[proofplan] We prove by induction on $[L : K]$ that the jumps of the upper-numbering filtration for a finite abelian extension $L/K$ of local fields are non-negative integers. The base case $[L : K] = 1$ is vacuous. For the inductive step, we choose a non-trivial normal subgroup $H \leq G = \operatorname{Gal}(L/K)$ (which exists since $G$ is abelian), set $F = L^H$, and apply the inductive hypothesis to the abelian extensions $L/F$ and $F/K$ separately. The [Compatibility of Upper Numbering with Quotients](/theorems/???) identifies the upper-numbering filtration of $F/K$ as the image of that of $L/K$, so a jump of $G^t(L/K)$ is either a jump of $G^t(F/K)$ (which is an integer by induction on $F/K$), or it occurs at a value $t$ where $G^t(L/K)$ shrinks within $H$, which after translation to the lower numbering of $L/F$ is shown to be an integer by induction on $L/F$. The abelianness is essential: it ensures every subgroup $H$ is normal, so both $F/K$ and $L/F$ are Galois and abelian, and the induction applies. [/proofplan] [step:Set up the induction and choose a non-trivial quotient] We proceed by strong induction on $n = [L : K]$. If $n = 1$, then $G = \operatorname{Gal}(L/K)$ is trivial, $G^t = \{1\}$ for all $t > -1$, and there are no jumps. The statement is vacuously true. Assume $n > 1$ and that the Hasse-Arf theorem holds for all finite abelian extensions of degree strictly less than $n$. Since $G = \operatorname{Gal}(L/K)$ is a non-trivial finite abelian group, it contains a non-trivial proper subgroup $H$. Indeed, by Cauchy's theorem applied to any prime $q$ dividing $|G|$, there exists an element of order $q$, generating a cyclic subgroup of order $q$; taking $H$ to be the complement (a subgroup of index $q$, which exists since $G$ is abelian and $q$ divides $|G|$) gives a proper normal subgroup. Set $F = L^H$, the fixed field of $H$. Since $G$ is abelian, every subgroup of $G$ is normal; in particular $H \trianglelefteq G$, so $F/K$ is Galois with $\operatorname{Gal}(F/K) \cong G/H$. The extension $L/F$ is Galois with $\operatorname{Gal}(L/F) = H$, which is abelian as a subgroup of the abelian group $G$. Both $[F : K] = |G/H| < n$ and $[L : F] = |H| < n$ (since $H$ is non-trivial and proper), so the inductive hypothesis applies to both $F/K$ and $L/F$. [/step] [step:Identify the jumps of $L/K$ using the quotient filtration] By the [Compatibility of Upper Numbering with Quotients](/theorems/???), for every $t \geq -1$: \begin{align*} \frac{G^t(L/K)}{H} = G^t(F/K), \end{align*} where we identify $\operatorname{Gal}(F/K) = G/H$. This compatibility — that the upper-numbering filtration of $G$ projects to the upper-numbering filtration of $G/H$ — is the defining advantage of the upper numbering over the lower numbering (for the lower numbering, the analogous statement fails in general). Let $t_0 \geq 0$ be a jump of the upper-numbering filtration of $L/K$, meaning $G^{t_0}(L/K) \supsetneq G^{t_0 + \varepsilon}(L/K)$ for all sufficiently small $\varepsilon > 0$. We split into two cases based on whether this drop is visible after projecting to $G/H$. **Case 1:** The jump is visible in the quotient filtration, i.e., $G^{t_0}(F/K) \supsetneq G^{t_0 + \varepsilon}(F/K)$ for all sufficiently small $\varepsilon > 0$. Then $t_0$ is a jump of the upper-numbering filtration of the abelian extension $F/K$, which has degree $|G/H| < n$. By the inductive hypothesis, $t_0 \in \mathbb{Z}_{\geq 0}$. **Case 2:** The jump is invisible in the quotient, i.e., $G^{t_0}(F/K) = G^{t_0 + \varepsilon}(F/K)$ for some $\varepsilon > 0$. Then the drop at level $t_0$ occurs entirely within $H$: since $G^{t_0}(L/K)/H = G^{t_0}(F/K) = G^{t_0+\varepsilon}(F/K) = G^{t_0+\varepsilon}(L/K)/H$, the quotient $G^{t_0}(L/K)/G^{t_0+\varepsilon}(L/K)$ maps directly into $G/H$, so it is contained in $H$. Therefore $G^{t_0}(L/K) \cap H \supsetneq G^{t_0 + \varepsilon}(L/K) \cap H$. [/step] [step:Reduce Case 2 to the upper-numbering filtration of $L/F$ and apply the inductive hypothesis] In Case 2, the compatibility again gives $G^t(L/K) \cap H = G^t(L/F)$ for the upper-numbering filtration of $L/F$. This identity follows from the [Compatibility of Upper Numbering with Quotients](/theorems/???) applied to the subextension $L/F$ within the tower $L/F/K$: the upper-numbering filtration of $L/K$, when intersected with the subgroup $H = \operatorname{Gal}(L/F) \leq G$, recovers the upper-numbering filtration of $L/F$. Concretely: from Case 2 we have $G^{t_0}(L/K) \cap H \supsetneq G^{t_0 + \varepsilon}(L/K) \cap H$ for all small $\varepsilon > 0$. Using the identity $G^t(L/K) \cap H = G^t(L/F)$, this becomes $G^{t_0}(L/F) \supsetneq G^{t_0 + \varepsilon}(L/F)$, i.e., $t_0$ is a jump of the upper-numbering filtration of the abelian extension $L/F$. Since $[L : F] = |H| < n$, the inductive hypothesis applies and gives $t_0 \in \mathbb{Z}_{\geq 0}$. In both cases, $t_0 \in \mathbb{Z}_{\geq 0}$. Since $t_0$ was an arbitrary jump of the upper-numbering filtration of $L/K$, all jumps are non-negative integers. This completes the inductive step, and by strong induction the Hasse-Arf theorem holds for all finite abelian extensions of local fields. [guided] The structure of the proof is elegant: we decompose the problem using a tower $L/F/K$ and the compatibility of the upper numbering with quotients, reducing to two smaller abelian extensions where induction applies. Why is abelianness essential? At the very first step: we need $H \trianglelefteq G$ so that $F/K$ is Galois and the quotient $G/H = \operatorname{Gal}(F/K)$ makes sense. In a non-abelian group, not every subgroup is normal, and the proof structure breaks down because $F/K$ may not be Galois. Indeed, the Hasse-Arf theorem is false for non-abelian extensions: there exist finite Galois extensions $L/K$ of local fields with non-abelian Galois group whose upper-numbering jumps are not integers. The abelianness assumption is genuinely necessary, not merely a convenience. The two cases correspond to two geometric pictures. In Case 1, the filtration of $G$ genuinely drops at level $t_0$ even after projecting to $G/H$, so the jump is "seen" by the smaller extension $F/K$. In Case 2, the filtration of $G/H$ is constant near $t_0$ but the filtration of $H$ drops, so the jump is "internal" to $H$ and detected by the smaller extension $L/F$. The key fact $G^t(L/K) \cap H = G^t(L/F)$ — that the upper-numbering filtration of $L/K$ restricts to the upper-numbering filtration of $L/F$ on the subgroup $H$ — is the content of the compatibility theorem applied in the reverse direction. This is where the upper numbering shows its advantage over the lower numbering: for the lower numbering, the intersection $G_s(L/K) \cap H$ does not in general equal $G_s(L/F)$, because the Herbrand change-of-variable function $\varphi$ is not compatible with restriction to subfields. The upper numbering was invented precisely to make this restriction property hold. Finally, let us verify the inductive step is complete: we have shown every jump $t_0$ of $G^t(L/K)$ satisfies $t_0 \in \mathbb{Z}_{\geq 0}$, which is what needed to be proved. Together with the base case $n = 1$ (vacuously true), the induction is complete. [/guided] [/step]