Prove that $G$ is nilpotent if $G/Z(G)$ is nilpotent.

I saw some answer is done by using central series or upper central series. But I want to attempt by using lower central series.
Define $\Gamma_1(G)=G$ and $\Gamma_{i+1}(G)=[\Gamma_i(G),G]$.

The first question I want to ask is for any normal subgroup $N$, is $$\Gamma_i(G/N)=\Gamma_i(G)/N$$ correct?

For $i=1$, $\Gamma_1(G/N)=G/N=\Gamma_1(G)/N$.
Suppose the result holds for some $i\geq 1$.
Then $\Gamma_{i+1}(G/N)=[\Gamma_i(G)/N,G/N]=[\Gamma_i(G),G]/N=\Gamma_{i+1}(G)/N$.
I saw that in some references they give $$\Gamma_i(G/N)=\Gamma_i(G)N/N$$ but I can't understand how they prove and use this and also the difference with the formula that I derived. Please check whether my following arguments is correct:

$\Gamma_i(G)N/N\\ =\{xnN\;|\;x\in\Gamma_i(G),n\in N\}\\ =\{xN\;|\;x\in \Gamma_i(G)\}\\ =\Gamma_i(G)/N$

If this is correct, since $G/Z(G)$ is nilpotent, then there exists an integer $r$ such that $$\Gamma_r(G)/Z(G)=\Gamma_r(G/Z(G))=\{Z(G)\}=\{1\}/Z(G)$$ Thus $\Gamma_r(G)=\{1\}$ and we are done. Also, if this attempt is correct, can it be used for other normal subgroup which is not $Z(G)$?

Furthermore, I try to use another method which involves homomorphism.
Define a natural map $\phi:G\rightarrow G/Z(G)$ where $\ker \phi=Z(G)$.
I had verified that $\phi(\Gamma_i(G))=\Gamma_i(G/Z(G))$.
So $\phi(\Gamma_r(G))=\{Z(G)\}$.
Thus $\Gamma_r(G)\leq \ker\phi=Z(G)$.
Then $\Gamma_{r+1}(G)=[\Gamma_r(G),G]\leq[Z(G),G]=\{1\}$
Thus $G$ is nilpotent.

I do apologize if this question sounds silly as I just start studying nilpotent group.

  • $\begingroup$ $\Gamma_i(G/N)=\Gamma_i(G)/N$ can't be right in general, because the right side isn't even defined unless $\Gamma_i(G) \supset N$. $\endgroup$ – Ted Nov 12 '16 at 8:44
  • $\begingroup$ @Ted So can I say that in term of coset they are equal. Just for a quotient group to be defined. $N$ should be normal subgroup of $\Gamma_i(G)$. So it is changed to $\Gamma_i(G)N$ to make it define? $\endgroup$ – Alan Wang Nov 12 '16 at 9:19
  • $\begingroup$ The correct statement is $\Gamma_i(G/N) = \Gamma_i(G)N/N$, which is isomorphic to $\Gamma_i(G)/(N \cap \Gamma_i(G))$.. $\endgroup$ – Derek Holt Nov 12 '16 at 9:54

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