How to write down the stabilizer and G-Orbits of x with the definition given?

Question

QUESTION: Defining $X=\{hxh^{-1} : h \in G\}$ as a conjugate set. We say that a group $G$ acts over $X$ if

$$ f\colon\begin{array}[t]{ >{\displaystyle}r >{{}}c<{{}} >{\displaystyle}l } G\times X &\to& X \\ (g, hxh^{-1}) &\mapsto& f(g, hxh^{-1})=ghxh^{-1}g^{-1} \end{array} $$

RIGHT ANSWER:

1. $O_{G}(x)=X$ is the G-orbit of $x$;
2. $G_{x}=\{g \in G: gxg^{-1}=x\}$ is the stabilizer of $x$ in $G$.

MY QUESTION: How to find this answers? I didn't understand how to apply this two following definitions to find such answers:

Definition (Stabilizer): Let $G$ be a finite group such that acts over a set X. Given an $x \in X$ (written $G_{x}$) the stabilizer of $x$ is the set $$G_{x}=\{g \in G: gx=x\}.$$

Definition (G-Orbit): Let $G$ be a finite group such that acts over a set X. We definie $G-\text{orbit}$ of $x$ (written $O_{G}(x)$) as the set $$O_{G}(x) = \{gx: g \in G\}.$$

Answer 1

I think it's misleading to call $X$ with this symbol, because one might be led to think of $x$ as a generic element of $X$, which is not the case. Rather, for a given $x \in G$, call:

$$S_x:=\{hxh^{-1}, h \in G\}\subseteq G \tag 1$$

So, for a given $x \in G$, your action is a $G$-action on $S_x$ and, for a given $s \in S_x$, orbit's general definition states that:

$$O_G(s)=\{f(g,s), g\in G\} \tag 2$$

which in your case reads:

$$O_G(s)=\{(gh_s)x(gh_s)^{-1}, g\in G\} \tag 3$$

where $h_s$ is such that $s=h_sxh_s^{-1}$. Now, the map $\varphi\colon G\to G$, defined by $g \mapsto \varphi(g):=gh_s$, is (in particular) surjective, since $\forall g'\in G, g'=\varphi(g'h_s^{-1})$; therefore, as $g$ spans $G$, $g'=gh_s$ does so, and thence $(3)$ becomes:

$$O_G(s):=\{g'xg'^{-1}, g'\in G\}=S_x \tag 4$$

(whence the action is transitive.)

Moreover, again for a given $s \in S_x$, by definition of stabilizer we get:

\begin{alignat}{1} \operatorname{Stab}_G(s)&=\{g\in G\mid f(g,s)=s\} \\ &=\{g\in G\mid (gh_s)x(gh_s)^{-1}=h_sxh_s^{-1}\} \\ &=\{g\in G\mid (h_s^{-1}gh_s)x(gh_s)^{-1}h_s=x\} \\ &=\{g\in G\mid (h_s^{-1}gh_s)x(h_s^{-1}gh_s)^{-1}=x\} \\ &=\{h_sg'h_s^{-1}\in G\mid g'xg'^{-1}=x\} \\ &=h_s\{g'\in G\mid g'xg'^{-1}=x\}h_s^{-1} \\ &=h_sC_G(x)h_s^{-1} \\ \tag 5 \end{alignat}

where $C_G(x)$ is the centralizer of $x$ in $G$, and finally ($s=x \Rightarrow h_s \in C_G(x)$):

$$\operatorname{Stab}_G(x)=C_G(x) \tag 6$$

Answer 2

1. For any $hxh^{-1} \in X$, is there some $g \in G$ that can make the equation $hxh^{-1} = f(g, x)$ true?

2. I think you are getting confused about the notation, because if we write the definition of a $G$-orbit more clearly: $$O_G(x) = \{f(g, x) : g \in G\} = \{gxg^{-1} : g \in G\},$$ Then we can see that $O_G(x) = X$, by definition.