generator for the mutiplicative group of a field
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 . There exists an element in such that is not root of , since the polynomial has degree less than . Let . We note that has order . In fact and .
Finally we choose the element . By the Theorem 1 here (http://planetmath.org/OrderOfElementsInFiniteGroups), we obtain that the order of is i.e. is a generator of the cyclic group .
|Title||generator for the mutiplicative group of a field|
|Date of creation||2013-03-22 16:53:17|
|Last modified on||2013-03-22 16:53:17|
|Last modified by||polarbear (3475)|