Show that G is nilpotent when the quotient group is nilpotent

[mathmari]

Hey!

I want to show that if $H\subseteq Z(G)$ and $G/H$ is nilpotent then $G$ is also nilpotent. 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 $G/H$, $$\phi (A)\mapsto A/H, \ A\in G$$

We define $\tilde{N}_i:=\phi^{-1}(N_i), \ 1\leq i\leq k-1$.
We have that $\phi^{-1}(1)=H, \phi^{-1}(N_k)=\phi^{-1}(G/H)=G$.

Therefore, we get the series of normal subgroups $$1\leq H\leq \tilde{N}_1\leq \tilde{N}_2\leq \cdots \leq \tilde{N}_k=G$$ with $\tilde{N}_{i+1}/\tilde{N}_i\subseteq Z(G/\tilde{N}_i), \ 1\leq i\leq k$, right/ (Wondering)

Is everything correct? (Wondering)

Or do we not take $1\leq H$ at the second series? (Wondering)

 

[jvicens]

Hello! Great question. Your approach looks correct so far. However, I would suggest that you clarify your notation a bit. Instead of using $\tilde{N}_i$, which can be confusing, perhaps use $N'_i$ or $M_i$ for the subgroups of $G$ that correspond to $N_i$ in $G/H$. Also, your last question about $1\leq H$ in the second series is a valid concern. It would be better to explicitly state that $1\leq H\leq G$ is included in the series. Keep up the good work!