I'm reading Neukirch's book "Class Field Theory" (from 1980, not the newly translated one including cohomology) and there's one statement that I can't wrap my mind around, and I suspect it stems from some "easy" fact I'm missing.
Let me set up some notation. $G$ is a profinite group and we will denote the closed subgroups of $G$ by $G_K$ and call the indices $K$ for fields. We say that $K$ is the fixed field of $G_K.$ We write $K \subset L$ if $G_L \subset G_K$ and we call $K \subset L$ a field extension. We say that it is a finite extension if $G_L$ is of finite index in $G_K$ and we say that it is Galois if $G_L$ is normal in $G_K.$ Further, let us say that we have a continuous multiplicative $G$-module $A.$ We mean with this a multiplicative abelian group $A$ on which $\sigma \in G$ act as automorphisms on the right and they act continuously. We let $A_K = \{a \in A | a^\sigma = a \forall \sigma\in G_K\}.$Clearly we then have that $A_K \subset A_L$ for any field extension $K \subset L.$ If $$K \subset L$$ is finite, then we have a norm map $$N{L|K}:A_L \rightarrow A_K,$$ $$N_{L|K}(a) = \Pi_\sigma a^\sigma$$ where $\sigma$ varies over a system of representatives of $G_K / G_L.$
Now, Neukirch claims on pg. 23 of this book: http://tomlr.free.fr/Math%E9matiques/Neukirch%20-%20Class%20Field%20Theory.pdf the proof of lemma 2.4 that if $K \subset L \subset L'$ are finite field extensions and if a congruence holds mod $N_{L'|K}(A_{L'})$ then a fortiori it holds mod $N_{L|K}(A_L).$ But this seems strange to me, but I suspect I'm making an easy mistake. We have that $A_L \subset A_{L'}$so it seems that if it holds mod $N_{L'|K}A_{L'}$ there's no reason for it to hold $mod N_{L|K} A_L.$ Surely it holds if $N_{L'|L}$ is surjective, since $N_{L'|K} = N_{L|K} N_{L'|L},$ but otherwise?
Any help would be most welcome!
The reason follows from the fact you note, that $N_{L'|K}=N_{L|K}N_{L'|L}$. Since $N_{L'|L}(A_{L'})\subseteq A_L$, we will have $N_{L'|K}(A_{L'})\subseteq N_{L/K}(A_L)$. Now two elements of $A_K$ are congruent mod $N_{L'|K}(A_{L'})$ if their difference lies in $N_{L'|K}(A_{L'})$. If this is the case, then the difference is also an element of the bigger set $N_{L|K}(A_L)$, which means that the two elements are congruent mod $N_{L|K}(A_L)$.