Inverse limit of Galois groups of a finite Galois extension $K/F$ let $K/F$ be a finite Galois extension. Suppose $Gal(K/F)$ is abelian.
Then we know that there are finite number of subfields $F=F_0\subseteq F_1\subseteq\cdots \subseteq F_n=K$.
Then $Gal(F/F)\geq Gal(F_1/F) \geq \cdots \geq Gal(K/F)$. This gives an inverse system.
So my problem is 

What is the $\varprojlim Gal(F_i/F)$.?

I think, $Gal(K/F)=\varprojlim Gal(F_i/F)$. But I am unable to write down a rigorous proof. Any help is appreciated.
 A: It's easy to verify from the definition, that for any category $\mathcal C$, if the base category $J$ of a diagram $D:J\to\mathcal C$ has an initial object $0$, then
$$\overleftarrow\lim D\ =\ D(0)\,.$$
In particular, the limit of a left-closed chain (in particular, of a finite chain) $A_0\to A_1\to\dots $ is always $A_0$.
A: Firstly, a correction. If $F \subset F_i \subset K$ is an intermediary field extension which is normal, you don't have an inverse system of subgroups. Instead, you have an inverse system of quotients: $$\text{Gal}(K/F) \overset{\varphi_i}{\to} \text{Gal}(F_i/F) \to \text{Gal}(F/F) = \{\text{Id}\}\,.$$
The quotient application $\varphi_i\colon\text{Gal}(K/F) \to \text{Gal}(F_i/F)$ is given by killing out the subgroup of all automorphisms of $K$ leaving $F_i$ fixed, the subgroup $\text{Gal}(K/F_i) \leq \text{Gal}(K/F)$. For an automorphism $\sigma \in \text{Gal}(K/F)$, it's image $\varphi_i(\sigma) = \sigma_i\colon F_i \to F_i$ is simply the restriction of $\sigma$ to $F_i$ (because the extension is a normal extension, it's image is still contained in $F_i$).
Given that $\text{Gal}(K/F)$ is abelian, every intermediary extension is normal, so that's not a worry. Your inverse limit is: $$\varprojlim_{F \subseteq F_i \subseteq K} \text{Gal}(F_i/F) = \left\{\left.(\sigma_i) \in \prod_{F\subset F_i\subseteq K} \text{Gal}(F_i/F) \right| \sigma_i(x) = \sigma_j(x) \text{, for all } x \in F_j\text{ and all fields } F_j \subseteq F_i\right\}
$$
There is a natural homomorphism $\phi\colon \text{Gal}(K/F) \to \prod_{F\subseteq F_i\subseteq K}\text{Gal}(F_i/F)$ sending $\sigma$ to $(\varphi_i(\sigma)) = (\sigma_i)$. Should be a straightfoward computation to check that it's image lands on $\varprojlim_{F\subseteq F_i\subseteq K} \text{Gal}(F_i/F)$, and that it's bijective over that set.
