Cayley's Theorem (Group Theory)

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Theorem

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

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


General Case

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 1

Let $H = \left\{{e}\right\}$.

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.

$\blacksquare$


Proof 2

Let $G$ be any arbitrary finite group whose identity is $e_G$.

Let $S$ be the symmetric group 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:

$\forall x \in G: \map \theta x = \lambda_x$

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 $\theta$:

$\map \theta x \circ \map \theta y = \map \theta {x y}$

demonstrating that $\theta$ is a homomorphism.


Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the kernel of $\theta$.

Let $x \in G$.

We have that:

\(\displaystyle x\) \(\in\) \(\displaystyle \map \ker \theta\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \theta x\) \(=\) \(\displaystyle e_S\) $\quad$ Definition of Kernel of Group Homomorphism $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lambda_x\) \(=\) \(\displaystyle I_G\) $\quad$ where $I_G$ is the identity mapping of $G$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map {\lambda_x} {e_G}\) \(=\) \(\displaystyle \map {I_G} {e_G}\) $\quad$ Definition of Identity Mapping $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle x\) \(=\) \(\displaystyle e_G\) $\quad$ Definitions of $\lambda_x$ and $I_G$ $\quad$

So $\map \ker \theta$ can contain no element other than $e_G$.

So, since clearly $e_G \in \map \ker \theta$, it follows that:

$\map \ker \theta = \set {e_G}$

By Kernel is Trivial iff Monomorphism, it follows that $\theta$ is a monomorphism.

By Monomorphism Image Isomorphic to Domain, we have that:

$\theta \sqbrk G \cong \Img \theta$

that is, $\theta$ is isomorphic to its image.

Hence the result.

$\blacksquare$


Also known as

It is also known as the Representation Theorem for Groups.


Also see

What this theorem tells us is that in order to study finite groups, it is necessary only to study subgroups of the symmetric groups on $n$ letters.


Source of Name

This entry was named for Arthur Cayley.


Sources