This might be a easy question, but I couldn't get the point.


Let $F$ be a p-adic field, $\bar{F}$ a separable algebraic closure of $F$. Set $\Omega\_F=Gal(\bar{F}/F)$. Use $F\_{\infty}\subset \bar{F}$ to denote the maximal unramified extension of $F$. Use $E\_{\infty} \supset F\_{\infty} $ to denote the maximal tamely ramified extension of $F$.


Let $\mathcal{I}=Gal(\bar{F}/F\_{\infty})$, $\mathcal{P}=Gal(\bar{F}/E\_{\infty})$. In the book 28.3.2 says that there is a isomorphism


$$t\_0:\mathcal{I}/\mathcal{P}\cong \prod\_{l\ne p}\mathbb{Z}\_l$$
where $l$ ranges over all prime numbers other than $p$, and $\mathbb{Z}\_l$ denotes the ring of $l$-adic integers.


The statement I'm confused is that in the book above proposition 28.3.3, the group $Gal(F\_{\infty}/F)$ acts on $Gal(E\_{\infty}/F\_{\infty})$ by conjugation, and moreover 


$$t\_0(\Phi\_F\sigma\Phi\_F^{-1})=q^{-1}t\_0(\sigma)$$ 
for any $\sigma\in \mathcal{I}/\mathcal{P}$, where $\Phi\_F$ is the geometric Frobenius in $Gal(F\_{\infty}/F)$, and $q$ is the number of elements in the residue field of $F$.


I couldn't see how $Gal(F\_{\infty}/F)$ acts on $Gal(E\_{\infty}/F\_{\infty})$ by conjugation,and thus don't know the meaning of $\Phi\_F\sigma\Phi\_F^{-1}$. 


Thanks a lot for any help, and sorry if the question is not appropriate here.



a question about a result in Bushnell-Henniart book 'the local Langlands conjecture for GL(2)'
Your real question is : When we have a short exact sequence of (multiplicatively written) groups $1\to C\to E\to G\to1$ in which $C$ is commutative, how does $G$ act on $C$ "by conjugation" ?


The answer is that you make $g\in G$ act on $C$ by $c\mapsto\tilde g c\tilde g^{-1}$, where $\tilde g\in E$ is any lift of $g$. It doesn't matter which lift you take, because any two lifts differ by an element of $C$, and the action of $C$ on itself by conjugation is trivial by the hypothesis that $C$ is commutative.


(By abuse of notation, one may write $gcg^{-1}$ for $\tilde g c\tilde g^{-1}$ because the thing only depends on $g$, not on the lift $\tilde g$).


