Suppose given a finite extension $L/K$ of number fields. I would like to develop a better intuition for the Verlagerung giving an embedding $Ver : Gal(K^{ab}/K) \to Gal(L^{ab}/L)$. 


For example, let $L'/L$ be a finite abelian (Galois) extension. Define $\phi:Gal(K^{ab}/K)\to Gal(L'/L)$ to be the composition of the Verlagerung $Gal(K^{ab}/K) \to Gal(L^{ab}/L)$ followed by the projection $Gal(L^{ab}/L) \to Gal(L'/L)$. Then there is a finite, abelian (Galois) extension $K' / K$ such that $Gal(K'/K) \cong Gal(K^{ab}/K)/Ker\phi$.


Is it true that $K' \subset L'$ or even $K' = K^{ab}\cap L'$? (we think of every field being embedded into $L^{ab}$)


Further I would be happy about literature dealing with the Verlagerung in this context.


Thank you very much in advance!


Summary:


As shown below, in the answer of GH plus comments, it is not alway true that $K' \subset L'$. But, if we restrict to strict ray class fields $L^{\mathfrak m}$ of $L$, for $\mathfrak m$ an ideal of $L$, one can show that for every $\mathfrak m$ there is an ideal $\widetilde {\mathfrak m}$ of $L$ with $\mathfrak m | \widetilde {\mathfrak m}$ such that $ K^{\widetilde{\mathfrak m}}$ $\subset L^{\widetilde{\mathfrak m}}$.


Moreover, it seems that it is generally not true (except for trivial cases) that $K^{\widetilde{\mathfrak m}} = L^{\widetilde{\mathfrak m}} \cap K^{ab}$.


But, nevertheless, the $K^{\mathfrak m}$ exhaust the maximal abelian extension $K^{ab}$ of $K$, i.e. $$\bigcup\_{\mathfrak m}K^{\mathfrak m} = \ \ \bigcup\_{\mathfrak m}K^{\widetilde{\mathfrak m}} \ = \ K^{ab}$$



Verlagerung made "explicit"
I am no expert, but let me share an idea. For a number field $M$ let us denote by $C\_M$ the idele class group of $M$. By class field theory, $K' \subset L'$ is the same as $N\_{L'/K}(C\_{L'})\subset N\_{K'/K}(C\_{K'})$, where $N$ stands for the norm map. Let us denote by $U$ the open subgroup $N\_{L'/L}(C\_{L'})$ of $C\_L$, and by $j$ the natural injection of $C\_K$ into $C\_L$. Then, by the transfer theorem, the previous relation can be rewritten as $N\_{L/K}(U)\subset j^{-1}(U)$, that is, $j(N\_{L/K}(U))\subset U$. For certain open subgroups $U$ of $C\_L$ this relation follows from ramification theory, and further local analysis might extend it to more general open subgroups.


**EDIT:** I think now that the conclusion $K' \subset L'$ fails when $L$ has two places $w$ and $w'$ above the same place $v$ of $K$ such that $L'/L$ is much more ramified at $w$ than at $w'$. Indeed, viewing $U$ as a subgroup of the ideles $J\_L$, this assumption implies that $U\_w:=U\cap L\_w^\times$ is much smaller than $U\_{w'}:=U\cap L\_{w'}^\times$. However, $j(N\_{L/K}(U))\subset U$ would imply that $N\_{L\_{w'}/K\_v}(U\_{w'})\subset U\_w$ which is false when $U\_w$ is sufficiently small in terms of $U\_{w'}$. Perhaps this argument can be rewritten in terms of the transfer map and ramification theory only, i.e. avoiding class field theory as a whole.


