Cyclotomic extension of $K$, $Gal_{K}{F}$ is isomorphic to a subgroup of $\mathbb Z_n^*$

Question

Let $n$ be a positive integer, $K$ be a field with $charK$ does not divide $n$, and $F$ be the cyclotomic extension of $K$ of order $n$.

Theorem says that $Gal_{K}{F}$ is isomorphic to a subgroup of $\mathbb Z_n^*$, i.e., the multiplicative group of units of $\mathbb Z_n$.

I don't know why $Gal_{K}{F}$ only isomorphic to a subgroup of $\mathbb Z_n^*$, but not necessarily the whole group. I know that for $K=\mathbb Q$, $Gal_{\mathbb Q}{F} \cong \mathbb Z_n^*$, but is there any field $K$ such that $Gal_{K}{F}$ is only a proper subgroup of $\mathbb Z_n^*$?

Answer 1

By "cyclotomic extension of order $n$," you mean $K(\zeta)$ for a primitive $n$th root of unity. What happens if $K$ already has some $d$th roots of unity for divisors nontrivial $d\mid n$? Indeed, in the most extreme case, if $K$ already has all of the $n$th roots of unity, then $K(\zeta)/K$ is trivial so ${\rm Gal}$ is the trivial subgroup of $U(n)=(\Bbb Z/n\Bbb Z)^\times$.

Answer 2

Yes. $K$ may intersect non-trivially with a cyclotomic extension of $\Bbb{Q}$. The first example that comes to mind is $n=5$, $K=\Bbb{Q}(\sqrt{5})$. Here $K\subseteq\Bbb{Q}(\zeta_5)$. Or more precisely, $K=\Bbb{Q}(\zeta_5)\cap \Bbb{R}$. So $F=K(\zeta_5)$ is only a degree two extension of $K$, and the Galois group is cyclic of order two. The complex conjugation in $F$ is its only non-trivial automorphism, so $Gal(K/F)=\{\overline{1},\overline{-1}\}\le\Bbb{Z}_5^*$.