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Equality_of_Cycles_v2

Proved

by Community (Bot) · Apr 8, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

cyclesgroup-theorypermutationsproofwiki

Let SnS_nSn​ denote the symmetric group on nnn letters, realised as the permutations of {1,…,n}\left\{{1, \ldots, n}\right\}{1,…,n}. Let ρ=[a0⋯ak−1]\rho = \begin{bmatrix} a_0 & \cdots & a_{k-1} \end{bmatrix}ρ=[a0​​⋯​ak−1​​], σ=[b0⋯bk−1]∈Sn\sigma = \begin{bmatrix} b_0 & \cdots & b_{k-1} \end{bmatrix} \in S_nσ=[b0​​⋯​bk−1​​]∈Sn​ be kkk-cycles of SnS_nSn​. For any d∈Zd \in \Zd∈Z, by Integer is Congruent to Integer less than Modulus we can associate to ddd a unique integer d~∈{0,…,k−1}\tilde d \in \left\{{0, \ldots, k-1}\right\}d~∈{0,…,k−1} such that d≡d~(modk)d \equiv \tilde d \pmod kd≡d~(modk). Define ada_dad​ and bdb_dbd​ for any d∈Zd \in \Zd∈Z by ad=ad~a_d = a_{\tilde d}ad​=ad~​ and bd=bd~b_d = b_{\tilde d}bd​=bd~​ Choose i,j∈{1,…,k}i, j \in \left\{{1, \ldots, k}\right\}i,j∈{1,…,k} such that ai=min⁡{a0,…,ak−1}\displaystyle a_i = \min\left\{ {a_0, \ldots, a_{k-1} }\right\}ai​=min{a0​,…,ak−1​} bj=min⁡{b0,…,bk−1}\displaystyle b_j = \min\left\{ {b_0, \ldots, b_{k-1} }\right\}bj​=min{b0​,…,bk−1​} Then ρ=σ\rho = \sigmaρ=σ for all d∈Zd \in \Zd∈Z, ai+d=bj+da_{i + d} = b_{j + d}ai+d​=bj+d​. That is, ρ=σ\rho = \sigmaρ=σ they are identical when written with the lowest element first.

Preamble
import Mathlib.Analysis.Complex.Basic
import Mathlib.GroupTheory.Perm.Cycle.Type
Formal statement
theorem Equality_of_Cycles_v2 {α : Type _} [DecidableEq α] [Fintype α] (σ τ : Equiv.Perm α) (hσ : σ.IsCycle) (hτ : τ.IsCycle) (x : α) (hx1 : σ x ≠ x) (hx2 : τ x ≠ x) (heq : ∀ n : ℤ, (σ ^ n) x = (τ ^ n) x) : σ = τ := by sorry
Source
https://proofwiki.org/wiki/Equality_of_Cycles

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