Let $F$ be a field, then the polynomial $x^n - 1$ has $n$ roots in $F$ if $F$ contains a multiplicative subgroup of order $n$.

Question

Is this following true? If $F$ is a field, then the polynomial $x^n - 1$ has $n$ roots in $F$ whenever $F$ contains a multiplicative subgroup of order $n$.

Answer 1

Yes. The order of every element of this subgroup divides $n$. This gives $n$ distinct roots of the polynomial $x^n-1$. Since $F$ is a field, there can't be more roots, for degree reasons.

Answer 2

Yes, it's true and the proof requires only the special case of Lagrange's Theorem that I like to call Lagrange's Little Theorem:

In a finite commutative group, the order of every element divides the order of the group.

Answer 3

Any finite subgroup of the multiplicative group of a field is cyclic, thus the given data means $\,\Bbb F\,$ contains a cyclic group $\,G\,$ of order $\,n\,$, and thus

$$a^n=1\;,\;\;\forall\,a\in G\,$$

and the above means that every $\,a\in G\,$ is a root of $\,x^n-1=0\,$ . Since a polynomial of degree $\,n\,$ over a field cannot have more than $\,n\,$ distinct roots, we have thus found them all!