I want to show that if $H\subseteq Z(G)$ and $G/H$ is nilpotent then $G$ is also nilpotent.
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I have done the following:
Since $G/H$ is nilpotent there is a series of normal subgroups $$1\leq N_1\leq N_2\leq \cdots \leq N_k=G/H$$ with $N_{i+1}/N_i\subseteq Z((G/H)/N_i)$.
From the correspondence theorem we have that there is a bijective mapping between the subgroups of $G$ that contain $H$ and the subgroups of $G/H$, right?
Using this theorem could we take the preimage of the mappings that map to $N_i, i=1, \dots , k$ to get the corresponding series of normal subgroups $$1\leq \tilde{N}_1\leq \tilde{N}_2\leq \cdots \leq \tilde{N}_k=G$$ with $\tilde{N}_{i+1}/\tilde{N}_i\subseteq Z(G/\tilde{N}_i)$ ?