Bourbaki

Definition 5

Tag 05D9, permanent. Cite it as [Bourbaki, Tag 05D9].

Algebra, chapter I, § 6. Extensions, solvable groups, nilpotent groups

Let G be a group. The lower central series of G is the sequence (Cn(G))n≥1(C^n(G))_{n \geq 1} of subgroups of G defined inductively by:

C1(G)=G,Cn+1(G)=(G,Cn(G)). C^1(G) = G, \quad C^{n+1}(G) = (G, C^n(G)).

Let f:G→G′f : G \to G' be a group homomorphism. It is seen, by induction on nn, that f(Cn(G))⊂Cn(G′)f(C^n(G)) \subset C^n(G') and that, if ff is surjective, f(Cn(G))=Cn(G′)f(C^n(G)) = C^n(G'). In particular, for all n≥1n \geq 1, C2(G)C^2(G) is a characteristic (and hence normal) subgroup of GG. For all n≥1n \geq 1, Cn(G)/Cn+1(G)C^n(G)/C^{n+1}(G) is contained in the centre of G/Cn+1(G)G/C^{n+1}(G).

Let (G1,G2,…)(G_1, G_2, \ldots) be a decreasing sequence of normal subgroups of GG such that (1) G1=GG_1 = G; (2) for all ii, Gi/Gi+1G_i/G_{i+1} is contained in the centre of G/Gi+1G/G_{i+1}. Then Ci(G)⊂GiC^i(G) \subset G_i, as is seen by induction on ii.

Now

(Cm(G),Cn(G))⊂Cm+n(G). (C^m(G), C^n(G)) \subset C^{m+n}(G).
For, if this relation is denoted by (Fm,n)(F_{m,n}), it follows from (Fm,n)(F_{m,n}), by no. 2, Proposition 5, that
(Cm(G),Cn+1(G))⊂(G,(Cm(G),Cn(G))).(Cn(G),(G,Cm(G)))⊂Cm+n+1(G).(Cm+1(G),Cn(G)). \begin{align*} (C^m(G), C^{n+1}(G)) &\subset (G, (C^m(G), C^n(G))).(C^n(G), (G, C^m(G))) \\ &\subset C^{m+n+1}(G).(C^{m+1}(G), C^n(G)). \end{align*}
Hence ((Fm,n) and (Fm+1,n))⇒(Fm,n+1)((F_{m,n}) \text{ and } (F_{m+1,n})) \Rightarrow (F_{m,n+1}). As (Fm,1)(F_{m,1}) and (F1,n)(F_{1,n}) are obvious (Fm,n)(F_{m,n}) follows by induction.

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

Cited by

Cites

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