Let $K$ be a field complete with respect to a valuation $v_K$ on it. Let $K<L$ be a finite Galois extension that is totally ramified, and let $G$ be its Galois group. Let $A_K$ be the ring of $v_K$ with the maximal ideal $p_K$, and $U_K$ be the units in $A_K$. Define $A_L, p_L, U_L, v_L$ similarly.
Let $G_i$ be the $i$th ramification group of $G$: $$G_i=\{s\in G:v_L(s(a)-a)\geq i+1,\ \forall a\in A_L\}$$ Let $U_L^i$ be the filtration of $U_L$: $$U_L^0=U_L,\quad U_L^i=1+p_L^i\ (i\geq1)$$
These definitions come from Serre's Local Fields. And I came across two problems reading it.
(1) Why are $G_i$ normal subgroups? I know it suffices to prove that $$\forall s\in G_i,\ \forall\tau\in G,\ \forall x\in A_L,\ \tau^{-1}s\tau(x)-x\in P_L^{i+1}$$ For this I write $$\tau^{-1}s\tau(x)-x=\tau^{-1}(s\tau(x)-\tau(x))\in\tau^{-1}(p_L^{i+1})$$ So I guess it remains to show $\tau^{-1}(p_L^{i+1})\subset p_L^{i+1}$. How can this be done?
(2) What are the canonical isomorphisms $U^i_L/U^{i+1}_L\cong p^i_L/p^{i+1}_L$? In the books it says to pass the correspondence $$x\in p_L^i\leftrightarrow 1+x\in U_L^i$$ to the quotients. Does it mean to consider the mapping $$p_L^i\to U_L^i/U_L^{i+1}\\x\mapsto(1+x)U_L^i$$ and then prove it induces an isomorphism?
I am new at studying algebraic number theory, and may not be familiar with some "well-known" facts. So I would appreciate it if you could provide more detailed explanations. Thanks in advance!
UPDATE: For part (2), I defined the mapping $$\phi:U_L^i=1+p_L^i\to p_L^i/p_L^{i+1}\\1+x\mapsto x+p_L^{i+1}$$ and view $U_L^i$ as a multiplicative group and $p_L^i/p_L^{i+1}$ an additive group. Then $$\phi((1+x)(1+y))=x+y+xy+p_L^{i+1}\\=x+y+p_L^{i+1}=(x+p_L^{i+1})+(y+p_L^{i+1})\\=\phi(1+x)+\phi(1+y)$$ means it is a homomorphism. And $\phi(1+x)=\bar0\iff x\in P_L^{i+1}\iff 1+x\in U^{i+1}_L$. This says $\ker\phi=U_L^{i+1}$. Hence we have an isomorphism $$\bar\phi:U_L^i/U_L^{i+1}\to p_L^i/p_L^{i+1}$$ Does this make sense?