**Proof plan.** Let $G$ act on itself by left multiplication; the associated permutation representation has trivial kernel, so the [First Isomorphism Theorem for Groups](/theorems/842) embeds $G$ into $\operatorname{Sym}(G)$. **Step 1: Define the action.** Let $G$ act on itself by left multiplication: \begin{align*} G \times G &\to G \\ (g, a) &\mapsto g \cdot a = ga. \end{align*} This is a [group](/page/Group) action: $e \cdot a = a$ and $g_1 \cdot (g_2 \cdot a) = g_1(g_2 a) = (g_1g_2)a = (g_1g_2) \cdot a$. **Step 2: Associated homomorphism.** This action defines a group homomorphism $\varphi : G \to \operatorname{Sym}(G)$ by $\varphi(g) = (a \mapsto ga)$. Each $\varphi(g)$ is a bijection with inverse $\varphi(g^{-1})$, so $\varphi$ is well-defined. **Step 3: $\ker(\varphi) = \{e\}$.** [claim: Trivial Kernel] The homomorphism $\varphi$ has trivial kernel. [/claim] [proof] If $g \in \ker(\varphi)$ then $\varphi(g)$ is the identity permutation, so $ga = a$ for all $a \in G$. Setting $a = e$ gives $g = e$. [/proof] **Step 4: Conclusion.** Since $\ker(\varphi) = \{e\}$, the [First Isomorphism Theorem for Groups](/theorems/842) gives \begin{align*} G \cong G/\{e\} \cong \operatorname{im}(\varphi) \leq \operatorname{Sym}(G). \qquad \square \end{align*}