Galois group of $\mathbb{Q}(\zeta_p)/\mathbb{Q}(\sqrt{p})$ I just started to study Galois theory and so I'm not too good with calculating Galois groups, I know that $\operatorname{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})=(\mathbb{Z}/p\mathbb{Z})^\times$ but what about the Galois group of $\mathbb{Q}(\zeta_p)/\mathbb{Q}(\sqrt{p})$? In particular I'm interested in the case $p=13$.
I can't find a solution online, maybe it's the same? Anyway what is it and how do you find it?
EDIT: Thinking about it given that $\operatorname{Gal}(\mathbb{\Bbb{Q}(\sqrt{13})}/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}$ it may be that the group I'm looking for is $\mathbb{Z}/6\mathbb{Z}$, am I right? And does this mean that the morphism are just $\zeta\to \zeta^k$ for $k\in \{1,\dots, 6\}$ (all in $\pmod{13}$ )?
 A: The fundamental theorem of Galois theory gives an order-reversing bijection between the subgroups of $\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q})$ and the subfields of $\Bbb{Q}(\zeta_p)$. In particular the subfield $\Bbb{Q}(\sqrt{p})$ of degree $2$ corresponds to a subgroup $H\subset(\Bbb{Z}/p\Bbb{Z})^{\times}$ of index $2$, and then
$$\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q}(\sqrt{p}))=H.$$
Of course $(\Bbb{Z}/p\Bbb{Z})^{\times}$ is cyclic of even order, so it has precisely one subgroup of index $2$.

More explicitly, the canonical isomorphism $\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q})\cong(\Bbb{Z}/p\Bbb{Z})^{\times}$ is given by
$$(\Bbb{Z}/p\Bbb{Z})^{\times}\ \longrightarrow\ \operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q}):\ k\ \longmapsto\ (\zeta_p\ \mapsto\ \zeta_p^k).$$
The unique subgroup of index $2$ in the multiplicative abelian group $(\Bbb{Z}/p\Bbb{Z})^{\times}$ is the subgroup of all squares, i.e. (nonzero) quadratic residues mod $p$. This shows that
$$\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q}(\sqrt{p}))=\left\{(\zeta_p\ \mapsto\ \zeta_p^{k^2}):\ k\in(\Bbb{Z}/p\Bbb{Z})^{\times}\right\}.$$
For small values of $p$, such as $p=13$, you can simply list all six nonzero quadratic residues, and compute the minimal polynomial of $\zeta_p$ over $\Bbb{Q}(\sqrt{p})$ as the product
$$f=\prod_{\sigma\in H}(X-\sigma(\zeta_p)),$$
where $H:=\operatorname{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q}(\sqrt{p}))$.
