Let $p > 2$ be a prime number such that $p-1$ is not divisible by $4$.
Consider $K = \mathbb{Q}_p$ and $L = \mathbb{Q}(\sqrt[4]{p},\zeta_4)$ where $\zeta_4 \in L$ is a primitive forth root of unity (i.e. roots of the polynomial $x^2 +1 \in K[x]$). If my arguments are correct, the extension $L/K$ is Galois and has degree $8$.
Question: What is the Galois group of $L/K$?
I know that $F=K(\zeta_4)$ is the maximal unramified subextension of $L/K$ with degree $2$. I think this is probably how I should approach this problem. I have the feeling that I could show that the Galois group is not the cyclic group of order $8$ but I am not sure how the argument works.
Let $F = \mathbb{Q}_p(\zeta_4)$. By Kummer theory we have that $L$ is Galois over $F$ and $\text{Gal}(L/F) = \mathbb{Z}/4\mathbb{Z}$, generated by $\sigma \mapsto \frac{\sigma\sqrt[4]{p}}{ \sqrt[4]{p}}$ (which is independent of the choice of 4th root).
So we have an exact sequence $$ 0 \to \mathbb{Z}/4\mathbb{Z} \to G \to \mathbb{Z}/2\mathbb{Z} \to 0. $$ It follows that $G$ is either the Dihedral group of order 8 or is abelian. But $G$ is not abelian since the subextension $\mathbb{Q}_p(\sqrt[4]{p})$ is not Galois over $\mathbb{Q}_p$.