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 $$\mathrm{ker} \left({\theta}\right) = \left\{{e}\right\}$$.

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 $$N = H$$.