# simply transitive

Let $G$ be a group acting on a set $X$. The action is said to be simply transitive^{} if it is transitive and $\forall x,y\in X$ there is a *unique* $g\in G$ such that $g.x=y$.

###### Theorem.

A group action is simply transitive if and only if it is free and transitive

###### Proof.

Necessity follows since $g.x=x$ implies that $g={1}_{G}$ because ${1}_{G}.x=x$ also. Now assume the action is free and transitive and we have elements ${g}_{1},{g}_{2}\in G$ and $x,y\in X$ such that ${g}_{1}.x=y$ and ${g}_{2}.x=y$. Then ${g}_{1}.x={g}_{2}.x\u27f9{g}_{2}^{-1}.{g}_{1}.x=({g}_{2}^{-1}{g}_{1}).x=x$ hence ${g}_{2}^{-1}{g}_{1}={1}_{G}$ because the action is free. Thus ${g}_{1}={g}_{2}$ and so the action is simply transitive. ∎

Title | simply transitive |
---|---|

Canonical name | SimplyTransitive |

Date of creation | 2013-03-22 14:37:41 |

Last modified on | 2013-03-22 14:37:41 |

Owner | benjaminfjones (879) |

Last modified by | benjaminfjones (879) |

Numerical id | 7 |

Author | benjaminfjones (879) |

Entry type | Definition |

Classification | msc 20M30 |

Related topic | GroupAction |