# generator for the mutiplicative group of a field

###### Proposition 1

The multiplicative group^{} ${K}^{\mathrm{*}}$ of a finite field^{} $K$ is cyclic.

Theorem 3.1 in the finite fields (http://planetmath.org/FiniteField) entry proves
this proposition^{} along with a more general result:

###### Proposition 2

If for every natural number^{} $d$, the equation ${x}^{d}\mathrm{=}\mathrm{1}$ has at most $d$ solutions in a finite group^{} $G$ then $G$ is cyclic. Equivalently, for any positive divisor^{} $d$ of $\mathrm{|}G\mathrm{|}$.

This last proposition implies that every finite subgroup of the multiplicative group of a field (finite or not) is cyclic.

We will give an alternative constructive proof^{} of Proposition 1:

We first factorize $q-1={\prod}_{i=1}^{n}{p}_{i}^{{e}_{i}}$. There exists an element ${y}_{i}$ in ${K}^{*}$ such that ${y}_{i}$ is not root of ${x}^{(q-1)/{p}_{i}}-1$, since the polynomial^{} has degree less than $q-1$ . Let ${z}_{i}={y}_{i}^{(q-1)/{p}_{i}^{{e}_{i}}}$. We note that ${z}_{i}$ has order ${p}_{i}^{{e}_{i}}$. In fact ${z}_{i}^{p_{i}{}^{{e}_{i}}}=1$ and ${z}_{i}^{p_{i}{}^{{e}_{i}-1}}={y}_{i}^{(q-1)/{p}_{i}}\ne 1$.

Finally we choose the element $z={\prod}_{i=1}^{n}{z}_{i}$. By the Theorem 1 here (http://planetmath.org/OrderOfElementsInFiniteGroups), we obtain that the order of $z$ is $q-1$ i.e. $z$ is a generator^{} of the cyclic group^{} ${K}^{*}$.

Title | generator for the mutiplicative group of a field |
---|---|

Canonical name | GeneratorForTheMutiplicativeGroupOfAField |

Date of creation | 2013-03-22 16:53:17 |

Last modified on | 2013-03-22 16:53:17 |

Owner | polarbear (3475) |

Last modified by | polarbear (3475) |

Numerical id | 16 |

Author | polarbear (3475) |

Entry type | Result |

Classification | msc 11T99 |

Classification | msc 12E20 |