Someone familiar with Neukirch's terminology can understand this post better. Unfortunately it is so much terminology to just explain it here.
My question is about what is marked in the picture:
Why is this map $\chi_a$ is a homomorphism? I see no apparent reason to have
(Recall that $\alpha^{\sigma-1}=\alpha^{\sigma}\alpha^{-1}$ by definition) \begin{equation} \alpha^{\sigma\tau}=\alpha^{\sigma}\alpha^{\tau}\alpha^{-1}. \end{equation}
Take $a\in \Delta$, and choose $\alpha$ such that $\wp(\alpha)=a$. Then by definition of $\wp$ as a $G$-morphism, you have for all $\tau\in G(L|K)$: $\wp(\tau(\alpha)) = \tau(\wp(\alpha)) = \tau(a) =a = \wp(\alpha)$. So $\wp(\alpha^{\tau-1}) = 1$, ie $\alpha^{\tau-1}\in \mu_\wp\subset K$. In particular, for all $\sigma\in G(L|K)$, $\sigma(\alpha^{\tau-1}) = \alpha^{\tau-1}$.
Now $\sigma(\tau(\alpha)) = \sigma(\alpha^{\tau-1}\alpha) = \alpha^{\tau-1}\sigma(\alpha) = \alpha^{\tau-1}\alpha^{\sigma-1}\alpha$, which gives $\chi_a(\sigma\tau) = \chi_a(\sigma)\chi_a(\tau)$.