# generator

If $G$ is a cyclic group^{} and $g\in G$, then $g$ is a generator^{} of $G$ if $\u27e8g\u27e9=G$.

All infinite cyclic groups have exactly $2$ generators. To see this, let $G$ be an infinite cyclic group and $g$ be a generator of $G$. Let $z\in \mathbb{Z}$ such that ${g}^{z}$ is a generator of $G$. Then $\u27e8{g}^{z}\u27e9=G$. Then $g\in G=\u27e8{g}^{z}\u27e9$. Thus, there exists $n\in \mathbb{Z}$ with $g={({g}^{z})}^{n}={g}^{nz}$. Therefore, ${g}^{nz-1}={e}_{G}$. Since $G$ is infinite^{} and $|g|=|\u27e8g\u27e9|=|G|$ must be infinity^{}, $nz-1=0$. Since $nz=1$ and $n$ and $z$ are integers, either $n=z=1$ or $n=z=-1$. It follows that the only generators of $G$ are $g$ and ${g}^{-1}$.

A finite cyclic group of order $n$ has exactly $\phi (n)$ generators, where $\phi $ is the Euler totient function. To see this, let $G$ be a finite cyclic group of order $n$ and $g$ be a generator of $G$. Then $|g|=|\u27e8g\u27e9|=|G|=n$. Let $z\in \mathbb{Z}$ such that ${g}^{z}$ is a generator of $G$. By the division algorithm^{}, there exist $q,r\in \mathbb{Z}$ with $$ such that $z=qn+r$. Thus, ${g}^{z}={g}^{qn+r}={g}^{qn}{g}^{r}={({g}^{n})}^{q}{g}^{r}={({e}_{G})}^{q}{g}^{r}={e}_{G}{g}^{r}={g}^{r}$. Since ${g}^{r}$ is a generator of $G$, it must be the case that $\u27e8{g}^{r}\u27e9=G$. Thus, $n=|G|=|\u27e8{g}^{r}\u27e9|=|{g}^{r}|={\displaystyle \frac{|g|}{\mathrm{gcd}(r,|g|)}}={\displaystyle \frac{n}{\mathrm{gcd}(r,n)}}$. Therefore, $\mathrm{gcd}(r,n)=1$, and the result follows.

Title | generator |
---|---|

Canonical name | Generator |

Date of creation | 2013-03-22 13:30:39 |

Last modified on | 2013-03-22 13:30:39 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 10 |

Author | Wkbj79 (1863) |

Entry type | Definition |

Classification | msc 20A05 |

Related topic | GeneratingSetOfAGroup |

Related topic | ProperGeneratorTheorem |