Let $G$ act on $S$ and fix some element $x_0\in S.$ Suppose that $\forall y\in S, \exists g\in G\:$ s.t. $g\cdot x_0=y.$ Prove that this action is transitive.
I'm not sure how to go about this? The only thing I can think of is somehow showing that $x_0=y$ but I'm not sure about this.
Also, I read somewhere that $GL_2(\mathbb{R}) \:\text{acting on}\: \mathbb{R}^2$ is not transitive but $GL_2(\mathbb{R}) \:\text{acting on}\: \mathbb{R}^2-\{0\}$ is. Can someone explain this?
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Hint: Take $y_1,y_2\in S$. There are $g_1,g_2\in G$ such that $g_1\cdot x_0=y_1$ and that $g_2\cdot x_0=y_2$. Can you see how to find a $g\in G$ such that $g\cdot y_1=y_2$?
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You have to show that for all $x, y\in S$, there exists a $g\in G$ such that $g\cdot x = y$. But you know already that for a $g\in G$ we have $g\cdot x_0 = y$ and we have an $f\in G$ such that $f\cdot x_0 = x$, which is equivalent to $x_0 = f^{-1}\cdot x$. Putting this all together, we obtain that $(gf^{-1})\cdot x = y$.
As for your second question, being transitive means that the action has exactly one orbit. But for every invertible homomorphism, the kernel is trivial and so the two orbits are $\mathbb{R}^2\backslash\{0\}$ and $\{0\}$.
A group action is called transitive if for every $x,y\in S$, there exists a $g\in G$ such that $g.x=y$. Now, in your case there exits $g_1,g_2\in G$ such that $g_1.x_0=x$ and $g_2.x_0=y$.
Set $g=g_2g_1^{-1}$. Then, $$ g.x=(g_2g_1^{-1}).x=g_2.(g_1^{-1}.x)=g_2.x_0=y $$ so the action is transitive.
As for the action of $GL_2(\mathbb{R})$, note that for any nonzero vector $v\in \mathbb{R}^2$, there exists an $A\in GL_2(\mathbb{R})$ such that $Ae_1=v$ (where $e_1$ is the first coordinate vector, and we take $A$ so the first column is $v$). Of course, elements of $GL_2(\mathbb{R})$ send nonzero vectors to nonzero vectors so the action is transitive on $\mathbb{R}^2\backslash\{0\}$.