Frobenius conjugacy over inertia group modulo wild inertia I'm reading "On the two-dimensional modular representations of $Gal(\overline{Q}/Q)$, the translation of the Serre's work.
On page 5, he says:
Let $s$ be an element of $Gal(\overline{Q}/Q)$ representing the Frobenius automorphism and let $u \in I$, then 
$$sus^{-1} \equiv u^p (mod I_p)$$
where $I_p$ is the wild inertia group.
I don't have idea how to do it. 
Thank you! :D
 A: Just some comments on Lukas's answer (because I don't have enough reputation to leave a comment). I think the crucial point is to show "Frobenius acts on $Gal(\mathbb{Q}_p^t/\mathbb{Q}_p)$ by $x \mapsto x^p$", and I can't think of a simple proof of that. But I think one plausible answer is given by the answer here: Galois group of tamely ramified extension. I would be appreciated if someone can elaborate on that.
I see. Thanks for reuns's comments. Consider $\mathbb{Q}_p(\zeta_{p^n-1}, p^{1/(p^n-1)})$. Denote $\zeta=\zeta_{p^n-1}, \omega=p^{1/(p^n-1)}$. Let $\sigma \in Gal(\mathbb{Q}_p^t/\mathbb{Q}_p)$ be a lift of Frobenius, say, $\sigma(\zeta)=\zeta^p, \sigma(\omega)=\zeta^l\omega$, for some integer $l$ (note elements in the Galois group must permutes roots of the polynomials $X^{p^n-1}-1$ and $X^{p^n-1}-p$). Let $\tau$ be an element in $Gal(\mathbb{Q}_p^t/\mathbb{Q}_p^{un})$, say $\tau(\zeta)=\zeta, \tau(\omega)=\zeta^m\omega$, for some integer $m$. Then we compute
$$\sigma\tau\sigma^{-1}(\omega)=\sigma\tau(\zeta^{-l/p}\omega)=\sigma(\zeta^{-l/p}\zeta^m\omega)=\zeta^{-l}\zeta^{mp}\zeta^l\omega=\zeta^{mp}\omega.$$ On the other hand, $$\tau^p(\omega)=\zeta^{mp}\omega.$$ So $\sigma\tau\sigma^{-1}=\tau^q$ on $\mathbb{Q}_p(\zeta_{p^n-1}, p^{1/(p^n-1)})$. By passing to limit, we see $\sigma\tau\sigma^{-1}=\tau^q$ on $\mathbb{Q}_p^t$. Does this make sense?
A: I checked in the paper and the proposition is actually concerned with the absolute Galois group of $\mathbb Q_p$.
Let $\mathbb Q_p^{nr}$ be the maximal unramified extension of $\mathbb Q_p$ and let $\mathbb Q_p^t$ be the maximal tamely ramified extension of $\mathbb Q_p$.
It is a known fact that we can write $$Gal(\mathbb Q_p^t/\mathbb Q_p) \cong Gal(\mathbb Q_p^t/\mathbb Q_p^{nr}) \rtimes Gal(\mathbb Q_p^{nr}/\mathbb Q_p),$$ where the action of the right group on the left group is induced by that of the Frobenius, which acts by conjugation. We know that the Frobenius, on $Gal(\mathbb Q_p^t/\mathbb Q_p)$, acts as $x \mapsto x^p$.
$$Gal(\mathbb Q_p^t/\mathbb Q_p) \cong Gal(\overline{ \mathbb Q}_p/\mathbb Q_p)/I_p,$$ so we are done.
