Bourbaki

Proposition 8

Tag 05BR, permanent. Cite it as [Bourbaki, Tag 05BR].

Algebra, chapter I, § 5. Groups operating on a set

Let E be a finite set. The group SE\mathfrak{S}_E is generated by the transpositions.

For every permutation σ\sigma let FσF_\sigma be the set of x∈Ex \in E such that σ(x)=x\sigma(x) = x. We show by descending induction on pp that every permutation σ\sigma such that Card(Fσ)=p\mathrm{Card}(F_\sigma) = p is a product of transpositions. If p≥Card(E)p \geq \mathrm{Card}(E), the permutation σ\sigma is the identity mapping of E; it is the product of the empty family of transpositions. If p<Card(E)p < \mathrm{Card}(E), suppose that the property is proved for every permutation σ′\sigma' such that Card(Fσ′)>p\mathrm{Card}(F_{\sigma'}) > p. Now E−Fσ≠∅E - F_\sigma \neq \varnothing; let x∈E−Fσx \in E - F_\sigma and y∈σ(x)y \in \sigma(x). Then y≠xy \neq x and y∈E−Fσy \in E - F_\sigma. Let σ′=τx,y⋅σ\sigma' = \tau_{x,y} \cdot \sigma. The set Fσ′F_{\sigma'} contains FσF_\sigma and xx and hence Card(Fσ′)>Card(Fσ)=p\mathrm{Card}(F_{\sigma'}) > \mathrm{Card}(F_\sigma) = p. By the induction hypothesis σ′\sigma' is a product of transpositions and hence σ=τx,y⋅σ′\sigma = \tau_{x,y} \cdot \sigma' is a product of transpositions.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Nothing.

Transcribed from the printed edition, 1998, Springer. Not translated.