For each number field $K$ with place $\mathfrak{p}$, prove that $K_{\mathfrak{p}}^{\mathrm{ab}} = K_{\mathfrak{p}} K^{\mathrm{ab}}$ As described in the title, I want to show that for each number field $K$ with place $\mathfrak{p}$, $K_{\mathfrak{p}}^{\mathrm{ab}} = K_{\mathfrak{p}} K^{\mathrm{ab}}$. Here $K_{\mathfrak{p}}$ is the completion of $K$ with respect to the place $\mathfrak{p}$, and the superscript ${}^{\mathrm{ab}}$ means taking maximal abelian extension.
I was given the hint to use (5.6) and (5.8) (in Chapter VI) in Neukirch's book Algebraic number theory.
(5.6) If $L|K$ is an abelian extension and $\mathfrak{p}$ is a place of $K$, then the diagram
\begin{array}{c}
K_{\mathfrak{p}}^{\times} & \xrightarrow{(-, L_{\mathfrak{p}}|K_{\mathfrak{p}})}
) & \mathrm{Gal}(L_{\mathfrak{p}}|K_{\mathfrak{p}}) \\
 \downarrow{[-]} & & \downarrow{\mathrm{incl.}} \\
 C_K = \mathbb{I}_K/K^{\times} & \xrightarrow{\varphi} & \mathrm{Gal}(L|K)
\end{array}
is commutative. Here the horizontal maps are norm residue symbols, and the left vertical one is the canonial injection $$[-]: K_{\mathfrak{p}}^{\times} \rightarrow C_K; \quad\quad a_{\mathfrak p} \mapsto \text{ the class of }(1, \ldots, 1, a_{\mathfrak p}, 1, \ldots, 1).$$
(5.8): For every finite abelian extension, one has $\mathcal{N}(C_L) \cap K_{\mathfrak{p}}^{\times} = \mathcal{N}_{\mathfrak{p}} L_{\mathfrak{p}}^{\times}$. Here we identify $K_{\mathfrak{p}}^{\times}$ with its image in $C_K$ under the map $[-]$. We use the abbreviations $\mathcal{N} = N_{L|K}$ and $\mathcal{N}_{\mathfrak{p}} = N_{L_{\mathfrak{p}} | K_{\mathfrak{p}}}$.
The two propositions provide the local-global compatibility of class field theories.
My question is how to prove $K_{\mathfrak{p}}^{\mathrm{ab}} = K_{\mathfrak{p}} K^{\mathrm{ab}}$.
My attempts: After discussing with a friend of mine, we have a vague idea: We start from any norm group $N$ (i.e. open normal subgroups of finite index) in $K_{\mathfrak{p}}^{\times}$, then via the local existence theorem, there exists an extension of $K_{\mathfrak{p}}$ with norm group $N$. Now applying $[-]$ to the global, we obtain a norm group in the global $C_K$, which by global existence theorem corresponds to an extension $L$ of $K$. Then maybe we can use (5.8) to say that the local field correspond to $N$ is the completion $L_{\mathfrak p}$. Then somehow we have a correspondence between the abelian extensions of $K_{\mathfrak p}$ and the abelian extensions of $K$. But we don't know how to carry on (or use (5.6)). Meanwhile, the above attempts are quite vague at some points. So how to clean this up?
Thank you all for answering and commenting! I'm a new learner on class field theory and sorry for such a possiblely trivial question.

By the way, I'm still quite puzzled how to really play with CFT well. Though I have gone through the proof of CFT via Milne's note, it seems that I have learnt nothing except calculating various cohomology groups and chasing on diagrams (Of course, Milne's note is awesome! It is only the silly of me). When given an algebraic number theory problem, I still cannot handle them with simply the statements of CFT. It makes me feel quite frastrated. :(

 A: The idea isn’t too far off, but you need to note that $[-]$ is far from being surjective, so it won’t map norm groups to norm groups… I really hope I got this right.
Clearly, if $L/K$ is finite abelian, any completion of $L$ at a place above $\mathfrak{p}$ is abelian over $K_{\mathfrak{p}}$. So what remains to be proved is that, for any finite abelian $L/K_{\mathfrak{p}}$, there exists an abelian extension $E/K$ such that $L \subset EK_{\mathfrak{p}}$.
Passing to norm subgroups, and using (5.8), what needs to be shown is that for every norm subgroup $N \leq K_{\mathfrak{p}}^{\times}$ (corresponding to $L$) there exists a finite index open subgroup $N’ \leq C_K$ such that $[-]^{-1}(N’) \subset N$. In other words, we have to show that there exists a finite index open subgroup $K^{\times} \leq N’ \leq \mathbb{A}_K^{\times}$ such that $[-]^{-1}(N’) \subset N$.
Assume $N \supset N_1 \cap N_2$ where we know the solution for $N_1$ and $N_2$ – ie we know satisfactory $N’_1$ and $N’_2$. Then $N’=N’_1 \cap N’_2$ clearly works for $N$.
Let $\pi \in K$ be such that $(\pi)$ is a nontrivial power of $\mathfrak{p}$. We know that $N$ contains $\{u\pi^n,\,  u \in 1+\mathfrak{p}^{\alpha}, \beta|n\}$ for $\alpha >0$, $\beta >0$. Let $N_1=\pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$, $N_2=\mathcal{O}_{K_{\mathfrak{p}}}^{\times}\pi^{\beta \mathbb{Z}}$. Then $N \supset N_1 \cap N_2$.
$N_2$ is the norm group of the non-ramified extension $K_{\mathfrak{p}}(\mu_{q^{\nu\beta}-1})$ (where $\pi$ has valuation $\nu$, the order of $\mathfrak{p}$ in the class group and $q$ is the cardinal of the residue field at $\mathfrak{p}$) which clearly comes from a global extension (with the associated $N’_2$). So it’s enough to solve the case $N=N_1$.
In other words, we want to construct a finite index open subgroup $N’$ of $\mathbb{A}_K^{\times}$ containing $K^{\times}$ such that its only elements with support in $K_{\mathfrak{p}}$ are in $\pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$.
Let $S$ be a finite set of finite places of $K$ containing $\mathfrak{p}$, $S’$ the other finite places, $S_{\infty}$ the infinite places. We will take $N’=K^{\times}F$, where $F=\prod_{v \in S_{\infty}}{K_v^{\times}} \prod_{v \in S’}{\mathcal{O}_{K_v}^{\times}}\prod_{v \in S}{H_v}$ where $H_v \subset \mathcal{O}_{K_v}^{\times}$, which is an open subgroup (then $N’$ is open and finite index essentially because the $\mathcal{O}_v^{\times}$ are compact and the class group is finite).
To show the statement, we need to see that $H_{\mathfrak{p}} \subset (1+\mathfrak{p}^{\alpha})$ and if $x \in K^{\times}$ is in $H_v$ for all $v \in S_0 :=S \backslash \{\mathfrak{p}\}$, then $x \in \pi^{\mathbb{Z}}(1+\mathfrak{p})^{\alpha})$.
Now by the assumption on $\pi$, $\mathcal{O}_K^{\mathfrak{p},\times}$ (the superscript means that we require invertibility everywhere but at $\mathfrak{p}$) is the direct product $\pi^{\mathbb{Z}}\times \mathcal{O}_K^{\times}$. So we want to show that if $x\in \pi^{\mathbb{Z}}\times \mathcal{O}_K^{\times} $ satisfies a certain “non-$\mathfrak{p}$-congruence then $x \in \pi^{\mathbb{Z}}(1+\mathfrak{p}^{\alpha})$.
It is enough to show that, for some integer $f$ (the index of the right subgroup), a non-$\mathfrak{p}$ congruence condition can cut out the set $fG$ of $f$-powers in $G=\pi^{\mathbb{Z}} \times \mathcal{O}_K^{\times}$.
Now, let $P$ be the product of all the unit groups of the residual fields at finite places not $\mathfrak{p}$. By (*) below, $G \rightarrow P$ is injective with a cokernel without $f$-torsion, ie $G/fG \rightarrow P/fP$ is injective, and thus ($G$ has finite rank) there is an injection $G/fG$ to $P’/fP’$ where $P’$ is the product of finitely many multiplicative groups of residue fields at $v\neq\mathfrak{p}$ — and this concludes.
Lemma(*): Let $S$ be a cofinite set of finite places of a number field $K$ and $x \in K^{\times}, f \geq 1$. Assume that $x$ is a $f$-power in every residue field at a place of $S$ – then $x$ is a $f$-th power in $K$.
Proof: We can always assume that $S$ contains only places coprime to $f$ and $x$.
Let $L=K(e^{2i\pi/f},x^{1/f})$. Then $L/K$ is Galois. Each place $w$ of $L$ above some $v \in S$ has a well-defined Frobenius in $Gal(L/K)$ acting on the residue field at $w$, and it fixes one of the roots mod $w$ of the minimal polynomial $\mu$ of $x^{1/f}$. As $L/K$ is unramified at $w$, it means that this Frobenius (acting on the global roots of $\mu$) has a fixed point. By Cebotarev, every element of $Gal(L/K)$ fixes one of the roots of $\mu$. But $Gal(L/K)$ acts transitively on them – and by Burnside lemma it follows that $\mu$ has only a root and $L=K$.
