Cayley's theorem

Step 1 — Associate a permutation to each group element. For each $g \in G$, define the left-translation map $\lambda_g : G \to G,\; x \mapsto g * x$. Because $\lambda_{g^{-1}}$ is its two-sided inverse, $\lambda_g$ is a bijection on $G$, so $\lambda_g \in \mathrm{Sym}(G)$.

**Step 2 — Verify that $\Lambda : G \to \mathrm{Sym}(G),\; g \mapsto \lambda_g$ is an injective homomorphism.** By associativity, $\lambda_{g * h}(x) = (g * h) * x = g * (h * x) = (\lambda_g \circ \lambda_h)(x)$ for all $x \in G$, so $\Lambda(g * h) = \Lambda(g) \circ \Lambda(h)$. If $\Lambda(g) = \mathrm{id}$, evaluating at the identity gives $\lambda_g(e) = g * e = g = e$, proving $\ker(\Lambda) = \{e\}$. Thus $G$ is isomorphic to its image $\Lambda(G) \le \mathrm{Sym}(G)$.