The strategy is to use the left regular action of $G$ on itself, show it is faithful (trivial kernel), then apply the [First Isomorphism Theorem](/theorems/791) to embed $G$ into $\operatorname{Sym}(G)$. **Step 1: Define the left regular action.** Let $G$ act on $X = G$ by left multiplication: $\rho(g, h) = gh$ for $g, h \in G$. This is an action since $\rho(g_1 g_2, h) = g_1 g_2 h = g_1(g_2 h) = \rho(g_1, \rho(g_2, h))$ and $\rho(e, h) = eh = h$. **Step 2: Construct the associated homomorphism.** The action induces a homomorphism: \begin{align*} \Phi : G &\to \operatorname{Sym}(G) \\ g &\mapsto \varphi_g, \end{align*} where $\varphi_g(h) = gh$ for all $h \in G$. **Step 3: Show the action is faithful.** [claim:Left Regular Action Is Faithful] $\ker(\Phi) = \{e\}$. [/claim] [proof] Suppose $\varphi_g = \varphi_e$, i.e., $gh = eh = h$ for all $h \in G$. Taking $h = e$ gives $g = e$. So $\ker(\Phi) = \{e\}$. [/proof] **Step 4: Apply the First Isomorphism Theorem.** By the [First Isomorphism Theorem](/theorems/791): \begin{align*} G / \{e\} \cong \operatorname{Im}(\Phi) \leq \operatorname{Sym}(G). \end{align*} Since $G / \{e\} \cong G$, we conclude $G \cong \operatorname{Im}(\Phi) \leq \operatorname{Sym}(G)$.