I wrote a proof down for the fact that if $G/Z (G)$ is nilpotent then $G$ is nilpotent and I wonder if this is correct. I would appreciate it if someone verifies this and/or suggests some alternative ways of doing this.
I intend to prove this using the upper central series. Let us denote $G/Z(G)$ as $\overline{G}$. Let
$$\{ Z(G) \}=Z_{0} (\overline{G}) \triangleleft Z_{1} (\overline{G})\triangleleft \ldots\triangleleft Z_{n} (\overline{G}) \triangleleft \ldots$$
be the upper central series for $\overline{G}$. Since each $Z_i(\overline{G})$ is a subgroup of $\overline G$, we can find subgroups $H_i$ of $G$ such that $Z_i (\overline G) =\frac{H_i}{Z(G)}$ by the Lattice Isomorphism Theorem for Groups. Again by the Lattice Isomorphism Theorem, we obtain
$$Z(G) = H_0 \triangleleft H_1 \triangleleft\ldots\triangleleft H_n \triangleleft \ldots$$
Let
$$\{ e \}=Z_{0} (G) \triangleleft Z_{1} (G)\triangleleft \ldots\triangleleft Z_{n} (G) \triangleleft \ldots$$
be the upper central series for $G$.
I claim that $H_i = Z_{i+1} (G)$ for each $k \in \mathbb{Z}$. I prove it using induction. The case where $i=0$ is immediate. Assume that $H_{i-1} = Z_{i} (G)$. Then we have
\begin{align*} \frac{Z_{i}(\overline{G})}{Z_{i-1}(\overline{G})} &= Z \left( \frac{\overline{G}}{Z_{i-1} (\overline{G})} \right) &\text{(by definition)}\\\frac{H_{i} / Z(G)}{H_{i-1} /Z(G)} &= Z \left( \frac{G/Z(G)}{H_{i-1}/Z (G)} \right) & \text{} \\ \frac{H_{i} / Z(G)}{Z_i (G) /Z(G)} &= Z \left( \frac{G/Z(G)}{Z_i(G)/Z (G)} \right) & \text{(by induction hypothesis)}\end{align*}
Since $Z_{i+1}(G)/Z_i (G) = Z(G/Z_i(G))$, we have from here that
$$Z \left( \frac{G/Z(G)}{Z_i(G)/Z (G)} \right) = \frac{Z_{i+1}(G) /Z(G)}{Z_i (G) /Z(G)}$$
From the previous two equations, we have
$$\frac{Z_{i+1}(G) /Z(G)}{Z_i (G) /Z(G)} = \frac{H_{i} / Z(G)}{Z_i (G) /Z(G)}$$
Applying the Lattice Isomorphism Theorem twice, we have that $Z_{i+1} (G) = H_i$. This completes the proof of my claim.
Since $\overline G$ is nilpotent, we have that $H_c = G$ for some $c \in \mathbb N$. Consequently from my claim, $Z_{c-1} (G) = G$ which shows that $G$ is nilpotent.
Best Answer
This looks fine, but a bit complicated. I am using $\gamma$'s and $\zeta$'s to denote the lower and upper central series respectively. Why don't you use the fact that $$\gamma_i(G/\zeta(G))=\gamma_i(G)\zeta(G)/\zeta(G).$$ This is easy to prove with induction on $i$. Once you have this, assume that $G/\zeta(G)$ is nilpotent of class $c$, so $\gamma_{c+1}(G/\zeta(G))=\overline{1}$. The above formula then yields $\gamma_{c+1}(G) \subseteq \zeta(G)$, implying $\gamma_{c+2}(G)=[G,\gamma_{c+1}(G)]=1$, so $G$ is nilpotent.