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Conjugates_of_Transpositions

Proved

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

conjugationgroup-theorypermutationsproofwikitranspositions

Let k1,k2,k3∈{1,2,…,n}k_1, k_2, k_3 \in \left\{{1, 2, \ldots, n}\right\}k1​,k2​,k3​∈{1,2,…,n}. Then (1):[k1k2]=[k3k2][k1k3][k3k2](1): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_1 & k_3 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix}(1):[k1​​k2​​]=[k3​​k2​​][k1​​k3​​][k3​​k2​​] (2):[k1k2]=[k3k1][k3k2][k3k1](2): \quad \begin{bmatrix} k_1 & k_2 \end{bmatrix} = \begin{bmatrix} k_3 & k_1 \end{bmatrix} \begin{bmatrix} k_3 & k_2 \end{bmatrix} \begin{bmatrix} k_3 & k_1 \end{bmatrix}(2):[k1​​k2​​]=[k3​​k1​​][k3​​k2​​][k3​​k1​​]

Preamble
import Mathlib.Analysis.Complex.Basic
Formal statement
theorem Conjugates_of_Transpositions {α : Type _} [DecidableEq α] (σ : Equiv.Perm α) (a b : α) : σ * Equiv.swap a b * σ⁻¹ = Equiv.swap (σ a) (σ b) := by sorry
Source
https://proofwiki.org/wiki/Conjugates_of_Transpositions

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