Completion of a Galois Extension $K | \mathbb{Q}$ w.r.t. to a non archimedean abs. value is a Galois extension As the title suggests, I'm interested in a proof of the following fact:

Let $K | \mathbb{Q}$ be a finite Galois extension, and let $\mathfrak{p}$ be a prime ideal of the ring of integers of $K$. Using the norm we can define a non-archimedean absolute value, and we can consider the completion of $K$ w.r.t this absolute value, call it $K_{\mathfrak{p}}$. Then $K_{\mathfrak{p}}|\mathbb{Q}_p$ is Galois.

Clearly $\mathfrak{p}\mid (p)$.
I've seen this fact used in some proportions (here, prop 7.7 for example) but I can't find a proof of it. 
Can someone give me some details-hints about how to do that?
 A: Let $L/K$ be any finite Galois extension of number fields, $G =\mathrm{Gal}(L/K)$, and let $\mathfrak P$ be a prime of $\mathcal O_L$ lying above a prime $\mathfrak p$ of $\mathcal O_K$. Then $G$ acts transitively on the set of primes of $\mathcal O_L$ lying above $\mathfrak p$ (see Thm 2.2. here).
Define the decomposition group$$D_{\mathfrak P/\mathfrak p}:=\{\sigma\in G :\sigma(\mathfrak P) = \mathfrak P\}$$
to be the subgroup of $G$ which fixes $\mathfrak P$.  We have $\#D_{\mathfrak P/\mathfrak p} = e_{\mathfrak P/\mathfrak p}f_{\mathfrak P/\mathfrak p}$.

Theorem: $L_\mathfrak P/K_\mathfrak p$ is a Galois extension with Galois group $D_{\mathfrak P/\mathfrak p}$.

Proof: Since $D_{\mathfrak P/\mathfrak p}$ fixes $\mathfrak P$, it fixes $|\cdot|_\mathfrak P$, and we get a map $$D_{\mathfrak P/\mathfrak p}\to\mathrm{Aut}_{K_\mathfrak p}(L_\mathfrak P)$$
which is injective as $L\hookrightarrow L_\mathfrak P$. Let $D$ be the image of this map. Then by Galois theory, $L_\mathfrak P/L_\mathfrak P^D$ is a Galois extension. But $$[L_\mathfrak P:L_\mathfrak P^D]=\#D = [L_\mathfrak P:K_\mathfrak p]$$so $$K_\mathfrak p =L_\mathfrak P^D.$$
