generator for the mutiplicative group of a field

Proposition 1

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

Theorem 3.1 in the finite fields (http://planetmath.org/FiniteField) entry provesthis proposition along with a more general result:

Proposition 2

If for every natural number $d$ , the equation $x^{d}=1$ has at most $d$ solutions in a finite group $G$ then $G$ is cyclic. Equivalently, for any positive divisor $d$ 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 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 $z_{i}^{{p_{i}}^{e_{i}-1}}=y_{i}^{(q-1)/p_{i}}\eq 1$ .
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 $K^{*}$ .

单词	GeneratorForTheMutiplicativeGroupOfAField
释义	generator for the mutiplicative group of a field Proposition 1 The multiplicative group $K^{}$ of a finite field $K$ is cyclic. Theorem 3.1 in the finite fields (http://planetmath.org/FiniteField) entry provesthis proposition along with a more general result: Proposition 2 If for every natural number $d$ , the equation $x^{d}=1$ has at most $d$ solutions in a finite group $G$ then $G$ is cyclic. Equivalently, for any positive divisor $d$ 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 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 $z_{i}^{{p_{i}}^{e_{i}-1}}=y_{i}^{(q-1)/p_{i}}\eq 1$ . 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 $K^{*}$ .
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