[proofplan] We study the kernel $K=\ker\varphi$ of the permutation representation. The [kernel is a normal subgroup](/theorems/788) of $G$, and simplicity of $G$ forces $K$ to be either $\{e_G\}$ or all of $G$. In the first case the action is faithful, while in the second case every element acts as the identity permutation, so the image of $\varphi$ is trivial. [/proofplan] [step:Form the kernel of the permutation representation and prove it is normal] Define \begin{align*} K := \ker\varphi = \{g \in G : \varphi(g)=\operatorname{id}_X\}. \end{align*} We prove that $K \trianglelefteq G$. Let $k \in K$ and let $g \in G$. Since $\varphi$ is a [group homomorphism](/page/Group%20Homomorphism), \begin{align*} \varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}. \end{align*} Because $k \in K$, we have $\varphi(k)=\operatorname{id}_X$, hence \begin{align*} \varphi(gkg^{-1})=\varphi(g)\operatorname{id}_X\varphi(g)^{-1}=\operatorname{id}_X. \end{align*} Thus $gkg^{-1}\in K$, so $K$ is normal in $G$. [guided] The relevant subgroup is the kernel of the homomorphism attached to the action. Define \begin{align*} K := \ker\varphi = \{g \in G : \varphi(g)=\operatorname{id}_X\}. \end{align*} Here $\operatorname{id}_X$ is the identity permutation of the set $X$. An element $g \in G$ lies in $K$ exactly when it acts as the identity permutation on $X$. We now verify directly that $K$ is normal in $G$. Take an arbitrary element $k \in K$ and an arbitrary element $g \in G$. Since $\varphi: G \to S_X$ is a group homomorphism, it preserves products and inverses, so \begin{align*} \varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}. \end{align*} Because $k \in K$, the definition of $K$ gives $\varphi(k)=\operatorname{id}_X$. Substituting this into the previous identity gives \begin{align*} \varphi(gkg^{-1})=\varphi(g)\operatorname{id}_X\varphi(g)^{-1}=\operatorname{id}_X. \end{align*} Therefore $gkg^{-1}\in K$. Since this holds for every $k \in K$ and every $g \in G$, the subgroup $K$ is invariant under conjugation by elements of $G$, which is exactly $K \trianglelefteq G$. [/guided] [/step] [step:Use simplicity to reduce the kernel to two cases] Since $G$ is simple and $K \trianglelefteq G$, the [normal subgroup](/page/Normal%20Subgroup) $K$ is either $\{e_G\}$ or $G$. If $K=\{e_G\}$, then the only element of $G$ acting as $\operatorname{id}_X$ is $e_G$. Hence the action is faithful. If $K=G$, then every $g \in G$ satisfies $\varphi(g)=\operatorname{id}_X$. Therefore \begin{align*} \varphi(G)=\{\operatorname{id}_X\}. \end{align*} [/step] [step:Conclude the dichotomy] The two alternatives forced by simplicity are exactly the two alternatives in the theorem: either $\varphi(G)=\{\operatorname{id}_X\}$, or the kernel of the action is $\{e_G\}$, which means the action is faithful. This proves the claimed dichotomy. [/step]