Nilpotent Quotient Group (6.1.6)

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

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). $\square$