$Z_{i+1}(G)/Z(G)=Z_i(G/Z(G))$ I want to show the following statement:
$$ G/Z(G) \ \text{nilpotent} \ \Rightarrow \ G \ \text{nilpotent} $$
I know that $Z_{i+1}(G)/Z(G)=Z_i(G/Z(G))$ clearly gives the desired result. But, how can I prove this?
 A: It follows directly that $G$ is nilpotent. Just write out what it means that $G/Z(G)$ is nilpotent, say of class $m$, and consider the canonical map $p: G\rightarrow G/Z(G)$. Then
$$
p([\ldots [[g_1,g_2],g_3],\ldots , g_m],g_{m+1}])=[...[[p(g_1),p(g_2)],p(g_3)],\ldots ,p(g_m)],p(g_{m+1})]=1
$$
for all $g_1,\ldots ,g_{m+1}\in G$. Hence $[\ldots [[g_1,g_2],g_3],\ldots , g_m],g_{m+1}]\in Z(G)$, so that 
$$
[\ldots [[g_1,g_2],g_3],\ldots , g_{m+1}],g_{m+2}]=1
$$
for all $g_1,\ldots ,g_{m+2}\in G$. This means that $G$ is nilpotent of class at most $m+1$.
A: Another way to do this is to use the lower central series $\gamma_i(.)$: if $G/Z(G)$ is nilpotent then for some natural number $n$, $\gamma_{n+1}(G/Z(G))=\bar{1}$. In general, for any normal subgroup $N$ of $G$, $\gamma_i(G/N)=\gamma_i(G)N/N$. So we get $\gamma_{n+1}(G)Z(G)/Z(G)=\bar{1}$, that is, $\gamma_{n+1}(G) \subseteq Z(G)$. This implies that $\gamma_{n+2}(G)=[\gamma_{n+1}(G),G]=1$, whence $G$ is nilpotent.
A: @Derek, all right: it can be proved with induction on $i$, for $i=1$ by definition we have $Z(G/Z(G))=Z_2(G)/Z(G)$. Let $i \gt 1$ and write overbar $\overline {.}$ for quotients modulo $Z(G)$. Observe that this is well-defined since $Z(G) \subseteq Z_{i}(G)$ for all $i\geq 1$. Then we have to prove that $Z_i(\overline{G})=\overline{Z_{i+1}(G)}$. Now $$\overline{x} \in Z_i(\overline{G}) \Leftrightarrow \forall \overline{g} \in \overline{G}:  [\overline{g},\overline{x}] \in Z_{i-1}(\overline{G})$$
$$\Leftrightarrow  \forall \overline {g} \in \overline{G}:  \overline{[g,x]} \in Z_{i-1}(\overline{G})$$ $$\Leftrightarrow (induction step) \forall \overline{g} \in \overline{G}:  \overline{[g,x]} \in \overline{Z_{i}(G)}$$
$$\Leftrightarrow \overline {x} \in \overline{Z_{i+1}(G)}.$$
A: Another alternative for a proof in the case of $G$ being finite: Since $\overline{G}:=G/Z(G)$ is nilpotent, any Sylow subgroup of $G$ is normal in $G$. Let $p$ be a prime dividing the order of $G$, and let $P$ be a Sylow $p$-subgroup of $G$. Then $H:=PZ(G)$ is normal in $G$, so by Frattini's argument, $G=N_G(P)H$. Since $H$ normalizes $P$, this implies $G=N_G(P)$, i.e. $P$ is normal in $G$.
Since this holds for all $p$, $G$ is itself nilpotent.
