$G$ has a cyclic normal subgroup $H$

Question

Please help me to prove the following result:

$\newcommand{\gal}{\operatorname{Gal}}$

Let $F\subset E$ be a Galois extension and assume that $E=F(\alpha)$ with $\alpha^n\in F$. Let $$G=\gal(E/F)$$.

Prove that $G$ has a cyclic normal subgroup $H$ of order dividing $n$ such that $G/H$ is abelian and has order dividing $\phi(n)$.

Thanks for help.

Answer 1

Hint: $E$ contains all $n$-th roots of unity because $E/F$ is Galois and $\alpha^n\in F$. Note that some of these roots may already be in $F$.

Answer 2

I think lhf already gave an excellent hint. Anymore on that is to give you almost the whole answer:

So $\;E\;$ contains all the roots of unity of order $\;n\;$ . Whether some of these roots were already in $\;F\;$ or not, $\;E\;$ contains all the roots of an irreducible polynomial (over $\;F\;$) dividing $\;x^n-1\in F[x]\;$ . We know the set of automorphisms (over $\;F\;$) permuting these roots is a cyclic one.

Now use the Fundamental Theorem of Galois Theory to deduce that the field that fixes these roots must be normal and, of course, corresponds to a normal cyclic (since any subgroup of a cyclic group is cyclic) subgroup of $\;G\;$ .

Finally, just remember that the degree of the cyclotomic polynomial of order $\;n\;$ is $\;\phi(n)\;$, which also gives the order of the corresponding extension, and that taking into account that some of those roots of unity may already be in $\;F\;$ , we then must get a divisor of $\;\phi(n)\;$ as the order of the corresponding group (or of the corresponding subextension).