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Sunday, March 31, 2013

Nilpotent Quotient Group (6.1.6)

Dummit and Foote Abstract Algebra, section 6.1, exercise 6:

MathJax TeX Test Page If G/Z(G) is nilpotent, prove G is nilpotent.

Proof: Lemma 1: For φ an isomorphism, we have φ(Z(G))=Z(φ(G)). Proof: For any x,yG, we observe φ(x)φ(Z(G))xZ(G)xy=yxφ(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)/Zn1(¯G)=Z(¯G/Zn1(¯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)=φ(Zn+1(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)=Zn1(¯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)=Zm1(¯G)¯G so Zm(G)G, therefore G is of class m+1 when G is nontrivial).

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