Proof: Lemma 1: For φ an isomorphism, we have φ(Z(G))=Z(φ(G)). Proof: For any x,y∈G, we observe φ(x)∈φ(Z(G))⇔x∈Z(G)⇔xy=yx⇔φ(xy)=φ(yx)⇔φ(x)φ(y)=φ(y)φ(x)⇔φ(x)∈Z(φ(G)) ◻
Now, let the overbar denote passage into G/Z(G).
Inductively prove that Zn(¯G)=¯Zn+1(G) for all nonnegative n. When n=0, we have Z0(¯G)=Z(G)/Z(G)=Z1(G)/Z(G)=¯Z1(G), so now validate the inductive step: Let φ:G/Zn(G)→¯G/¯Zn(G) be an isomorphism by φ(gZn(G))=¯g¯Zn(G) (the inverse of the natural isomorphism by the Third Isomorphism Theorem), and we have:
Zn(¯G)/Zn−1(¯G)=Z(¯G/Zn−1(¯G))⇒
Zn(¯G)/¯Zn(G)=Z(¯G/¯Zn(G))⇒
Zn(¯G)/¯Zn(G)=Z(φ(G/Zn(G)))⇒
Zn(¯G)/¯Zn(G)=φ(Z(G/Zn(G)))⇒
Zn(¯G)/¯Zn(G)=¯Zn+1(G)/¯Zn(G).
Since Zn+1(G) contains Zn(G) contains Z(G), we have ¯Zn+1(G) contains ¯Zn(G). Further, Zn(¯G) contains ¯Zn(G)=Zn−1(¯G). The Lattice Isomorphism Theorem describes a bijection between the subgroups of a parent group containing the normal subgroup and the subgroups of the quotient group, so we are left with Zn(¯G)=¯Zn+1(G).
Now, set n=m where Zm(¯G)=¯G and we have Zm(¯G)=¯Zm+1(G)=¯G, so similarly as above Zm+1(G)=G, therefore G is nilpotent (and ¯Zm(G)=Zm−1(¯G)≠¯G so Zm(G)≠G, therefore G is of class m+1 when G is nontrivial). ◻
No comments:
Post a Comment