Processing math: 5%

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,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))~~~\square

Now, let the overbar denote passage into G/Z(G).

Inductively prove that Z_{n}(\overline{G})=\overline{Z_{n+1}(G)} for all nonnegative n. When n=0, we have Z_0(\overline{G})=Z(G)/Z(G)=Z_1(G)/Z(G)=\overline{Z_1(G)}, so now validate the inductive step: Let φ : G/Z_n(G)→\overline{G}/\overline{Z_n(G)} be an isomorphism by φ(gZ_n(G))=\overline{g}\overline{Z_n(G)} (the inverse of the natural isomorphism by the Third Isomorphism Theorem), and we have:
Z_n(\overline{G})/Z_{n-1}(\overline{G})=Z(\overline{G}/Z_{n-1}(\overline{G}))⇒
Z_n(\overline{G})/\overline{Z_n(G)}=Z(\overline{G}/\overline{Z_n(G)})⇒
Z_n(\overline{G})/\overline{Z_n(G)}=Z(φ(G/Z_n(G)))⇒
Z_n(\overline{G})/\overline{Z_n(G)}=φ(Z(G/Z_n(G)))⇒Z_n(\overline{G})/\overline{Z_n(G)}=φ(Z_{n+1}(G)/Z_n(G))⇒
Z_n(\overline{G})/\overline{Z_n(G)}=\overline{Z_{n+1}(G)}/\overline{Z_n(G)}.
Since Z_{n+1}(G) contains Z_{n}(G) contains Z(G), we have \overline{Z_{n+1}(G)} contains \overline{Z_n(G)}. Further, Z_n(\overline{G}) contains \overline{Z_n(G)}=Z_{n-1}(\overline{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 Z_n(\overline{G})=\overline{Z_{n+1}(G)}.

Now, set n=m where Z_m(\overline{G})=\overline{G} and we have Z_{m}(\overline{G})=\overline{Z_{m+1}(G)}=\overline{G}, so similarly as above Z_{m+1}(G)=G, therefore G is nilpotent (and \overline{Z_{m}(G)}=Z_{m-1}(\overline{G})≠\overline{G} so Z_{m}(G)≠G, therefore G is of class m+1 when G is nontrivial). \square

No comments:

Post a Comment