# example of a Jordan-Hölder decomposition

A group that has a composition series^{} will often have several different composition series.

For example, the cyclic group^{} ${C}_{12}$ has $(E,{C}_{2},{C}_{6},{C}_{12})$, and $(E,{C}_{2},{C}_{4},{C}_{12})$, and $(E,{C}_{3},{C}_{6},{C}_{12})$ as different composition series.
However, the result of the Jordan-Hölder Theorem is that any two composition series of a group are equivalent^{}, in the sense that the sequence of factor groups in each series are the same, up to rearrangement of their order in the sequence ${A}_{i+1}/{A}_{i}$. In the above example, the factor groups are isomorphic^{} to $({C}_{2},{C}_{3},{C}_{2})$, $({C}_{2},{C}_{2},{C}_{3})$, and $({C}_{3},{C}_{2},{C}_{2})$, respectively.

This is taken from the http://en.wikipedia.org/wiki/Solvable_groupWikipedia article on solvable groups^{}.

Title | example of a Jordan-Hölder decomposition |
---|---|

Canonical name | ExampleOfAJordanHolderDecomposition |

Date of creation | 2013-03-22 14:24:33 |

Last modified on | 2013-03-22 14:24:33 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 10 |

Author | mathcam (2727) |

Entry type | Example |

Classification | msc 20E15 |

Synonym | example of Jordan-Holder decomposition |