Bourbaki

Theorem 1

Tag 0A3H, permanent. Cite it as [Bourbaki, Tag 0A3H].

Lie Groups and Lie Algebras, chapter II, § 2. Free Lie algebras

Let HH be a Hall set relative to XX and Ψ\Psi the canonical mapping of M(X)M(X) into the free Lie algebra L(X)L(X). The restriction of Ψ\Psi to HH is a basis of the module L(X)L(X).

For every element ww of HH we write wˉ=Ψ(w)\bar{w} = \Psi(w).

(A) The case where XX is finite.

If XX is empty, so are M(X)M(X) and therefore HH and L(X)L(X) is zero. If XX has a single element xx, H∩Mn(X)H \cap M^n(X) is empty for n≥2n \geq 2 (Proposition 12 (a)). Therefore, H={x}H = \{x\}; we know also (no. 2, Remark) that the module L(X)L(X) is free and has basis {xˉ}\{\bar{x}\}. The theorem is therefore true when XX has at most one element.

Suppose henceforth that XX has at least two elements; choose sequences (wp)(w_p) and (Pp)(P_p) with the properties stated in Proposition 13. For every integer p≥0p \geq 0, let LpL_p denote the submodule of L(X)L(X) generated by the elements wˉi\bar{w}_i with 0≤i<p0 \leq i < p and gpg_p the Lie subalgebra of L(X)L(X) generated by the family (uˉ)u∈Pp(\bar{u})_{u \in P_p}.

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

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