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 $K$.
 A: 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.
