proof of Cayley’s theorem

Let $G$ be a group, and let $S_G$ be the permutation group of the underlying set $G$. For each $g\in G$, define ${\rho }_g:G\to G$ by ${\rho }_g(h)=gh$. Then ${\rho }_g$ is invertible with inverse ${\rho }_{g^{-1}}$, and so is a permutation of the set $G$.

Define $\mathrm{\Phi }:G\to S_G$ by $\mathrm{\Phi }(g)={\rho }_g$. Then $\mathrm{\Phi }$ is a homomorphism, since

$$(\mathrm{\Phi }(gh))(x)={\rho }_{gh}(x)=ghx={\rho }_g(hx)=({\rho }_g\circ {\rho }_h)(x)=((\mathrm{\Phi }(g))(\mathrm{\Phi }(h)))(x)$$

And $\mathrm{\Phi }$ is injective, since if $\mathrm{\Phi }(g)=\mathrm{\Phi }(h)$ then ${\rho }_g={\rho }_h$, so $gx=hx$ for all $x\in X$, and so $g=h$ as required.

So $\mathrm{\Phi }$ is an embedding of $G$ into its own permutation group. If $G$ is finite of order $n$, then simply numbering the elements of $G$ gives an embedding from $G$ to $S_n$.

Title	proof of Cayley’s theorem
Canonical name	ProofOfCayleysTheorem
Date of creation	2013-03-22 12:30:50
Last modified on	2013-03-22 12:30:50
Owner	Evandar (27)
Last modified by	Evandar (27)
Numerical id	5
Author	Evandar (27)
Entry type	Proof
Classification	msc 20B99