# proof that all subgroups of a cyclic group are cyclic

The following is a proof that all subgroups^{} of a cyclic group^{} are cyclic.

###### Proof.

Let $G$ be a cyclic group and $H\le G$. If $G$ is trivial, then $H=G$, and $H$ is cyclic. If $H$ is the trivial subgroup, then $H=\{{e}_{G}\}=\u27e8{e}_{G}\u27e9$, and $H$ is cyclic. Thus, for the of the proof, it will be assumed that both $G$ and $H$ are nontrivial.

Let $g$ be a generator^{} of $G$. Let $n$ be the smallest positive integer such that ${g}^{n}\in H$.

Claim: $H=\u27e8{g}^{n}\u27e9$

Let $a\in \u27e8{g}^{n}\u27e9$. Then there exists $z\in \mathbb{Z}$ with $a={({g}^{n})}^{z}$. Since ${g}^{n}\in H$, we have that ${({g}^{n})}^{z}\in H$. Thus, $a\in H$. Hence, $\u27e8{g}^{n}\u27e9\subseteq H$.

Let $h\in H$. Then $h\in G$. Let $x\in \mathbb{Z}$ with $h={g}^{x}$. By the division algorithm^{}, there exist $q,r\in \mathbb{Z}$ with $$ such that $x=qn+r$. Thus, $h={g}^{x}={g}^{qn+r}={g}^{qn}{g}^{r}={({g}^{n})}^{q}{g}^{r}$. Therefore, ${g}^{r}=h{({g}^{n})}^{-q}$. Recall that $h,{g}^{n}\in H$. Hence, ${g}^{r}\in H$. By choice of $n$, $r$ cannot be positive. Thus, $r=0$. Therefore, $h={({g}^{n})}^{q}{g}^{0}={({g}^{n})}^{q}{e}_{G}={({g}^{n})}^{q}\in \u27e8{g}^{n}\u27e9$. Hence, $H\subseteq \u27e8{g}^{n}\u27e9$.

This proves the claim. It follows that every subgroup of $G$ is cyclic. ∎

Title | proof that all subgroups of a cyclic group are cyclic |
---|---|

Canonical name | ProofThatAllSubgroupsOfACyclicGroupAreCyclic |

Date of creation | 2013-03-22 13:30:47 |

Last modified on | 2013-03-22 13:30:47 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 10 |

Author | Wkbj79 (1863) |

Entry type | Proof |

Classification | msc 20A05 |

Related topic | ProofThatEverySubringOfACyclicRingIsACyclicRing |