# subgroups of finite cyclic group

Let $n$ be the order of a finite cyclic group^{} $G$. For every positive divisor (http://planetmath.org/Divisibility) $m$ of $n$, there exists one and only one subgroup^{} of order $m$ of $G$. The group $G$ has no other subgroups.

*Proof.* If $g$ is a generator^{} of $G$ and $n=mk$, then ${g}^{k}$ generates the subgroup $\u27e8{g}^{k}\u27e9$, the order of which is equal to the order of ${g}^{k}$, i.e. equal to $m$. Any subgroup $H$ of $G$ is cyclic (see http://planetmath.org/node/4097this entry). If $|H|=m$, then $H$ must have a generator of order $m$; thus apparently $H=\u27e8{g}^{\pm k}\u27e9=\u27e8{g}^{k}\u27e9$.

Title | subgroups of finite cyclic group |
---|---|

Canonical name | SubgroupsOfFiniteCyclicGroup |

Date of creation | 2013-03-22 18:57:13 |

Last modified on | 2013-03-22 18:57:13 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 20A05 |