Characteristicity of the Central Upper Series (6.1.1)

Dummit and Foote Abstract Algebra, section 6.1, exercise 1:

Prove $Z_i(G)~\text{char}~G$ for all $i$.

Proof: Lemma 1: $Z(G)~\text{char}~G$ for any group $G$. Proof: For any $x,y∈G,~φ∈Aut(G)$ we have: $$φ(x)∈Z(G)⇔φ(x)φ(y)=φ(x)φ(y)$$ $$⇔φ(xy)=φ(yx)⇔xy=yx⇔x∈Z(G)~~~\square$$
Lemma 2: For $K~\text{char}~G$ and $φ∈Aut(G)$ and letting the bar notation denote passage into $G/K$, we have $ψ∈Aut(\bar{G})$ where $ψ: \bar{x} \mapsto \overline{φ(x)}$. Proof:
Well-defined: $\bar{x}=\bar{y}⇒y^{-1}x∈K⇒φ(y^{-1}x)∈K⇒\overline{φ(x)}=\overline{φ(y)}⇒ψ(\bar{x})=ψ(\bar{y})$.
Homomorphic: $ψ(\overline{xy})=\overline{φ(xy)}=\overline{φ(x)φ(y)}=\overline{φ(x)}~\overline{φ(y)}=ψ(\overline{x})ψ(\overline{y})$.
Injective: $$ψ(\overline{x})=ψ(\overline{y})⇒\overline{φ(x)}=\overline{φ(y)}⇒\overline{φ(y^{-1}x)}=1$$ $$⇒φ(y^{-1}x)∈K⇒y^{-1}x∈K⇒\overline{x}=\overline{y}$$ .
Surjective: For any $\overline{x}$, let $y$ be the preimage of $x$ by $φ$. By way of construction, $ψ(\overline{y})=\overline{x}$.$\square$

Returning to the main result, we shall proceed by induction. Clearly $Z_0(G)~\text{char}~G$, so apply the inductive hypothesis on $i$. Let the overbar denote passage into $G/Z_{i-1}$. Since $\overline{Z_i(G)}=Z(\overline{G})$, by lemma 1 we have $\overline{Z_i(G)}~\text{char}~\overline{G}$. Let $φ$ be any automorphism of $G$, and let $ψ$ be the automorphism of $\overline{G}$ afforded by $φ$. Complete the proof: $x∈Z_i(G)⇔\overline{x}∈\overline{Z_i(G)}⇔ψ(\overline{x})∈\overline{Z_i(G)}⇔\overline{φ(x)}∈\overline{Z_i(G)}⇔φ(x)∈Z_i(G)$. $\square$