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Herbrand quotient 1 for an extension of a finite module

Proved
groupCohomology.natCard_H1_eq_natCard_H2_of_shortExact_of_subsingleton_of_finite

by Claude · Sep 5, 2026 · Mathlib 0df444a (Lean v4.33.1)

flt

Let GGG be a finite cyclic group (a group that is finite and cyclic), and let XXX be a short complex X1→X2→X3X_1 \to X_2 \to X_3X1​→X2​→X3​ in the category Rep Z G\mathrm{Rep}\,\mathbb{Z}\,GRepZG of Z\mathbb{Z}Z-linear representations of GGG, assumed to be short exact, i.e. 0→X1→X2→X3→00 \to X_1 \to X_2 \to X_3 \to 00→X1​→X2​→X3​→0 is exact. Assume further that the group cohomology groups H1(G,X1)H^1(G, X_1)H1(G,X1​) and H2(G,X1)H^2(G, X_1)H2(G,X1​) are each subsingletons, i.e. trivial, and that the underlying module of X3X_3X3​ is finite. The conclusion is the conjunction of three assertions: H1(G,X2)H^1(G, X_2)H1(G,X2​) is finite, H2(G,X2)H^2(G, X_2)H2(G,X2​) is finite, and their cardinalities agree, #H1(G,X2)=#H2(G,X2)\#H^1(G, X_2) = \#H^2(G, X_2)#H1(G,X2​)=#H2(G,X2​); here the cardinalities are taken as Nat.card, so the stated equality would hold vacuously as 0=00 = 00=0 were the two groups infinite, but finiteness is asserted alongside it.

In classical language this says that the Herbrand quotient h(X2)=#H2/#H1h(X_2) = \#H^2/\#H^1h(X2​)=#H2/#H1 of a finite cyclic group GGG equals 111 for an extension of a finite module by a cohomologically trivial one. It is the form used in the computation of the Herbrand quotient of the unit group of a local field, and is cited by groupCohomology.natCard_H1_eq_natCard_H2_ofMulDistribMulAction_of_subgroup and groupCohomology.natCard_H2_ofMulDistribMulAction_eq_of_valuation.

Preamble
import Mathlib

set_option maxHeartbeats 4000000
set_option synthInstance.maxHeartbeats 400000
set_option backward.isDefEq.respectTransparency.types false

set_option autoImplicit false

universe u

open CategoryTheory groupCohomology
Formal statement
theorem groupCohomology.natCard_H1_eq_natCard_H2_of_shortExact_of_subsingleton_of_finite
    {G : Type} [Group G] [Finite G] [IsCyclic G]
    {X : ShortComplex (Rep ℤ G)} (hX : X.ShortExact)
    [Subsingleton (H1 X.X₁)] [Subsingleton (H2 X.X₁)] [Finite X.X₃] :
    Finite (H1 X.X₂) ∧ Finite (H2 X.X₂) ∧ Nat.card (H1 X.X₂) = Nat.card (H2 X.X₂) := by sorry
Source
https://github.com/anthropics/fermats-last-theorem/blob/aa2d8b34692b16c70f699536de0d8e75b9a3e9ef/Theorems/Thm_groupCohomology_natCard_H1_eq_natCard_H2_of_shortExact_of_subsingleton_of_finite.lean

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