Cayley's Representation Theorem/Proof 2

Theorem
Let $S_n$ denote the symmetric group on $n$ letters.

Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.

Proof
Let $G$ be any arbitrary group with a finite number of elements, whose identity is $e_G$.

Let $S$ be the group of permutations 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:
 * $\forall x \in G: \theta \left({x}\right) = \lambda_x$

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 $\theta$:
 * $\theta \left({x}\right) \circ \theta \left({y}\right) = \theta \left({x y}\right)$

demonstrating that $\theta$ is a homomorphism.

Having established that fact, we can now consider $\ker \left({\theta}\right)$, where $\ker$ denotes the kernel of $\theta$.

Let $x \in G$.

We have that:

So $\ker \left({\theta}\right)$ can contain no element other than $e_G$.

So, since clearly $e_G \in \ker \left({\theta}\right)$, it follows that:
 * $\ker \left({\theta}\right) = \left\{{e_G}\right\}$

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

By Monomorphism Image Isomorphic to Domain, we have that:
 * $\theta \left({G}\right) \cong \operatorname{Im} \left({\theta}\right)$

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

Hence the result.