If $G$ is nilpotent then so is $G/Z(G)$. The Problem: If $G$ is nilpotent then so is $G/Z(G)$.
My Background: Chapter 1-5 of Abstract Algebra $\mathit{3^{rd}}$ edition by Dummit and Foote.

For any group $G$ define the following subgroups inductively:
$Z_0(G)=1, Z_1(G)=Z(G)$, and $Z_{i+1}(G)$ is the subgroup of $G$
containing $Z_i(G)$ such that $Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))$ (i.e.,
$Z_{i+1}(G)$ is the complete preimage in $G$ of the center of
$G/Z_i(G)$ under the natural projection). The chain of subgroups
$$Z_0(G)\leq Z_1(G)\leq\dots$$ is called the upper central series of
$G$. A group is called nilpotent if $Z_c(G)=G$ for some
$c\in\mathbb{Z}$.

My Attempt: Taking hint from this post, I tried to show that $Z_i(\overline{G})=Z_{i+1}(G)/Z(G)$, where $\overline{G}=G/Z(G)$. We proceed by induction on $i$.
Base Case: $i=1$, then $Z_1(\overline{G})=Z(\overline{G})=Z(G/Z(G))=Z_2(G)/Z_1(G)=Z_2(G)/Z(G)$.
Induction Hypothesis: Suppose $Z_{i-1}(\overline{G})=Z_i(G)/Z(G)$.
Now, we have
$$
Z_i(\overline{G})/Z_{i-1}(\overline{G})=Z(\overline{G}/Z_{i-1}(\overline{G}))=Z(G/Z(G)/[Z_i(G)/Z(G)]),\tag1
$$
where the first equality is by the definition of the upper central series of $\overline{G}$ and the last equality is by the induction hypothesis. By the Third Isomorphism Theorem, $G/Z(G)/[Z_i(G)/Z(G)]\cong G/Z_i(G)$, hence $Z(G/Z(G)/[Z_i(G)/Z(G)])\cong Z(G/Z_i(G))$, thus $(1)$ becomes
$$
Z_i(\overline{G})/Z_{i-1}(\overline{G})\cong Z(G/Z_i(G))=Z_{i+1}(G)/Z_i(G),\tag2
$$
where the last equality is by the definition of the upper central series of $G$. Screeching halt.
I realize that the isomorphism I derived did not really help. But thinking of $Z_i(\overline{G})$ as the complete preimage in $\overline{G}$ of the center of $\overline{G}/Z_{i-1}(\overline{G})$ did not quite offer any insight on its relation with $Z_{i+1}(G)/Z(G)$ either, which is supposed to be an equality. Any help would be greatly appreciated.
 A: You need to remember exactly what the isomorphism is.
The isomorphism
$$\frac{G/N}{M/N} \cong \frac{G}{M}$$
is induced by the map $G/N \to G/M$ given by $xN \longmapsto gM$.
If we let $\overline{G}=G/Z(G)$, and we assume we already know that $Z_{k}(\overline{G}) = Z_{k+1}(G)/Z(G)$, then to prove that $Z_{k+1}(\overline{G}) = Z_{k+2}(G)/Z(G)$ we need to show that $Z_{k+2}(G)/Z(G)$ are precisely the elements of $\overline{G}$ that map to the center of
$$\frac{\overline{G}}{Z_k(\overline{G})} = \frac{G/Z(G)}{Z_{k+1}(G)/Z(G)}.$$
Let $g\in G$. Then:
$$\begin{align*}
gZ(G)\in Z_{k+2}(G)/Z(G) &\iff g\in Z_{k+2}(G)\\
&\iff gZ_{k+1}(G)\in Z(G/Z_{k+1}(G))\\
&\stackrel{*}{\iff} (gZ(G))\left(\frac{Z_{k+1}(G)}{Z(G)}\right) \in Z\left(\frac{G/Z(G)}{Z_{k+1}(G)/Z(G)}\right) \\
&\iff \left(gZ(G)\right)Z_{k}(\overline{G})\in Z\left(\frac{\overline{G}}{Z_k(\overline{G})}\right)\\
&\iff gZ(G)\in Z_{k+1}(\overline{G}).
\end{align*}$$
Thus, $Z_{k+2}(G)/Z(G) = Z_{k+1}(\overline{G})$.
The step tagged with a $*$ above the biconditional is the one that you are having trouble with. This comes because the isomorphism between $G/Z_{k+1}(G)$ and $\frac{G/Z(G)}{Z_{k+1}(G)/Z(G)}$ is induced as noted above: starting in $G/Z(G)$, you map $xZ(G)$ to $xZ_{k+1}(G)$. That means the induced isomorphism is
$$(xZ(G))\left(\frac{Z_{k+1}(G)}{Z(G)}\right) \longmapsto xZ_{k+1}(G).$$
In particular, $x\left(\frac{Z_{k+1}(G)}{Z(G)}\right)$ lies in the center  of the corresponding group if and only if $xZ_{k+1}(G)$ lies in the center of $G/Z_{k+1}(G)$. That gives the equivalence in question, with $x=gZ(G)$.
