My question pertains to page 235 in Grillet's 'Abstract Algebra', Second Edition. The setup is as follows:
Let $R$ be a formally real field, such that
(i) every positive element of $R$ is a square in $R$
(ii) every polynomial of odd degree in $R[X]$ has a root in $R$
where 'formally real means that there is a total order relation on $R$ that makes it an ordered field.
From the algebraic closure of $R$, pick $i$ as a square root of $-1$ and consider
$$C := R(i)$$
It can be shown that every element of $C$ is a square root and that $C$ has no irreducible polynomial of degree 2 (which implies that $C$ has no extension of degree 2).
Wanting to show that $C$ is algebraically closed, pick some element $\alpha \in \overline C$.
Along with its $R$-conjugates, $\alpha$ generates a finite Galois extension $E$ of $C$, which is also a finite Galois extension of $R$.
Now it can be shown that $G:=\text{Gal}(E/R)$ is of even order. So we can pick a Sylow 2-subgroup $S$ of $G$ and consider its fixed field $F:=\text{Fix}_E(S)$.
I know that $E/F$ is Galois and that $S = \text{Gal}(E/F)$. However, Grillet states the equation
$$[F:R] = [G:S]$$
which I can only come up with, if $F/R$ is a Galois extension ($\iff \text{Gal}(E:F)$ is a normal subgroup of $G$), using that in such a case, $[F:R] = \text{Gal}(F/R) \cong \text{Gal}(E/R)\,/\,\text{Gal}(E/F)$.
But I don't know how to show this. Or is there another way to arrive at the above equation $[F:R]=[G:S]$?
Any help would be appreciated. I have 'self-studied' the basics of Galois theory for progressing with this theorem, so maybe I'm just missing some theorems/propositions here.
Note:
It's very simple, as Jef Laga pointed out in a comment on my question.
We have $$ |\text{Gal}(E/R)| = [E:R] = [E:F][F:R] \overset{E/F \,\text{finite, Galois}}= |\text{Gal}(E/F)|[F:R] $$
and hence
$$ [G:S] = [\text{Gal}(E/R) : \text{Gal}(E/F)] = \frac{|\text{Gal}(E/R)|}{|\text{Gal}(E/F)|} = [F : R] $$