# 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 $\rho_{g}:G\rightarrow G$ 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 $\Phi:G\rightarrow S_{G}$ by $\Phi(g)=\rho_{g}$. Then $\Phi$ is a homomorphism, since

 $(\Phi(gh))(x)=\rho_{gh}(x)=ghx=\rho_{g}(hx)=(\rho_{g}\circ\rho_{h})(x)=((\Phi(% g))(\Phi(h)))(x)$

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

So $\Phi$ 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 ProofOfCayleysTheorem 2013-03-22 12:30:50 2013-03-22 12:30:50 Evandar (27) Evandar (27) 5 Evandar (27) Proof msc 20B99