Proposition 5
Tag 0A32, permanent. Cite it as [Bourbaki, Tag 0A32].
Lie Groups and Lie Algebras, chapter II, § 2. Free Lie algebras
Let g be a Lie algebra and P a submodule of g. We define the submodules Pn of g by the formulae P1=P and Pn+1=[P,Pn] for n≥1. Then
[Pm,Pn]⊂Pm+n, Pn=p=1∑n−1[Pp,Pn−p]for n≥2. We prove (18) by induction on
m. The case
m=1 is obvious. By the Jacobi identity,
[[P,Pm],Pn]⊂[Pm,[P,Pn]]+[P,[Pm,Pn]], that is
[Pm+1,Pn]⊂[Pm,Pn+1]+[P,[Pm,Pn]]. The induction hypothesis implies that
[Pm,Pn+1]⊂Pm+n+1 and
[Pm,Pn]⊂Pm+n, whence
[Pm+1,Pn]⊂Pm+n+1+[P,Pm+n]=Pm+n+1. By formula (18),
Pn⊃∑p=1n−1[Pp,Pn−p]⊃[P1,Pn−1]=Pn, whence (19).
When we take P=g, the sequence (Pn) is the lower central series (Cng) of g (Chapter I, § 1, no. 5). Hence:
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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