This question concerns Yoshida's proof of the Hasse-Arf theorem in https://arxiv.org/abs/math/0606108 (page 16).
For a totally ramified extension $K^\prime/K$ of local fields define the ramification groups $G_n := \lbrace \sigma \in \operatorname{Gal}(K^\prime/K) : v(\sigma(\pi) – \pi) \geq n + 1\rbrace$ with $\pi$ a uniformiser of $K^\prime$. Call $G := \operatorname{Gal}(K^\prime/K)$ and $i(\sigma) := v(\sigma(\pi)-\pi)$ and define
$$\phi_G(n) := -1 + \frac{1}{|G|}\sum_{\sigma \in G}\min\lbrace i(\sigma), n+1\rbrace$$
for $n \in \Bbb R_{\geq 0}$.
Theorem: (Hasse-Arf) If $G$ is abelian, $n\in\Bbb Z_{\geq 0}$ and $G_n \neq G_{n + 1}$ then $\phi_G(n) \in \Bbb Z_{\geq 0}$.
Suppose $G \neq G_1$ and let $H := G_1$ and $|G/H| = e_0$.
Yoshida claims in the course of his proof of Hasse-Arf that: $\phi_{G/H}(n) = n/e_0$ "by definition".
I can't quite see why this is the case but here are some thoughts:
Dumping $G/H$ into the definition of $\phi_G(n)$ above we get
$$\phi_{G/H}(n) = -1 + \frac{1}{e_0} \sum_{\overline{\sigma} \in G/H}\min\lbrace i(\overline{\sigma}), n+ 1\rbrace.$$
We have that $i(\overline\sigma) = \frac{1}{|G_1|}\sum_{\tau \in G_1} i(\sigma\tau)$ and by definition of $G_1$ we have $i(\tau) \geq 2$. In addition,
\begin{align}
i(\sigma\tau) &= v(\sigma\tau(\pi) – \pi)\\
&= v(\sigma\tau(\pi) – \pi + \tau(\pi) – \tau(\pi))\\
&\geq \min\lbrace v(\sigma\tau(\pi) – \tau(\pi)), i(\tau).\rbrace
\end{align}
Since $\tau(\pi) \in \mathcal{O}_{K^\prime}$ we have $v(\sigma(\tau(\pi)) – \tau(\pi)) \geq 1$ so $i(\sigma\tau) \geq 1$. What can I conclude from this?
Alternatively, we have
$$\phi_G = \phi_{G/H} \circ \phi_H$$
and that $\phi_G$ is a homeomorphism $[0, \infty) \to [0, \infty)$ for any $G$, is it possible that $\phi_G \circ \phi_H^{-1} = \phi_{G/H}$ will give me something?
Best Answer
I think your issue is mainly forgetting that you're in the tame case. Since we are in the tame case, either $i(\bar\sigma) = 1$ or $i(\bar\sigma) = +\infty$.
I'll drop the bars for simplicity.
Our definition is this: $$\phi_{G/H}(n) = -1 + \frac 1 {e_0}\sum_{\sigma\in G/H} \min\{i(\sigma), n+1\}$$ and so by the remark above, we see that the min is always $1$ except when $\sigma = e$, and in that case the min is $n+1$. That leaves us with $$=-1 + \frac 1 {e_0} (|G/H| - 1) + \frac 1 {e_0}(n+1),$$ $$= \frac {n}{e_0} + \left(\frac 1{e_0} |G/H| - 1\right),$$ which is just $n/e_0$, as the term in parentheses is $0$ since $|G/H| = e_0$.