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