Notes on group actions and homomorphisms

Let $G$ be a group, $X$ a non-empty set and $S_X$ the symmetric group of $X$ , i.e. the group of all bijective maps on $X$ . $\cdot$ may denote a left group action of $G$ on $X$ .

1. For each $g \in G$ and $x \in X$ we define
      .
   Since $f_{g^-1}(f_g(x)) =g^{-1} \cdot (g \cdot x) =x$ for each $x \in X$ , $f_{g^-1}$ is the inverse of $f_g$ . so $f_g$ is bijective and thus element of $S_X$ . We define $F: G \to S_X, F(g) =f_g$ for all $g \in G$ . This mapping is a group homomorphism: Let $g,h \in G, x \in X$ . Then
   
   for all $x\in X$ implies $F(gh) =F(g) \circ F(h)$ . -- The same is obviously true for a right group action.
2. Now let $F: G \to S_x$ be a group homomorphism, and let $f: G \times X \to X, (g,x) \mapsto F(g)(x)$ satisfy
   1. $f(1_G, x) =F(1_g)(x) =x$ for all $x \in X$ and
   2. $f(gh, x) =F(gh)(x) =(F(g) \circ F(h)(x) =F(g)(F(h)(x)) =f(g, f(h,x))$ ,
   so $f$ is a group action induced by $F$ .

Characterization of group actions

For the trivial operation of $G$ on $X$ given by $g \cdot x =x, \; \forall g \in G$ the stabilizer subgroup $G_x$ is $G$ for all $x \in X$ , and thus

If the operation of $G$ on $X$ is free, then $G_x =\{1_G\}, \; \forall x \in X$ , thus the kernel of $F$ is $\{1_G\}$ -like for a faithful operation. But:

Let $X=\{1, \ldots, n\}$ and $G=S_n$ . Then the operation of $G$ on $X$ given by