# When $T_x: G \longrightarrow \mathrm{orb}(x)$ is injective?

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?

• 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 $x$" – user636532 Oct 16 '19 at 5:36
• 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? – user636532 Oct 18 '19 at 4:32

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 $Stab_G(x)=\{e \}$.

• 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
• 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 $x$" – user636532 Oct 16 '19 at 5:36

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 $$1-1$$.