Nilpotent Groups

Lemma 7.9 LetG = G 0 G 1 ··· G n = 1be a central series for G. Thenforalli:γ i+1 (G) G i and Z i (G) G n−i .Proof: First observe that γ 1 (G) =G = G 0 . Suppose that γ i (G) G i−1for some i. Ifx ∈ γ i (G) andy ∈ G, thenG i x ∈ G i−1 /G i Z(G/G i ),so G i x commutes with G i y.Thereforeso [x, y] ∈ G i .HenceG i [x, y] =(G i x) −1 (G i y) −1 (G i x)(G i y)=G i ,γ i+1 (G) =[γ i (G),G] G i .Thus, by induction, the first inclusion holds.Now, Z 0 (G) =1 = G n . Suppose that Z i (G) G n−i . Since (G i )isacentral series for G,Thus if x ∈ G n−i−1 and y ∈ G, thenG n−i−1 /G n−i Z(G/G n−i ).G n−i x and G n−i y commute; i.e., [x, y] ∈ G n−i .Hence [x, y] ∈ Z i (G), so Z i (G)x and Z i (G)y commute. Since y is an arbitraryelement of G, wededucethatZ i (G)x ∈ Z(G/Z i (G)) = Z i+1 (G)/Z i (G)for all x ∈ G n−i−1 .ThusG n−i−1 Z i+1 (G) andthesecondinclusionholdsby induction.□We have now established the link between a general central series andthe behaviour of the lower and the upper central series.Theorem 7.10 The following conditions are equivalent for a group G:(i) γ c+1 (G) =1 for some c;(ii) Z c (G) =G for some c;(iii) G has a central series.Thus these are equivalent conditions for a group to be nilpotent.86