# order of products

If $a$ and $b$ are elements of a group, then both $ab$ and $ba$ have always the same order.

*Proof.* Let $e$ be the indentity element of the group. For $n>1$, we have the
equivalent^{} (http://planetmath.org/Equivalent3) conditions

$$e={(ab)}^{n}=\underset{n}{\underset{\u23df}{(ab)(ab)\mathrm{\cdots}(ab)}}=a{(ba)}^{n-1}b,$$ |

$${a}^{-1}{b}^{-1}={(ba)}^{n-1},$$ |

$${(ba)}^{-1}={(ba)}^{n-1},$$ |

$$e={(ba)}^{n}.$$ |

As for the infinite order, it makes the conditions false.

Note. More generally, all elements of any conjugacy class^{} have the same order.

Title | order of products |
---|---|

Canonical name | OrderOfProducts |

Date of creation | 2013-03-22 18:56:43 |

Last modified on | 2013-03-22 18:56:43 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 20A05 |

Related topic | InverseFormingInProportionToGroupOperation |