# Prove a group action is transitive if an element is “fixed”

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?

• $GL_2(\Bbb R)$ fixes the origin – Lord Shark the Unknown Oct 4 '18 at 17:37
• 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 $y$. – John Hughes Oct 4 '18 at 17:45
• "Fix" may have been entirely the wrong word for the lecturer to use here. With hindsight, I'd probably write "select some particular element $x_0$." – Chessanator Oct 4 '18 at 17:49
• @JohnHughes yes, I think that's what was confusing me. That's how the question was worded though. Thanks. – Tomás Palamás Oct 4 '18 at 19:01
• @Chessanator I agree. – Tomás Palamás Oct 4 '18 at 19:01

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 $$\mathbb{R}^2\backslash\{0\}$$.
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 $$g\cdot y_1=y_2$$?
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 $$\{0\}$$.