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

Question

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?

Answer 1

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

Answer 2

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 $1-1$.