Cayley's Representation Theorem

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

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

Corollary
Let $G$ be a group and let $p$ be the smallest prime such that $p \backslash \left|{G}\right|$.

If $\exists H: H \le G$ such that $\left|{H}\right| = p$, then $H \triangleleft G$.

Proof 1
Let $H = \left\{{e}\right\}$. Thus we can apply Permutation of Cosets to $H$ so that $\mathbb S = G$ and $\ker \left({\theta}\right) = \left\{{e}\right\}$.

The result follows by the First Isomorphism Theorem.

Proof 2
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.

Proof of Corollary
Apply Permutation of Cosets: Corollary to $H$ to find some $N \triangleleft G$ such that $\left[{G : N}\right] \backslash p!$.

Since $\left[{G : N}\right] \backslash \left|{G}\right|$, it divides $\gcd \left\{{\left|{G}\right|, p!}\right\}$.

Since $p$ is the smallest prime dividing $\left|{G}\right|$, $\gcd \left\{{\left|{G}\right|, p!}\right\} = p$.

Thus $\left[{G : N}\right] = p = \left[{G : H}\right]$.

Since $N \subseteq H$, it must follow that $N = H$.

Comment
What this theorem tells us is that in order to study groups, it is necessary only to study subgroups of the symmetric groups.

It is also known as the Representation Theorem for Groups.