Cayley's Representation Theorem/General Case

Theorem
Let $\struct {G, \cdot}$ be a group.

Then there exists a permutation group $P$ on some set $S$ such that:


 * $G \cong P$

That is, such that $G$ is isomorphic to $P$.

Proof
Let $G$ be a group and let $a \in G$.

Consider the left regular representation $\lambda_a: G \to G$ defined as:


 * $\map {\lambda_a} x = a \cdot x$

From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation.

Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as:


 * $\map {\lambda_b} x = b \cdot x$

From the Cancellation Laws it follows that $\lambda_a \ne \lambda_b \iff a \ne b$.

Let $H = \set {\lambda_x: x \in G}$.

Consider the mapping $\Phi: G \to H$ defined as:
 * $\forall a \in G: \map \Phi a = \lambda_a$

From the above we have that $\Phi$ is a bijection.

Let $a, b \in G$.

From Composition of Regular Representations we have that:
 * $\lambda_a \circ \lambda_b = \lambda_{a \cdot b}$

where $\circ$ denotes composition of mappings.

That is, $\Phi$ has the morphism property.

Thus $\Phi$ is seen to be a group isomorphism.

We also have that:
 * $\paren {\lambda_a}^{-1} = \lambda_\paren {a^{-1} }$

because:
 * $\lambda_a \circ \paren {\lambda_a}^{-1} = \lambda_\paren {a \cdot a^{-1} }$

Hence the set of left regular representations $\set {\lambda_x: x \in G}$ is a group which is isomorphic to $G$.