TITLE: Equivalent conditions for the ramification group(s) in a finite Galois extension
QUESTION [7 upvotes]: Let $L|K$ be a finite Galois extension and $v$ a discrete normalized valuation on $L$ such that its restriction to $K$ extends uniquely to $L$.
(1) Why is $G_1=\{\sigma\in G(L|K)\mid v(\sigma(x)-x)\ge 2\ \ \forall x\in\mathcal{O}\}$ equal to the ramification group $R(L|K)=\{\sigma\in G(L|K)\mid v\left(\frac{\sigma(x)}{x}-1\right)>0\ \ \forall x\in L^*\}$? (Here $\mathcal{O}$ is the valuation ring of $L$).
(2) Is it true that in general $G_s$ equals $\{\sigma\in G(L|K)\mid v(\sigma(\pi)-\pi)\ge s+1\}$, where $\pi$ is any element of $L^*$ such that $v(\pi)=1$? (The $G_s$'s are the higher ramification groups, that is $G_s=\{\sigma\in G(L|K)\mid v(\sigma(x)-x)\ge s+1\ \ \forall x\in\mathcal{O}\}$).
This question arises from Chapter II, $\S 10$ of Neukirch's Algebraic Number Theory.

REPLY [3 votes]: For (1): In $\S$ II.9 Neukirch shows that $R(L/K)$ is the (normal, hence unique) Sylow $p$-subgroup of the inertia group $I(L/K)$.  But (Proposition II.10.2) the quotient $G_0/G_1$ is cyclic of order prime to $p$, while for each $i \geq 1$, $G_i/G_{i+1}$ is a $p$-group.  Since $G_i = \{1\}$ for sufficiently large $i$, $G_1$ must be the Sylow $p$-subgroup of $I(L/K)$.  (Admittedly Proposition II.10.2 comes shortly after Neukirch asserts that "clearly" $R(L/K) = G_1$, so this argument is not what he has in mind.  But it is an argument, at least.)
For (2): no.  $L/K$ could be unramified, and then we can take $\pi \in K$, so $v(\sigma(\pi)-\pi) = v(0) = \infty \geq s+1$ holds for any $\sigma$, so the latter group is always equal to $G$, while $G_s = \{1\}$ for all $s$ in this case.