[proofplan] We define $\phi: G_s(L/K) \to U_L^{(s)}/U_L^{(s+1)}$ by $\sigma \mapsto \sigma(\pi_L)/\pi_L \bmod U_L^{(s+1)}$ and verify four properties: (1) the image lies in $U_L^{(s)}$, so $\phi$ is well-defined; (2) the map is independent of the choice of uniformizer $\pi_L$; (3) $\phi$ is a group homomorphism; (4) $\ker \phi = G_{s+1}(L/K)$, which simultaneously shows injectivity on the quotient and normality of $G_{s+1}$ in $G_s$. [/proofplan] [step:Show $\sigma(\pi_L)/\pi_L \in U_L^{(s)}$ for $\sigma \in G_s(L/K)$] Let $\sigma \in G_s(L/K)$, so $v_L(\sigma(x) - x) \geq s + 1$ for all $x \in \mathcal{O}_L$. Applying this with $x = \pi_L$: \begin{align*} v_L(\sigma(\pi_L) - \pi_L) \geq s + 1. \end{align*} Write $\sigma(\pi_L) = \pi_L + \pi_L^{s+1} c$ for some $c \in \mathcal{O}_L$ (using $v_L(\sigma(\pi_L) - \pi_L) \geq s + 1$ and $v_L(\pi_L) = 1$). Then \begin{align*} \frac{\sigma(\pi_L)}{\pi_L} = 1 + \pi_L^s c \in U_L^{(s)}. \end{align*} Since $v_L(1 + \pi_L^s c - 1) = v_L(\pi_L^s c) \geq s$, the quotient lies in $U_L^{(s)}$. [guided] An element $\sigma$ of the $s$-th ramification group $G_s(L/K)$ satisfies $i_{L/K}(\sigma) \geq s + 1$, meaning $v_L(\sigma(x) - x) \geq s + 1$ for every $x \in \mathcal{O}_L$. In particular, for $x = \pi_L$, we get $v_L(\sigma(\pi_L) - \pi_L) \geq s + 1$. Since $\sigma$ is a $K$-automorphism preserving $v_L$, and $v_L(\pi_L) = 1$, we have $v_L(\sigma(\pi_L)) = 1$. So $\sigma(\pi_L)$ is also a uniformizer. We can write $\sigma(\pi_L) - \pi_L = \pi_L^{s+1} c$ for some $c \in \mathcal{O}_L$ (since the difference has $v_L$-value $\geq s + 1$, while $v_L(\pi_L^{s+1}) = s + 1$). Therefore \begin{align*} \frac{\sigma(\pi_L)}{\pi_L} = 1 + \pi_L^s c, \end{align*} and $v_L(\pi_L^s c) = s + v_L(c) \geq s$, confirming $\sigma(\pi_L)/\pi_L \in U_L^{(s)}$. [/guided] [/step] [step:Verify independence of the choice of uniformizer modulo $U_L^{(s+1)}$] Let $\pi_L' = \pi_L u$ for some unit $u \in \mathcal{O}_L^\times$ be another uniformizer. Since $\sigma \in G_s(L/K) \subseteq G_0(L/K) = I(L/K)$ (the inertia group), we have $v_L(\sigma(u) - u) \geq s + 1$. Write $\sigma(u) = u + \pi_L^{s+1} d$ for some $d \in \mathcal{O}_L$. Then \begin{align*} \frac{\sigma(\pi_L')}{\pi_L'} = \frac{\sigma(\pi_L)\sigma(u)}{\pi_L u} = \frac{\sigma(\pi_L)}{\pi_L} \cdot \frac{\sigma(u)}{u} = (1 + \pi_L^s c)\left(1 + \pi_L^{s+1} \frac{d}{u}\right). \end{align*} The second factor satisfies $v_L(1 + \pi_L^{s+1}d/u - 1) = v_L(\pi_L^{s+1}d/u) \geq s + 1$, so $\sigma(u)/u \in U_L^{(s+1)}$. Therefore \begin{align*} \frac{\sigma(\pi_L')}{\pi_L'} \equiv \frac{\sigma(\pi_L)}{\pi_L} \pmod{U_L^{(s+1)}}, \end{align*} and the map $\phi$ is independent of the choice of uniformizer. [guided] Why should the quotient $\sigma(\pi_L)/\pi_L$ modulo $U_L^{(s+1)}$ be independent of which uniformizer we pick? Any other uniformizer has the form $\pi_L' = \pi_L u$ for a unit $u \in \mathcal{O}_L^\times$. The key is that $\sigma \in G_s$ acts "close to the identity" on units. Specifically, $\sigma \in G_0 = I(L/K)$ means $\sigma$ fixes the residue field, and $\sigma \in G_s$ means $v_L(\sigma(x) - x) \geq s + 1$ for all $x \in \mathcal{O}_L$. In particular, $v_L(\sigma(u) - u) \geq s + 1$, so $\sigma(u)/u = 1 + \pi_L^{s+1}(d/u) \in U_L^{(s+1)}$. Computing: \begin{align*} \frac{\sigma(\pi_L u)}{\pi_L u} = \frac{\sigma(\pi_L)}{\pi_L} \cdot \frac{\sigma(u)}{u}. \end{align*} The second factor lies in $U_L^{(s+1)}$, so the two ratios are congruent modulo $U_L^{(s+1)}$. This means $\phi(\sigma)$ is well-defined as an element of $U_L^{(s)}/U_L^{(s+1)}$. [/guided] [/step] [step:Verify $\phi$ is a group homomorphism] For $\sigma, \tau \in G_s(L/K)$, note that $\tau(\pi_L)$ is a uniformizer (since $\tau \in I(L/K)$ preserves the valuation). Compute: \begin{align*} \phi(\sigma\tau) = \frac{(\sigma\tau)(\pi_L)}{\pi_L} = \frac{\sigma(\tau(\pi_L))}{\tau(\pi_L)} \cdot \frac{\tau(\pi_L)}{\pi_L}. \end{align*} By the independence result (previous step), the first factor $\sigma(\tau(\pi_L))/\tau(\pi_L)$ equals $\phi(\sigma) \bmod U_L^{(s+1)}$ (since $\tau(\pi_L)$ is an equally valid uniformizer). The second factor is $\phi(\tau)$. Therefore \begin{align*} \phi(\sigma\tau) = \phi(\sigma) \cdot \phi(\tau) \quad \text{in } U_L^{(s)}/U_L^{(s+1)}. \end{align*} [guided] To verify the homomorphism property, we expand $(\sigma\tau)(\pi_L)/\pi_L$ by inserting $\tau(\pi_L)$ in the numerator and denominator: \begin{align*} \frac{(\sigma\tau)(\pi_L)}{\pi_L} = \frac{\sigma(\tau(\pi_L))}{\tau(\pi_L)} \cdot \frac{\tau(\pi_L)}{\pi_L}. \end{align*} The second factor is $\phi(\tau)$ by definition. For the first factor: since $\tau \in I(L/K)$, the element $\tau(\pi_L)$ is a uniformizer of $L$. We showed in the previous step that $\phi(\sigma)$ is independent of the uniformizer modulo $U_L^{(s+1)}$, so $\sigma(\tau(\pi_L))/\tau(\pi_L) \equiv \sigma(\pi_L)/\pi_L = \phi(\sigma) \pmod{U_L^{(s+1)}}$. Combining: $\phi(\sigma\tau) = \phi(\sigma)\phi(\tau)$ in $U_L^{(s)}/U_L^{(s+1)}$. [/guided] [/step] [step:Identify the kernel as $G_{s+1}(L/K)$] The inclusion $G_{s+1}(L/K) \subseteq \ker \phi$ is immediate: if $\sigma \in G_{s+1}$, then $v_L(\sigma(\pi_L) - \pi_L) \geq s + 2$, so $\sigma(\pi_L)/\pi_L = 1 + \pi_L^{s+1} c' \in U_L^{(s+1)}$, meaning $\phi(\sigma) = 1$ in $U_L^{(s)}/U_L^{(s+1)}$. For the reverse inclusion, suppose $\sigma \in \ker \phi$, so $v_L(\sigma(\pi_L) - \pi_L) \geq s + 2$. We must show $v_L(\sigma(x) - x) \geq s + 2$ for all $x \in \mathcal{O}_L$. Since $\sigma \in G_s \subseteq G_0 = I(L/K)$, the automorphism $\sigma$ acts directly on the residue field $k_L$. By the [Teichmüller Map](/theorems/???), every element $x \in \mathcal{O}_L$ admits a Teichmüller expansion \begin{align*} x = \sum_{n=0}^{\infty} [\bar{x}_n] \pi_L^n, \quad \bar{x}_n \in k_L, \end{align*} where $[\bar{x}_n]$ denotes the Teichmüller representative of $\bar{x}_n$. Since $\sigma$ fixes the residue field, $\sigma([\bar{x}_n]) = [\bar{x}_n]$ for all $n$. Write $\sigma(\pi_L) = \pi_L + \pi_L^{s+2} y$ for some $y \in \mathcal{O}_L$ (using $v_L(\sigma(\pi_L) - \pi_L) \geq s + 2$). Then \begin{align*} \sigma(x) - x = \sum_{n=1}^{\infty} [\bar{x}_n]\bigl(\sigma(\pi_L)^n - \pi_L^n\bigr). \end{align*} For each $n \geq 1$, factor: \begin{align*} \sigma(\pi_L)^n - \pi_L^n = (\sigma(\pi_L) - \pi_L)\sum_{j=0}^{n-1} \sigma(\pi_L)^j \pi_L^{n-1-j}. \end{align*} Since $v_L(\sigma(\pi_L)) = v_L(\pi_L) = 1$, each term in the sum has $v_L$-value $\geq n - 1$, so $v_L(\sigma(\pi_L)^n - \pi_L^n) \geq (s+2) + (n-1) = n + s + 1$. Therefore $v_L([\bar{x}_n](\sigma(\pi_L)^n - \pi_L^n)) \geq n + s + 1 \geq s + 2$ for $n \geq 1$, and \begin{align*} v_L(\sigma(x) - x) \geq \min_{n \geq 1}(n + s + 1) = s + 2. \end{align*} So $\sigma \in G_{s+1}(L/K)$, establishing $\ker \phi = G_{s+1}(L/K)$. Since $\ker \phi = G_{s+1}(L/K)$, the subgroup $G_{s+1}$ is normal in $G_s$ (as the kernel of a homomorphism), and $\phi$ induces an injective homomorphism $G_s/G_{s+1} \hookrightarrow U_L^{(s)}/U_L^{(s+1)}$. [guided] The containment $G_{s+1} \subseteq \ker \phi$ is straightforward: if $\sigma \in G_{s+1}$, then $v_L(\sigma(\pi_L) - \pi_L) \geq s + 2$, so $\sigma(\pi_L)/\pi_L \in U_L^{(s+1)}$, meaning $\phi(\sigma) = 1$ in the quotient. The reverse direction is more substantial. Suppose $\sigma \in G_s$ with $\phi(\sigma) = 1$, i.e., $v_L(\sigma(\pi_L) - \pi_L) \geq s + 2$. We need to show $v_L(\sigma(x) - x) \geq s + 2$ for all $x \in \mathcal{O}_L$, not just $x = \pi_L$. The strategy uses the Teichmüller expansion. Since $\sigma \in I(L/K)$, it fixes the residue field, hence fixes Teichmüller representatives: $\sigma([\bar{a}]) = [\bar{a}]$ for $\bar{a} \in k_L$. Every $x \in \mathcal{O}_L$ can be written as $x = \sum_{n=0}^\infty [\bar{x}_n]\pi_L^n$ (by the [Teichmüller Map](/theorems/???)), and $\sigma$ acts only on the powers of $\pi_L$: \begin{align*} \sigma(x) = \sum_{n=0}^\infty [\bar{x}_n]\sigma(\pi_L)^n. \end{align*} The $n = 0$ term is unchanged, so $\sigma(x) - x = \sum_{n=1}^\infty [\bar{x}_n](\sigma(\pi_L)^n - \pi_L^n)$. For each $n \geq 1$, the algebraic identity $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$ gives \begin{align*} v_L(\sigma(\pi_L)^n - \pi_L^n) \geq v_L(\sigma(\pi_L) - \pi_L) + (n-1) \cdot v_L(\pi_L) \geq (s + 2) + (n - 1) = n + s + 1. \end{align*} (Here we used $v_L(\sigma(\pi_L)) = v_L(\pi_L) = 1$ for each term in the geometric sum.) Since $v_L([\bar{x}_n]) = 0$ for $\bar{x}_n \neq 0$, each summand has $v_L \geq n + s + 1 \geq s + 2$. By the ultrametric property, $v_L(\sigma(x) - x) \geq s + 2$, so $\sigma \in G_{s+1}$. Since $G_{s+1} = \ker \phi$, it is a normal subgroup of $G_s$, and the first isomorphism theorem gives the injective homomorphism $G_s/G_{s+1} \hookrightarrow U_L^{(s)}/U_L^{(s+1)}$. [/guided] [/step]