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We suppose that the group $G$ acts on the set $X\neq \emptyset$. For $x\in X$, we define the map $$T_x:G\longrightarrow \mathrm{ orb}(x),\ g \longmapsto T_x(g):=g*x.$$

We want to find necessary and sufficient condition such that $T_x$ is injective.

My first thought is to claim that $$T_x \text{ is injective } \iff G= \mathrm{Stab}_G(x)$$

But the only obvious releation that I can see is this: $$T_x(g_1)=T_x(g_2)\iff g_1*x=g_2*x\iff x=g_1^{-1}g_2*x\iff h:=g_1^{-1}g_2\in \mathrm{Stab}_G(x).$$

Is this in the right way? Any ideas please?

Chris
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    The map $G\to{\rm Orb}(x)$ factors through the quotient map $G\to G/{\rm Stab}(x)$, and the orbit-stabilizer theorem says $G/{\rm Stab}(x)\to{\rm Orb}(x)$ is a bijection, so this should tell us the map is injective iff ${\rm Stab}(x)=1$. – anon Apr 16 '18 at 16:57
  • Chris and @anon What's the term for this, if any please? I was thinking something like "free at element $x$" or something and then "free" if "free at element $x$" for every $x$" –  Oct 16 '19 at 05:36
  • I think we can $T_x$ is surjective, so we have $\tilde{T_x}: G \to X$ is injective iff $T_x: G \to \text{orb}(x)$ is bijective iff $\text{Stab}_G(x) = {e}$. Right? –  Oct 18 '19 at 04:32

2 Answers2

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I think you are almost correct. Yes, $T_x$ injective iff $h:=g_1^{-1}g_2 \in Stab_G(x)$ iff $g_2 \in g_1 Stab_G(x)$. You could see that this holds iff $Stab_G(x)=\{e \}$.

Bryan Shih
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  • Thank you for your answer. Could you please explain why $g_1^{-1}g_2 \in Stab_G(x) \iff g_2\in g_1 Stab_G(x)$ and the last one relation? – Chris Apr 16 '18 at 17:41
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    You might want to right out what each side means. For example, for RHS, $g_1Stab_G(x) = { g_1y , : , y \in Stab_G(x) }$ So $g_2=g_1y$ for some $y \in Stab_G(x)$. – Bryan Shih Apr 16 '18 at 17:45
  • write* my bad, the irony... – Bryan Shih Apr 16 '18 at 17:58
  • Thank you very much, it's ok. – Chris Apr 16 '18 at 18:26
  • Bryan Shih, what's the term for this, if any please? I was thinking something like "free at element $x$" or something and then "free" if "free at element $x$" for every $x$" –  Oct 16 '19 at 05:36
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We will show that $$\boxed{T_x \text{ is }1-1 \iff \mathrm{Stab}_G(x)=\{1_G\}}.$$

"$\Longrightarrow$" We suppose that $T_x$ is $1-1$. Let's take an element $g\in \mathrm{Stab}_G(x)$. Then, $$g\in \mathrm{Stab}_G(x)\iff g*x=x\iff g*x=1_G*x\implies g=1_G.$$ So, $ \mathrm{Stab}_G(x)=\{1_G\}$.

"$\Longleftarrow$" We suppose that $ \mathrm{Stab}_G(x)=\{1_G\}$. Let's take $g_1,g_2\in G$. Then, \begin{align*} T_x(g_1)=T_x(g_2) & \iff g_1*x=g_2*x \\ & \iff x=g_1 ^{-1}g_2 *x \\ & \iff g_1^{-1}g_2\in \mathrm{Stab}_G(x)=\{1_G\} \\ & \iff g_1^{-1}g_2=1_G \\ & \iff g_1=g_2. \end{align*}

So, $T_x$ is $1-1$.

Chris
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