Proof of Cayley's theorem by left multiplication

Proof

We use the action of $G$ on itself by left multiplication. Let $e$ be the identity of $G$, and for $g \in G$ let $\ell_g \colon G \to G$, $\ell_g(x) = g x$.

1. For $g, h, x \in G$, associativity gives $\ell_g(\ell_h(x)) = g(hx) = (gh)x = \ell_{gh}(x)$. So $\ell_g \circ \ell_h = \ell_{gh}$.
2. Also $\ell_e(x) = ex = x$, so $\ell_e$ is the identity map of $G$.
3. By steps 1 and 2, $\ell_g \circ \ell_{g^{-1}} = \ell_e$ and $\ell_{g^{-1}} \circ \ell_g = \ell_e$. So $\ell_g$ is a bijection with inverse $\ell_{g^{-1}}$, and $\ell_g \in \operatorname{Sym}(G)$, the symmetric group of $G$.
4. Let $\varphi \colon G \to \operatorname{Sym}(G)$, $\varphi(g) = \ell_g$. The operation of $\operatorname{Sym}(G)$ is composition, so step 1 says $\varphi(gh) = \varphi(g) \circ \varphi(h)$. Thus $\varphi$ is a group homomorphism.
5. If $\varphi(g) = \varphi(h)$, then $g = \ell_g(e) = \ell_h(e) = h$. So $\varphi$ is injective.
6. Let $K = \set{\ell_g : g \in G}$ be the image of $\varphi$. By steps 1 to 3, $K$ contains the identity $\ell_e$, is closed under composition, and contains the inverse $\ell_{g^{-1}}$ of each $\ell_g$. So $K$ is a subgroup of $\operatorname{Sym}(G)$.
7. As a map $G \to K$, $\varphi$ is a homomorphism, injective by step 5 and onto by the definition of $K$. So it is an isomorphism, and $G$ is isomorphic to the subgroup $K$ of $\operatorname{Sym}(G)$.