Cayley's Representation Theorem/Proof 2

Proof
Let $G$ be any arbitrary finite group whose identity is $e_G$.

Let $S$ be the symmetric group on the elements of $G$, where $e_S$ is the identity of $S$.

For any $x \in G$, let $\lambda_x$ be the left regular representation of $G$ with respect to $x$.

From Regular Representations in Group are Permutations, $\forall x \in G: \lambda_x \in S$.

So, we can define a mapping $\theta: G \to S$ as:
 * $\forall x \in G: \map \theta x = \lambda_x$

From Composition of Regular Representations, we have:
 * $\forall x, y \in G: \lambda_x \circ \lambda_y = \lambda_{x y}$

where in this context $\circ$ denotes composition of mappings.

Thus by definition of $\theta$:
 * $\map \theta x \circ \map \theta y = \map \theta {x y}$

demonstrating that $\theta$ is a homomorphism.

Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the kernel of $\theta$.

Let $x \in G$.

We have that:

So $\map \ker \theta$ can contain no element other than $e_G$.

So, since clearly $e_G \in \map \ker \theta$, it follows that:
 * $\map \ker \theta = \set {e_G}$

By Kernel is Trivial iff Monomorphism, it follows that $\theta$ is a monomorphism.

By Monomorphism Image Isomorphic to Domain, we have that:
 * $\theta \sqbrk G \cong \Img \theta$

that is, $\theta$ is isomorphic to its image.

Hence the result.