Let $G=\operatorname{Gal}(L/K)$ and let
$$
M=L^{[G,G]}
$$
be the fixed field of the commutator subgroup. Then $M/K$ is the maximal abelian subextension of $L/K$, with
$$
\operatorname{Gal}(M/K)\cong G^{\mathrm{ab}}.
$$
The <nonabelian norm-residue kernel theorem> identifies the kernel of the global reciprocity map
$$
C_K\longrightarrow G^{\mathrm{ab}}
$$
with $N_{L/K}C_L$. Applied to the abelian extension $M/K$, the <Artin reciprocity law> identifies the kernel of the same map with $N_{M/K}C_M$. Therefore
$$
N_{L/K}C_L=N_{M/K}C_M,
$$
so the norm group of the Galois extension $L/K$ is the norm group of the abelian extension $M/K$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 123 / 4 / 2 / d / Solution

For a finite Galois extension $L/K$ with group $G$, the kernel of the global reciprocity map $C_K\to G^{\mathrm{ab}}$ is $N_{L/K}C_L$. It follows that this norm group equals the norm group from the maximal abelian subextension $L^{[G,G]}/K$.


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