Galois group of tamely ramified extension

Let $$p$$ be a prime and let $$K$$ be a finite extension of $$\mathbb{Q}_p$$. Suppose $$L/K$$ is a tamely ramified Galois extension. I want to show that if $$\sigma$$ is a lift of the Frobenius element of the Galois group of the residue field extension and $$\tau$$ is a generator of $$\textrm{Gal}(L/E)$$ where $$E$$ is the maximal unramified subextension of $$L$$, then $$\sigma\tau\sigma^{-1} = \tau^q$$ where $$q$$ is the order of the residue field of $$K$$.

I know that $$L = E(\lambda^{1/e}$$) where $$\lambda$$ is some uniformizer for $$E$$ and $$e$$ is the ramification index of $$L/K$$. If $$\sigma(\lambda^{1/e})$$ is another $$e$$-th root of $$\lambda$$, then I can show the desired relation holds. However, it seems to me that we can't be sure this is the case, because $$\lambda$$ may not be in $$K$$.

• The title doesn't correspond to the question. (To answer the implied question in the title, the Galois group is not always a semidirect product of the inertia group with a lift of Frobenius.) Apr 29, 2020 at 11:42
• @user760870 I think it is a semidirect product provided it's a tamely ramified extension a $p$-adic field? Using OP's notation, Galois group is generated by $\sigma, \tau$, subjected to $\sigma\tau\sigma^{-1} = \tau^q$. Am I missing something? Apr 29, 2020 at 12:18
• @pisco Yes you are missing something. I suggest thinking about Galois extensions $K/\mathbf{Q}_p$ of degree $4$ (when $p$ is odd) . Apr 29, 2020 at 16:54
• @user760870 Yes, you're right. I had the wrong impression that $\tau$ and $\sigma$ generates disjoint subgroups, which is not always the case. Thank you very much for pointing this out. Apr 29, 2020 at 18:48
• @user760870 So are you saying this relation may not always hold, or just that it may not be enough to determine the Galois group? Apr 29, 2020 at 23:42

Let $$\pi$$ be a uniformizer of the totally ramified extension $$L/E$$. We have a map: $$\theta: \text{Gal}(L/E) \to U_L/U^{(1)}_L\qquad \tau_0\mapsto \tau_0(\pi)/\pi$$ where $$U_L$$ is the unit group of $$\mathcal{O}_L$$, and $$U^{(1)}_L = 1+\mathfrak{p}_L$$. This map is independent of the unifromizer $$\pi$$ chosen. Its kernel is the wild ramification group, which is trivial since $$L/K$$ is assumed to be tamely ramified. Hence $$\theta$$ is injective.
Denote $$\tau(\pi)\equiv a\pi \pmod{\mathfrak{p}_L^2}$$. Then $$\tag{1}\sigma\tau(\pi) \equiv \sigma(a)\sigma(\pi) \equiv a^q\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$ because $$\sigma$$ acts like Frobenius. Moreover, since $$\theta$$ is independent of unifromizer, we have $$\tau(\sigma(\pi))\equiv a\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$, so $$\tau^2(\sigma(\pi))\equiv \tau(a)\tau(\sigma(\pi))\equiv \tau(a)a\sigma(\pi) \equiv a^2\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$ the last equality follows from the fact that $$\tau$$ acts on residual field trivially. Induction shows $$\tau^q(\sigma(\pi))\equiv a^q\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$
Comparing with $$(1)$$ shows that $$\sigma\tau$$ and $$\tau^q\sigma$$ have same image under $$\theta$$. Injectivity of $$\theta$$ proves your claim.