[proofplan] We unpack the associated homomorphism $\rho$ and identify its kernel with the set of group elements that fix every point of $X$. Faithfulness says precisely that the only element acting as the identity on all of $X$ is the identity element $e$. Therefore the action is faithful exactly when this kernel is the singleton subgroup $\{e\}$. [/proofplan] [step:Identify the kernel as the elements acting as the identity on $X$] Let $\operatorname{id}_X: X \to X$ denote the identity map on $X$. By definition of the kernel of the [group homomorphism](/page/Group%20Homomorphism) $\rho: G \to \operatorname{Sym}(X)$, \begin{align*} \ker \rho = \{g \in G : \rho(g)=\operatorname{id}_X\}. \end{align*} For $g \in G$, the equality $\rho(g)=\operatorname{id}_X$ is equality of functions $X \to X$, so it holds if and only if $\rho(g)(x)=x$ for every $x \in X$. Since $\rho(g)(x)=g \cdot x$, this gives \begin{align*} \ker \rho = \{g \in G : g \cdot x = x \text{ for every } x \in X\}. \end{align*} [guided] Let $\operatorname{id}_X: X \to X$ be the identity map on $X$. The kernel of the homomorphism \begin{align*} \rho: G \to \operatorname{Sym}(X) \end{align*} is, by definition, the preimage of the identity element of $\operatorname{Sym}(X)$. The identity element of $\operatorname{Sym}(X)$ is $\operatorname{id}_X$, so \begin{align*} \ker \rho = \{g \in G : \rho(g)=\operatorname{id}_X\}. \end{align*} Now $\rho(g)$ and $\operatorname{id}_X$ are both functions from $X$ to $X$. Equality of functions means equality at every point of the domain. Thus \begin{align*} \rho(g)=\operatorname{id}_X \end{align*} holds if and only if, for every $x \in X$, \begin{align*} \rho(g)(x)=\operatorname{id}_X(x)=x. \end{align*} Using the definition of the associated action homomorphism, $\rho(g)(x)=g \cdot x$, so this condition is exactly \begin{align*} g \cdot x = x \text{ for every } x \in X. \end{align*} Therefore \begin{align*} \ker \rho = \{g \in G : g \cdot x = x \text{ for every } x \in X\}. \end{align*} This is the key translation: the algebraic kernel records exactly the group elements that act as the identity on the whole set. [/guided] [/step] [step:Translate faithfulness into triviality of that kernel] By definition, the action of $G$ on $X$ is faithful if and only if the only element of $G$ that fixes every point of $X$ is the identity element $e$. Using the kernel description from the previous step, this is equivalent to \begin{align*} \ker \rho = \{e\}. \end{align*} This proves both implications and completes the proof. [/step]