$G/N$ 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
 A: No need for any of the book's lemmas!  Instead, understand what's going on. Intuitively, what you're being asked to prove is that whenever $\bigcap_{i\geq 0} G^i$ is nontrivial (and thus $G$ is not nilpotent), you can make $G$ into a nilpotent group by quotienting out by $\bigcap_{i\geq 0} G^i$.  In fact, $\bigcap_{i\geq 0} G^i$ is the smallest such subgroup so that this quotient is nilpotent.
What do you need to do to prove this?


*

*What is a nilpotent group?



 This one depends on what definition you're using, as there are a bunch of equivalent definitions, but I assume you're using this one: a group $K$ is nilpotent iff $K^n=1$ for some $n$, where $K^n$ (as in your notation) is the $n$th term in the lower central series.



*

*What is $\bigcap_{i\geq 0} G^i$?



 Surely $G^{i+1}\subseteq G^i$.  (Prove this real quick.)  Thus, if $\bigcap_{i\geq 0} G^i$ is not trivial, there must be some $m$ for which $G^m= G^k$ for every $k\geq m$, and so $G^m=\bigcap_{i\geq 0} G^i$.



*

*Determine the relationship between $G^i$ and $(G/N)^i$.



 What does it say about $G^i$ when $(G/N)^i$ is trivial?  What does it say about $(G/N)^i$ when $G^i\leqslant N$?   (If you're still stuck: can you rewrite $(G/N)^i$ in terms of $G^i$ and $N$)?

Can you connect the dots for me?
A: *

*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 $G^{(j)}=\bigcap_{\ell\ge0} G^{(\ell)}\le N$.
