For a finite group G, I need to prove that the order of an element in G is a divisor of the order of the group. I'm not sure what this exaclty means, but I think you have to use cyclic groups such that G={a^0,a^1,.....,a^n) where n+1 is the order of the group and a^0 is the identity element. So I think I need to use Lagrage's thm stating that the order of a subgroup is a divisor of the order of the group. So to find a subgroup of G seems to be a problem. Could I use a^m as a subgroup, where m < n+1?? What is exaclty meant by order of the element. Thanks.(adsbygoogle = window.adsbygoogle || []).push({});

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Groups Theory Proof

Loading...

Similar Threads - Groups Theory Proof | Date |
---|---|

A Tensor symmetries and the symmetric groups | Feb 9, 2018 |

I What are the groups for NxNxN puzzle cubes called? | Dec 24, 2017 |

I Addition of exponents proof in group theory | Sep 2, 2017 |

Group theory proofs | Feb 20, 2006 |

Help with a group theory proof | Aug 20, 2005 |

**Physics Forums - The Fusion of Science and Community**