(i) Let α,β be in R. There exists a biadditive mapping
ϕαβ:grα(G)×grβ(G)→grα+β(G)
which maps (xGα+,yGβ+) onto (x,y)Gα+β+.
(ii) Let ϕ be the biadditive mapping of gr(G)×gr(G) into gr(G) whose restriction to grα(G)×grβ(G) is ϕαβ for every ordered pair (α,β). The mapping ϕ gives gr(G) a Lie Z-algebra structure.
(i) Recall the identity
(xx′,y)=(x,y)x′(x′,y)
for x,x′,y in G (Algebra, Chapter I, § 6, no. 2, formula (4 bis)).
For x∈Gα and y∈Gβ, the class modulo Gα+β+ of the element (x,y) of Gα+β will be denoted by f(x,y). For a in Gα+β and x′ in G, a−1.ax′=(a,x′)∈Gα+β+; in particular f(x,y) is equal to the class modulo Gα+β+ of (x,y)x′. Formula (15) therefore implies
f(xx′,y)=f(x,y)f(x′,y).
Now (y,x)=(x,y)−1, whence
f(y,x)=f(x,y)−1.
From (16) and (17) we deduce
f(x,yy′)=f(x,y)f(x,y′).
We have to prove that the mapping f:Gα×Gβ→grα+β(G) defines on taking quotients a mapping ϕαβ:grα(G)×grβ(G)→grα+β(G). By (16) and (18) it suffices to prove that f(x,y)=0 if x∈Gα+ or y∈Gβ+, which follows from (13').
(ii) As (x,x)=e, it follows from (17) that ϕ is an alternating Z-bilinear mapping. Hence it remains to prove that, for u∈grα(G), v∈grβ(G) and w∈grγ(G),
ϕ(u,ϕ(v,w))+ϕ(v,ϕ(w,u))+ϕ(w,ϕ(u,v))=0.
Let x∈Gα,y∈Gβ and z∈Gγ be elements representing respectively u,v and w. We know that xy and x are two elements of Gα which are congruent modulo Gα+ and hence xy is a representative of u in Gα; as (y,z) is a representative of ϕ(v,w) in Gβ+γ, we see that (xy,(y,z)) is a representative of ϕ(u,ϕ(v,w)) in Gα+β+γ. By cyclic permutation, we see that (yz,(z,x)) and (zx,(x,y)) represent respectively ϕ(v,ϕ(w,u)) and ϕ(w,ϕ(u,v)) in Gα+β+γ. Relation (19) is then a consequence of the following identity (Algebra, Chapter I, § 6, no. 2, formula (15)):
(xy,(y,z)).(yz,(z,x)).(zx,(x,y))=e.
The Lie algebra gr(G) over Z defined in Proposition 1 is called the graded Lie algebra associated with the filtered group G.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.