Cayley's Theorem (Group Theory)/General Case

From ProofWiki
Jump to: navigation, search

Theorem

Let $\left({G, \cdot}\right)$ 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:

$\lambda_a \left({x}\right) = a \cdot x$

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:

$\lambda_b \left({x}\right) = b \cdot x$

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


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

Consider the mapping $\Phi: G \to H$ defined as:

$\forall a \in G: \Phi \left({a}\right) = \lambda_a$

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:

$\left({\lambda_a}\right)^{-1} = \lambda_\left({a^{-1}}\right)$

because:

$\lambda_a \circ \left({\lambda_a}\right)^{-1} = \lambda_\left({a \cdot a^{-1}}\right)$


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

$\blacksquare$


Also known as

This result is also known as the Representation Theorem for Groups.


Source of Name

This entry was named for Arthur Cayley.


Sources