# Proof: The orbit of any element of a group is a subgroup

Following is a proof that, if $G$ is a group and $g\in G$, then $\u27e8g\u27e9\le G$. Here $\u27e8g\u27e9$ is the orbit of $g$ and is defined as

$$\u27e8g\u27e9=\{{g}^{n}:n\in \mathbb{Z}\}$$ |

Since $g\in \u27e8g\u27e9$, then $\u27e8g\u27e9$ is nonempty.

Let $a,b\in \u27e8g\u27e9$. Then there exist $x,y\in \mathbb{Z}$ such that $a={g}^{x}$ and $b={g}^{y}$. Since $a{b}^{-1}={g}^{x}{({g}^{y})}^{-1}={g}^{x}{g}^{-y}={g}^{x-y}\in \u27e8g\u27e9$, it follows that $\u27e8g\u27e9\le G$.

Title | Proof: The orbit of any element of a group is a subgroup^{} |
---|---|

Canonical name | ProofTheOrbitOfAnyElementOfAGroupIsASubgroup |

Date of creation | 2013-03-22 13:30:58 |

Last modified on | 2013-03-22 13:30:58 |

Owner | drini (3) |

Last modified by | drini (3) |

Numerical id | 6 |

Author | drini (3) |

Entry type | Proof |

Classification | msc 20A05 |

Related topic | Group |

Related topic | Subgroup |

Related topic | ProofThatEveryGroupOfPrimeOrderIsCyclic |

Related topic | ProofOfTheConverseOfLagrangesTheoremForCyclicGroups |

Defines | orbit |