# 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 $G$.

$\blacksquare$

## Also known as

**Cayley's Representation Theorem** is also known as the **Representation Theorem for Groups**.

Some sources refer to it as just **Cayley's theorem**, but as there is more than one result so named, it is better to use the more specific form.

## Source of Name

This entry was named for Arthur Cayley.

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.6$: Theorem $1.6$ - 2010: Steve Awodey:
*Category Theory*(2nd ed.) ... (previous) ... (next): $\S 1.5$: Theorem