If $H\leq Z (G)$ and $G/H$ is nilpotent, then $G$ is nilpotent. Let $G$ be a group. If $H\leq Z(G)$ and $G/H$ is nilpotent, then $G$ is nilpotent. To prove this proposition, first, I have tried to show that $G/Z (G)$ has non-trivial center. I can see that all cosets of $H$ in $Z (G)$ belongs to $Z (G/H)$ but this does not satisfy $G/Z (G)$ has non-trivial center. Can any one give me a hint for the proposition. Thanks.
 A: I have the solution in my old notes:
$G/H$ is nilpotent, for example of class $m$, so $L_{m+1}(G/H)=H/H$. We can probe this fact that $L_{m+1}(G/H)=L_{m+1}(G)H/H$ so,$$L_{m+1}(G)H/H=H/H$$ or $L_{m+1}(G)\subset H$. Since $H\subset Z(G)$ then $L_{m+1}(G)=[L_m(G),G]\leq[H,G]=\{e\}$. This is what we need
A: So you want to show that $G/Z(G)$ has a non-trivial center. First, assume that $G$ is not abelian, or you would be done anyway. Next, use that $G/Z(G)$ is a quotient of $G/H$ since $H$ is contained in $Z(G)$. This tells you that $G/Z(G)$ is abelian. Now you should be able to complete the proof.
A: Hint: Consider the upper central series of $G$ and the upper central series of $G/H$.  What is the connection between $Z_{i}(G/H)$ and $Z_i(G)$?  Can you show that $Z_r(G)=G$ for some $r$, knowing that $H\leqslant Z(G)$?

 If you aren't familiar, the upper central series of a group $K$ is $$1=Z_0\lhd Z_1 \lhd \cdots \lhd Z_i \lhd \cdots $$ where $Z_i/Z_{i-1}=Z(K/Z_{i-1})$.  $K$ is nilpotent if and only if $Z_r=K$ for some $r\in \mathbb{N}$.

A: I'll make a try..Please let me know if I'm correct:
Let $1=L_0\lhd ... \lhd L_n=\dfrac{G}{H},(1)$ be a central series of $G/H$
.Then, from correspondence theorem we have $L_i=\dfrac{A_i}{H}$ with $A_i\lhd A_{i+1}$ and $A_i\leq G$.
Let $1\lhd H\lhd A_1\lhd ... \lhd A_n=G,(2)$.Then
$\bigg[\dfrac{A_{i+1}}{H},\dfrac{G}{H}\bigg]\leq\dfrac{A_i}{H}$ since $(1)$ is a central series of $G/H$
But $\bigg[\dfrac{A_{i+1}}{H},\dfrac{G}{H}\bigg]=\dfrac{[A_{i+1},G]H}{H}\leq\dfrac{A_i}{H}\Rightarrow [A_{i+1},G]\leq A_i \forall i,$ and $[H,G]\leq[Z(G),G]=1\leq1 \checkmark$ 
So $(2)$ is a central series of $G$
