$G/N$ is nilpotent $\Leftrightarrow$ $N$ contains the intersection of all subgroups in the lower central series. I want to solve the following exercise:

Let $G$ be a group. $G^{0}\geq G^{1} \geq G^{2}\dots$ is a lower central series, s.t. $G^{0} = G$ and $G^{i+1} = [G,G^{i}]$. Let $N$ be a normal subgroup of $G$.
a) Then $G/N$ nilpotent $\Rightarrow$ $\bigcap\limits_{i\geq 0}G^{i} \subset N$.
b) With $G$ finite, $\bigcap\limits_{i\geq 0}G^{i} \subset N$ $\Rightarrow$ $G/N$ nilpotent.


So far I just have ideas to a):
I found a Corrolary that says: $N$ normal, $G/N$ nilpotent and $N\leq Z_{i}(G)$ for some $i$ (where it notates the upper central series), then $G$ is nilpotent.
I thought, if I could show that $N\leq Z_{i}(G)$, then with $G$ nilpotent, the lower central series terminates and the intersection would be trivial (? is that right?) and so contained in $N$.
Is that the right way? Or should I better approach that exercise differently?
Thanks and best,
Sara

*

*Sub-exercise 1. Show for any subset $A$ and subgroup $B$ of $G$, $AB\le B\implies A\le B$.

*Sub-exercise 2. Show that for any $L,\,M\le G$ and $N\trianglelefteq G$, we have $$\left[\frac{L}{N},\frac{M}{N}\right]=\frac{[L,M]N}{N}.$$

*Sub-exercise 3. By induction and the above, show that $(G/N)^{(i)}=G^{(i)}N/N$.
As corollary, deduce part (a) by relating $(G/N)^{(i)}$, $G^{(i)}$ and $N$ together using $(G/N)^{(i)}=N/N$, which comes from the hypothesis that $G/N$ is nilpotent.


*

*Sub-exercise 4. Argue that if $G$ is finite, then $G^{(j)}=G^{(j+1)}=\cdots$ for some $j$. Subsequently, prove that this final term $G^{(j)}$ is equal to the intersection $\bigcap_{\ell\ge0}G^{(\ell)}$.


Thus deduce part (b) as corollary by considering $(G/N)^{(j)}$ in view of $G^{(j)}=\bigcap_{\ell\ge0} G^{(\ell)}\le N$.
