Let $\mathbb{E} = \mathbb{Q}[\sqrt{p_1}, \dots , \sqrt{p_n}]$, with $p_1, \dots, p_n$ distinct prime integers. Show that the group of Galois of the extension $G = Gal(\mathbb{E} | \mathbb{Q})$ has at least $2^n - 1$ subgroups of index $2$.
The extension $\mathbb{E} | \mathbb{Q}$ is Galois of degree $2^n$ which is equal to the order of $G$. I can see, for any $i = 1, \dots , n$, $Gal(\mathbb{E} | \mathbb{Q}[\sqrt{p_i}])$ being a subgroup of order $2^{n-1}$ of $G$, thus of index $2$ for Lagrange. However it's far away from the $2^n-1$ claimed in the text. Any hint about it?
The Set of subgroups of index $2$ is in bijection with the set of degree-2 extensions $L/\Bbb Q$ such that $\Bbb Q \subset L \subset \Bbb Q[\sqrt p_1, \dots, \sqrt p_n]$. Any (non-empty) choice of a subset $I \subset \{1, \dots, n\}$ yields such a degree 2 extension via $L = \Bbb Q[\sqrt \pi]$, where $\pi = \prod_{i \in I} p_i$. On the other hand, given such an extension $L$, let $D$ be the discriminant of $L/\Bbb Q$. Now we can reconstruct $\pi$ via $$ \pi = \begin{cases} \prod_{p \mid D} p \quad & \text{if } 4 \nmid D \\ \prod_{4p \mid D} p \quad & \text{if } 4 \mid D.\end{cases} $$ By theory of quadratic number fields we find $L = \Bbb Q[\sqrt \pi]$, and as $L \subset \Bbb Q[\sqrt p_1, \dots, \sqrt p_n]$ we find that $\pi$ only has prime divisors among the $p_i$ (hence is of the form $\pi = \prod_{i \in I} p_i$ for some subset $I$).