generator for the mutiplicative group of a field

Proposition 1

The multiplicative groupMathworldPlanetmath K* of a finite fieldMathworldPlanetmath K is cyclic.

Theorem 3.1 in the finite fields ( entry proves this propositionPlanetmathPlanetmath along with a more general result:

Proposition 2

If for every natural numberMathworldPlanetmath d, the equation xd=1 has at most d solutions in a finite groupMathworldPlanetmath G then G is cyclic. Equivalently, for any positive divisorMathworldPlanetmathPlanetmath d of |G|.

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 proofMathworldPlanetmath of Proposition 1:
We first factorize q-1=i=1npiei. There exists an element yi in K* such that yi is not root of x(q-1)/pi-1, since the polynomialPlanetmathPlanetmath has degree less than q-1 . Let zi=yi(q-1)/piei. We note that zi has order piei. In fact zipiei=1 and zipiei-1=yi(q-1)/pi1.
Finally we choose the element z=i=1nzi. By the Theorem 1 here (, we obtain that the order of z is q-1 i.e. z is a generatorPlanetmathPlanetmathPlanetmath of the cyclic groupMathworldPlanetmath 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