# 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 $q-1=\prod_{i=1}^{n}p_{i}^{e_{i}}$. 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 $z_{i}^{{p_{i}}^{e_{i}-1}}=y_{i}^{(q-1)/p_{i}}\neq 1$.
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 GeneratorForTheMutiplicativeGroupOfAField 2013-03-22 16:53:17 2013-03-22 16:53:17 polarbear (3475) polarbear (3475) 16 polarbear (3475) Result msc 11T99 msc 12E20