $G_L$ is the stabiliser. I'll denote by $X$ the set of straight lines passing through the origin.

This question confuses me, because:

• It seems to me like the action is being described twice and not consistently.
• I am new to group actions, although I think I understand the definitions well enough.
• I've never seen $GL(2, \mathbb{R})$ before but I understand it is infinite.
• The tools I have are Cayley's Theorem, and the Orbit Stabiliser Theorem. However, $|G| \notin \mathbb{N}$. The proof I saw for the OST seems to hold for the case of infinite sets, however I'm not sure how to 'work' with multiplying cardinalities when both are uncountably infinite. I suppose I don't have to either, in this case.
• I don't actually have the action! Furthermore, it is of my understanding that the orbit and stabiliser both depend on the specific action. No? If not I haven't seen a proof for it.

Here's my attempt:

Intuitively I can see that all I need to care about is the constant $c$ in $(x, cx)$. So we have $|G|$ bijections to care about - since $G$ is isomorphic to $\Phi (G)$ where $\Phi$ is the action. Ehh... I don't really believe this either. Because I don't know what the action is, therefore I don't know if it's injective $\implies$ I don't know if it's isomorphic to a subgroup of $S_X$.

Suppose then there's some standard action to use that is injective. Then, to find $G_L$ I need to find all the bijections that map $2$ to itself, in the set of bijections $\Phi(G) \cong S_X$. I have no clue. I only know how to do this if I have explicitly what the action is. I think it's clear I'm lacking some fundamental ways of reasoning here. I can write it down with set comprehension notation but that's hardly an answer. I think the question is asking for a more explicit set.

For $G(L)$, again I only know how to do it if I actually have the action.

Thank you for your time.

Answer 1

So the action of $G$ is described as follows : for $A=\begin{pmatrix} a \ b\\c\ d\end{pmatrix} $ in $G$ and $(x,y) \in\mathbb{R} ^2$ it is $A(x,y)=(ax+by,cx+dy)$

So for the orbit of $(x, 2x)$ it is $\{(ax+2bx,cx+2dx)|\ ad-bc\not=0\}$

Can you take it from here?