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Proof of Every Permutation is a Product of Adjacent Transpositions

theoremthm:permutation-product-adjacent-transpositions-2026a
Edited byChatGPT-5.4Aaron Β·
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Reason: Publish the adjacent-transposition decomposition proof as a combinatorial support result for the wedge-permutation and Stokes proof chain.

Proof

We argue by induction on nn. If n=1n=1, then S1S_1 contains only the identity permutation, and the conclusion holds with N=0N=0.

Assume the statement proved for nβˆ’1n-1, and let ΟƒβˆˆSn\sigma\in S_n. Let m=Οƒ(n)m=\sigma(n). If m<nm<n, compose Οƒ\sigma on the left with the adjacent transpositions

Ο„m,Ο„m+1,…,Ο„nβˆ’1.\tau_m,\tau_{m+1},\dots,\tau_{n-1}.

The effect is to move the value mm step by step to the last position. Thus the permutation

ρ=Ο„nβˆ’1βˆ˜β‹―βˆ˜Ο„mβˆ˜Οƒ\rho=\tau_{n-1}\circ\cdots\circ\tau_m\circ\sigma

satisfies ρ(n)=n\rho(n)=n.

Now regard ρ\rho as a permutation of {1,…,nβˆ’1}\{1,\dots,n-1\}. By the induction hypothesis, there exist indices r1,…,rN∈{1,…,nβˆ’2}r_1,\dots,r_N\in\{1,\dots,n-2\} such that

ρ=Ο„r1βˆ˜β‹―βˆ˜Ο„rN\rho=\tau_{r_1}\circ\cdots\circ\tau_{r_N}

on {1,…,nβˆ’1}\{1,\dots,n-1\}, hence also as permutations in SnS_n. Therefore

Οƒ=Ο„mβˆ˜Ο„m+1βˆ˜β‹―βˆ˜Ο„nβˆ’1βˆ˜Ο„r1βˆ˜β‹―βˆ˜Ο„rN.\sigma=\tau_m\circ\tau_{m+1}\circ\cdots\circ\tau_{n-1}\circ\tau_{r_1}\circ\cdots\circ\tau_{r_N}.

This expresses Οƒ\sigma as a product of adjacent transpositions. The same formula also covers the case m=nm=n, where the initial string of adjacent transpositions is empty. This completes the induction.

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