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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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    $\begingroup$ $GL_2(\Bbb R)$ fixes the origin $\endgroup$ – Lord Shark the Unknown Oct 4 '18 at 17:37
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    $\begingroup$ Your title is exactly wrong :). By that I mean that if some element is fixed (which usually means that $g \cdot a = a$ for all $g \in G$) then $G$ is not transitive. The hypothesis you give in the question is rather different: it says that there's some special element $x_0$ of $S$ that can be sent to every element $y \in S$ by the action of $G$. You now have to show that for *any pair if points $z, y \in S$, there's an element of $G$ sending $z$ to $y$. $\endgroup$ – John Hughes Oct 4 '18 at 17:45
  • $\begingroup$ "Fix" may have been entirely the wrong word for the lecturer to use here. With hindsight, I'd probably write "select some particular element $x_0$." $\endgroup$ – Chessanator Oct 4 '18 at 17:49
  • $\begingroup$ @JohnHughes yes, I think that's what was confusing me. That's how the question was worded though. Thanks. $\endgroup$ – Tomás Palamás Oct 4 '18 at 19:01
  • $\begingroup$ @Chessanator I agree. $\endgroup$ – Tomás Palamás Oct 4 '18 at 19:01
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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 $\mathbb{R}^2\backslash\{0\}$.

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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 $\mathbb{R}\backslash\{0\}$ and $\{0\}$.

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