I am trying to answer this question: Suppose $p$ is an odd prime and $K/\mathbb{Q}$ is the extension of $\mathbb{Q}$ obtained by adjoining a primitive $p$th root of 1 in $\mathbb{C}$. (a) Show that $K$ contains a unique quadratic extension $F$ of $\mathbb{Q}.$ (b) What is Gal$(K/F)$?
For part (a), I can let $\zeta_p$ be a primitive $p$th root of 1. Then $K=\mathbb{Q}(\zeta_p).$ I know that Gal$(K/\mathbb{Q})=(\mathbb{Z}/p\mathbb{Z})^{\times},$ which is a cyclic group of order $p-1,$ which is even. Thus 2 divides $p-1$, and Gal$(K/\mathbb{Q})$ has a unique subgroup of order 2. By the Galois correspondence, this implies $K$ contains a unique quadratic extension $F$ of $\mathbb{Q}$. (b) I am not sure how to go about finding Gal$(K/F).$