This question concerns Yoshida's proof of the Hasse-Arf theorem in the local class field theory in https://arxiv.org/abs/math/0606108 (page 16).
For a totally ramified extension $K′/K$ of local fields, write $G = \mathrm{Gal}(K'/K)$. For a normal subgroup $H$ of $G$, write $K'' = K'^H$, so $H = \mathrm{Gal}(K'/K'')$, $G/H = \mathrm{Gal}(K''/K)$. For any $n\in\mathbb{R}_{\ge 0}$, define the ramification groups $G_n = \{\sigma\in \mathrm{Gal}(K'/K):v(\sigma(\pi)-\pi)\ge n+1\}$ for $K'/K$, where $\pi$ being a uniformizer of $K'$; and similarly the ramification groups $H_n$ and $(G/H)_n$. Define
$$\varphi_G(n) = -1 + \dfrac{1}{|G|}\sum_{\sigma\in G}\min\{v(\sigma(\pi)-\pi),n+1\},$$
and similarly $\varphi_H(n)$, $\varphi_{G/H}(n)$. We have:
Proposition 6.9 (Herbrand's theorem). $G_nH/H = (G/H)_{\varphi_H(n)}$.
Lemma 6.10. $\varphi_G(n) = \dfrac{1}{|G|}\displaystyle{\sum^{n}_{i=1}} |G_i|$ for $n\in\mathbb{Z}_{\ge 0}$, and $\varphi_G = \varphi_{G/H}\circ\varphi_H$.
Now Yoshida begins the Theorem
Theorem 6.11 (Hasse-Arf theorem). If $G$ is abelian, $n\in\mathbb{Z}_{\ge 0}$ and $G_n = G_{n+1}$, then $\varphi_G(n)\in \mathbb{Z}_{\ge 0}$.
Proof. First assume $G = G_1$. Then $G\cong \displaystyle{\bigoplus^j_{i=1}} \mathbb{Z}/p^{m_i}\mathbb{Z}$, and we proceed by induction of $j$. When $j=1$, … (I'm omitting this part) For $j>1$, if $G_n\neq G_{n+1}$ we can find $H$ with $G/H\cong \mathbb{Z}/p^{m_i}\mathbb{Z}$, and $G_nH/H = G_{n+1}H/H$. We have $\varphi_H(n)\in\mathbb{Z}_{\ge 0}$ be inductive hypothesis, and $(G/H)_{\varphi_H(n)} \neq (G/H)_{\varphi_H(n+1)} = (G/H)_{\varphi_H(n)+1}$ by Proposition 6.9. As $G/H$ is cyclic, we see $\varphi_{G/H}(\varphi_H(n))\in \mathbb{Z}_{\ge 0}$ by Lemma 6.10.
I have no problem at all with Proposition 6.9 and Lemma 6.10, and with the case $j=1$ in the proof of Hasse-Arf. What I have doubt is finding $H$ such that $G_nH/H \neq G_{n+1}H/H$ and $\varphi_H(n)\in\mathbb{Z}_{\ge 0}$. To use inductive hypothesis to obtain $\varphi_H(n)\in\mathbb{Z}_{\ge 0}$ we need to have $H_n \neq H_{n+1}$, namely $G_n \cap H \neq G_{n+1}\cap H$; but it seems to me that such $H$ that $G_nH \neq G_{n+1}H$ and that $G_n \cap H \neq G_{n+1}\cap H$ cannot always be found! For example, if $G_n$ is cyclic of order $p$, $G_{n+1}$ is trivial (I believe that this is the real hard case to handle), then such $H$ cannot exist: either $G_n\subset H$, so $G_nH = G_{n+1}H = H$; either $G_n\cap H = e$, so $G_n \cap H = G_{n+1}\cap H = e$.
I was thinking about not inducting on the rank of $G_1$ but on the size of $G_1$. We can proceed if there is some $H$ such that
$$ G_n\supsetneq H\supset G_{n+1}, H\neq e$$
Then we have $H_0 = \cdots = H_n = H$ in this case so $\varphi_H(n) = n\in\mathbb{Z}_{\ge 0}$ (even if we may have $H_n = H_{n+1}$), and $G_nH/H = G_n/H \neq G_{n+1}H/H = e$, so we can use the inductive hypothesis because $|G/H| < |G|$. But the problem occurs when $G_n$ is cyclic of order $p$ and $G_{n+1} = e$: this is the only case where such $H$ does not exist.
So I would like to ask that: is there a way to fix the problem of choosing $H$, or was I mistaking for something in the original proof?
Best Answer
I would like to say that the original proof of Serre (in Corps Locaux, V §7, p. 102) is really easy and elegant, so I'm going to present my understanding of it here for anyone who does not pursue Yoshida's method.
Suppose that $G_n \neq G_{n+1}$, then $G_n/G_{n+1}\neq e$. There is a subgroup $H\subset G/G_{n+1}$ such that $G_n/G_{n+1}\not\subset H$ and that $(G/G_{n+1})/H$ is cyclic ($\star$$\star$). By Herbrand's theorem we have $$(G/G_{n+1})_{\varphi_{G_{n+1}}(n)} = G_nG_{n+1}/G_{n+1} = G_n/G_{n+1},$$ $$(G/G_{n+1})_{\varphi_{G_{n+1}}(n+\varepsilon)} = G_{n+\varepsilon}G_{n+1}/G_{n+1} = e, \forall \varepsilon>0;$$ and $$((G/G_{n+1})/H)_{\varphi_H(\varphi_{G_{n+1}}(n))} = (G_n/G_{n+1})_{\varphi_{G_{n+1}}(n)}H/H = (G_n/G_{n+1})H/H\neq e,$$ $$((G/G_{n+1})/H)_{\varphi_H(\varphi_{G_{n+1}}(n+\varepsilon))} = (G_n/G_{n+1})_{\varphi_{G_{n+1}}(n+\varepsilon)}H/H = e, \forall \varepsilon>0.$$
By continuity of $\varphi_H\circ\varphi_{G_{n+1}}$ we have $\varphi_H(\varphi_{G_{n+1}}(n))\in\mathbb{Z}_{\ge 0}$. By the case of cyclic groups we have $\varphi_{(G/G_{n+1})/H}(\varphi_H(\varphi_{G_{n+1}}(n)))\in\mathbb{Z}_{\ge 0}$, and by Lemma 6.10, $$\varphi_{(G/G_{n+1})/H}\circ \varphi_H\circ\varphi_{G_{n+1}} = \varphi_{G/G_{n+1}}\circ\varphi_{G_{n+1}} = \varphi_G,$$ and we are done.
($\star$$\star$) Lemma. Suppose that $G$ is a finitely-generated abelian group, $G'\neq e$ is subgroup of $G$, then there is a subgroup $H$ of $G$ such that $G'\not\subset H$ and that $G/H$ is cyclic.
Proof. By the structure theorem of finitely-generated abelian group we have $G = C_1\times \cdots \times C_r$ for cyclic $C_i$. For $1\neq g\in G'$, write $g = g_1\cdots g_r$ for $g_i\in C_i$, then $g_{i_0}\neq 1$ for some $i_0$. Pick $H = \displaystyle{\prod_{i\neq i_0}} C_i$, then $G/H\cong C_{i_0}$ is cyclic, and $g\notin H$ because $gH = g_{i_0}H\neq H$.