TITLE: Subgroups of Index $2$ of $(\mathbb{Z}_{2})^{\aleph_{0}}$
QUESTION [5 upvotes]: I am studying infinite Galois theory and I proved that if 
$$
L=\mathbb{Q}(\sqrt{p}:\,\text{$p$ is prime )}
$$
That is $L$ is the composition of all fields of the form $$\mathbb{Q}(\sqrt{p})$$ where $p$ is a prime
then
$$
Gal(L/\mathbb{Q})\cong(\mathbb{Z}_{2})^{\aleph_{0}}
$$
The notes I read claim that there are only $\aleph_{0}$ subextensions
$E$ s.t 
$$
\mathbb{Q\subseteq E\subseteq L}
$$
and 
$$
[E:\mathbb{Q}]=2
$$
but $\aleph$ subgroups of index $2$ for $(\mathbb{Z}_{2})^{\aleph_{0}}$.
The first part is clear since $L$ is countable and each such extension
is of the form $\mathbb{E=Q}(\sqrt{\alpha})$ for $\alpha\in L$.
Can someone help me understand why $(\mathbb{Z}_{2})^{\aleph_{0}}$
have $\aleph$ subgroups of index $2$ ?

REPLY [3 votes]: Take any non-empty set $X$ of $\mathbb{N}$ forms :
$$e_X:=(\chi_X(x))_{x\in\mathbb{N}}\in(\mathbb{Z}_2)^{\mathbb{N}} $$
Here $\chi_X(x)$ is $0$ if $x\notin X$ and $1$ if $x\in X$. 
Each $e_X$ allows you to construct a dual form $e_X^*$. Denote the kernel of each $X$ : $K_X$. 
Since $K_X$ is the kernel of a non-trivial group morphism from $(\mathbb{Z}_2)^{\mathbb{N}}$ to $\mathbb{Z}_2$, it is of index $2$.
Basically what you need to show is that $X\mapsto K_X$ is injective. Then you are done.