Galois extension and quotient group of multiplicative groups I am struggling with the problem related to the Galois extension of fields.

Let $E/F$ be a finite Galois extension. Show that if the quotient group $E^\times / F^\times$ has an element of order $n$, then $E^\times$ has an element of order $n$.

There are several facts that I tried to use: (1) every finite subgroup of the multiplicative group of a field is cyclic; (2) by the primitive element theorem, $E/F$ is a simple extension.
However, I have no idea how to approach this problem.
Any ideas or comments are welcome.
 A: Suppose $E,F$ are fields with $F\subseteq E$ such that

*

*$E/F$ is a Galois extension.$\\[4pt]$

*The quotient group $E^\times / F^\times$ has an element of order $n$.

Our goal is to show that $\;E^\times$ has an element of order $n$.

Proof:

By assumption there is a nonzero element $w\in E$ such that

*

*$w^n\in F$.$\\[4pt]$

*There is no positive integer $m < n$ for which $w^m\in F$.

Let $a=w^n$, and let $f\in F[x]$ be given by $f=x^n-a$.

Let $g\in F[x]$ be the product of the distinct monic irreducible factors of $f$ in $F[x]$ having a root in $E$.

Since $E/F$ is a Galois extension, it follows that
$$
g=(x-r_1)\cdots (x-r_m)
$$
where

*

*$r_1,...,r_m\in E$.$\\[4pt]$

*$r_1,...,r_m$ are distinct.$\\[4pt]$

*$\{r_1,...,r_m\}=\{r\in E{\,\mid\,}f(r)=0\}$.

Since $g\in F[x]$, it follows that
$
{\displaystyle{
\prod_{i=1}^m r_i
}}
\in F$.

Since $g{\,\mid\,}f$, it follows that $m\le n$.

Claim:$\;m=n$.

Suppose instead that $m < n$.

For all $i\in\{1,...,m\}$ we have ${\large{\frac{w^2}{r_i}}}\in E$, and
$$
\Bigl(
\frac{w^2}{r_i}
\Bigr)
^n
=
\frac{w^{2n}}{r_i^n}
=
\frac{a^2}{a}
=
a
$$
hence ${\large{\frac{w^2}{r_i}}}\in\{r_1,...,r_m\}$.

Then the map from the set $\{r_1,...,r_m\}$ to itself defined by
$$
r_i
\mapsto
\frac{w^2}{r_i}
$$
is injective, hence bijective.

It follows that
\begin{align*}
&
\prod_{i=1}^m r_i
=
\prod_{i=1}^m \frac{w^2}{r_i}
\\[4pt]
\implies\;&
w^{2m}
=
\left(
\prod_{i=1}^m r_i
\right)^{\!2}
\\[4pt]
\implies\;&
w^m
=
\pm
\prod_{i=1}^m r_i
\\[4pt]
\implies\;&
w^m\in F
\\[4pt]
\end{align*}
contradiction, since $m < n$.

Hence $m=n$, as claimed.

Returning now to the main proof . . .

Let $h\in F[x]$ be given by $h=x^n-1$.

For each $i\in\{1,...,n\}$, let $s_i=r_i/w$.

Since $r_1,...,r_n\in E$, and $w\in E$, it follows that $s_1,...,s_n\in E$, and since $r_1,...,r_n$ are distinct, it follows that $s_1,...,s_n$ are distinct.

Since
$r_i^n=a$
for all $i$, and $w^n=a$, it follows that
$s_i^n=1$
for all $i$.

Thus each $s_i$ is a root of $h$, and since $s_1,...,s_n$ are distinct, it follows that
$s_1,...,s_n$
are all the roots of $h$.

Let $S=\{s_1,...,s_n\}$.

For all $i,j\in\{1,...,n\}$, we have
$$
(s_is_j)^n=s_i^ns_j^n=1
$$
so $S$ is closed under multiplication, hence $S$ is a finite subgroup of $E^\times$.

It follows that $S$ is a cyclic group, hence since $|S|=n$, there must be some $s_i\in S$ such that $s_i$ has order $n$.

This completes the proof.
