# 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 |