Multiplicative group of a field contains maximal n-1 elements with order n

Question

Let $F$ be a field and $n\in \mathbb N,n>1$.

I want to show that the multiplicative group $K$\ $\{0\}$ contains maximal $n-1$ elements with order $n$.

I actually don't have any ideas how to solve this one. So I will be thankful for any kind of help (ideas at first and if it does not help me, then full solution...)

Thanks !:)

Answer 1

If char$\,K\mid n\;$, say $\;n=pr\;,\;\;p\;$ a prime, then $\;x^n-1=x^{pr}-1=(x^r-1)^p\;$ and this polynomial has at most $\;r<n\;$ different roots in $\;K^*\;$ and we're done.

Otherwise, define

$$W_n:=\left\{\;a\in K\;;\;w^n=1\;,\;\;n\in\Bbb N\;\right\}$$

It's easy to check $\;W_n\;$ is a subgroup of $\;K^*\;$ , and since it is finite it is then cyclic.

Since we know that there are exactly $\;\varphi(n)\;$ generators of this subgroup ,$\;\varphi=$ Euler's Function, we now get what we want by observing that $\;\varphi(n)\le n-1\;\;\forall\,n\in\Bbb N\;$