[proofplan] We first verify that each group element acts by a bijection of $X$, so that $\rho(g)$ really lies in $\operatorname{Sym}(X)$. Then we use the associativity axiom of the action to prove that $\rho(gh)=\rho(g)\circ\rho(h)$, which is exactly the homomorphism property. Finally, we identify the kernel of $\rho$ with the elements fixing every point of $X$, and the faithful-injective equivalence follows directly from this identification. [/proofplan] [step:Show that each group element defines a permutation of $X$] Fix $g \in G$. Define the map \begin{align*} \rho(g): X &\to X \\ x &\mapsto g \cdot x. \end{align*} Define also \begin{align*} \tau_g: X &\to X \\ x &\mapsto g^{-1} \cdot x. \end{align*} For every $x \in X$, the action axioms give \begin{align*} (\rho(g) \circ \tau_g)(x) &= \rho(g)(g^{-1} \cdot x) \\ &= g \cdot (g^{-1} \cdot x) \\ &= (gg^{-1}) \cdot x \\ &= e \cdot x \\ &= x. \end{align*} Similarly, \begin{align*} (\tau_g \circ \rho(g))(x) &= \tau_g(g \cdot x) \\ &= g^{-1} \cdot (g \cdot x) \\ &= (g^{-1}g) \cdot x \\ &= e \cdot x \\ &= x. \end{align*} Thus $\tau_g$ is the inverse map of $\rho(g)$, so $\rho(g)$ is a bijection $X \to X$. Hence $\rho(g) \in \operatorname{Sym}(X)$ for every $g \in G$, and the map $\rho: G \to \operatorname{Sym}(X)$ is well-defined. [/step] [step:Use the action axiom to prove the homomorphism identity] Let $g,h \in G$. To prove equality in $\operatorname{Sym}(X)$, we compare the two maps $\rho(gh)$ and $\rho(g)\circ\rho(h)$ pointwise. For every $x \in X$, \begin{align*} \rho(gh)(x) &= (gh) \cdot x \\ &= g \cdot (h \cdot x) \\ &= \rho(g)(\rho(h)(x)) \\ &= (\rho(g)\circ\rho(h))(x). \end{align*} Therefore $\rho(gh)=\rho(g)\circ\rho(h)$ for all $g,h \in G$. Since composition is the group operation in $\operatorname{Sym}(X)$, the map $\rho: G \to \operatorname{Sym}(X)$ is a [group homomorphism](/page/Group%20Homomorphism). [guided] The goal is to prove that $\rho$ preserves the group operation. The operation in $G$ is multiplication, while the operation in $\operatorname{Sym}(X)$ is composition of bijections. Therefore the required identity is \begin{align*} \rho(gh)=\rho(g)\circ\rho(h) \end{align*} for all $g,h \in G$. Fix $g,h \in G$. Since $\rho(gh)$ and $\rho(g)\circ\rho(h)$ are both maps $X \to X$, it is enough to evaluate them at an arbitrary point $x \in X$. By the definition of $\rho$, \begin{align*} \rho(gh)(x) = (gh)\cdot x. \end{align*} The associativity axiom for the left action says that acting by $gh$ is the same as first acting by $h$ and then acting by $g$. Hence \begin{align*} (gh)\cdot x = g \cdot (h \cdot x). \end{align*} Using the definition of $\rho(h)$ and then the definition of $\rho(g)$, this becomes \begin{align*} g \cdot (h \cdot x) &= \rho(g)(\rho(h)(x)) \\ &= (\rho(g)\circ\rho(h))(x). \end{align*} Combining the equalities gives \begin{align*} \rho(gh)(x) = (\rho(g)\circ\rho(h))(x) \end{align*} for every $x \in X$. Therefore $\rho(gh)=\rho(g)\circ\rho(h)$ as elements of $\operatorname{Sym}(X)$. This proves that $\rho$ is a group homomorphism. [/guided] [/step] [step:Identify the kernel with the elements fixing every point of $X$] Define the identity permutation \begin{align*} \operatorname{id}_X: X &\to X \\ x &\mapsto x. \end{align*} By definition of the kernel of the homomorphism $\rho: G \to \operatorname{Sym}(X)$, \begin{align*} \ker \rho &= \{g \in G : \rho(g)=\operatorname{id}_X\}. \end{align*} For $g \in G$, the equality $\rho(g)=\operatorname{id}_X$ holds if and only if \begin{align*} \rho(g)(x)=x \end{align*} for every $x \in X$. By the definition of $\rho(g)$, this is equivalent to \begin{align*} g \cdot x = x \end{align*} for every $x \in X$. Therefore \begin{align*} \ker \rho = \{g \in G : g \cdot x = x \text{ for every } x \in X\}. \end{align*} This set is precisely the kernel of the action, namely the subgroup of elements of $G$ that fix every point of $X$. [/step] [step:Conclude the faithful-injective equivalence] By definition, the action is faithful exactly when the only element of $G$ fixing every point of $X$ is the identity element $e$. Using the kernel identification above, this condition is \begin{align*} \ker \rho = \{e\}. \end{align*} For a group homomorphism, having kernel $\{e\}$ is equivalent to being injective. Hence the action is faithful if and only if $\rho$ is injective. [/step]