$\def\cU{{\cal U}} \let\lra\longrightarrow$ I'm studying the Hazewinkel article "Local class field is easy" (a bit presomptuous as far as i'm concerned !). Readable here https://www.sciencedirect.com/science/article/pii/0001870875901565
$K$ is a local field, $L,E/K$ totally ramified abelian. $\pi_L,\pi_E$ uniformizers for $L,E$ and $\pi_K^L=N_L(\pi_L)$ (id for $E$) uniformizers for $K$.
$K_n/K$ the unramified of degree $n$.
A morphism $\theta_L$ (may be a local Artin ) was constructed $\theta_L:\cU_K\lra Gal(L/K)$ which is onto, with kernel $N_L(\cU_L)$. idem for $E$
Now we define $\rho_L:K^\times\lra Gal(K_nL/K)$ as
$\rho_L(u)=\theta_L(1/u)$ for $u\in\cU_K$ and $\rho_L(\pi_K^L)=Frob$ the Frobenius $\in Gal(K_nL/L)$. Which is sufficiant because $K^\times=<\pi_K^L>\cU_K$.
Same for $E$.
Now it's assumed $K_nE=K_nL$ (that is the trick !).
It has been proved that $N_E(\cU_E)$ is the kernel of $K^\times\stackrel{\rho_L}\lra Gal(K_nL/K)\lra Gal(E/K)$.
I have to prove $\rho_L=\rho_E$
so $\rho$ does not depend on the totally ramified extension. Hazewinkel gives no hint just write it (p. 173) as corollary (without proof) of the last result about the kernel.
I have NO idea at all !
Because of the common part (the $Frob$ on $K_n/K$) i was able to show $\rho_L(\pi_K^L)=\rho_E(\pi_K^L)$ (and i was very proud of that !) but i have no idea for the $\cU$ part.
If i understood well the problem : As $Gal(K_nL/K)=Gal(K_n/K)\times Gal(L/K)$ (because $K_n\cap L=K$) we have
$\rho_L(u)=(id,\theta_L(u))$ then i restrict this application to $E$ and this one is $=\rho_E(u)=\theta_E(u)$!
If someone has read or knows the article or recognize the property ...
Thanks for each tiny hint