upper central series, taking intersection with subgroup ($\zeta_i G \cap H \leq \zeta_i H$) Let $\zeta_i G$ be the upper central series of $G$. Show that for 'any' subgroup $H$, we have 
$$\zeta_i G \cap H \leq \zeta_i H$$
where $\zeta_i H$ is the upper central series of $H$.

I tried induction. The case $n=0$ was easy. Then things get messy. 
My question stems from a small step in a proof. There they also presuppose that $G$ is nilpotent of class $\geq 2$ and $H= \left < a, [G,G] \right >$ with $a \in G$. 
However, I do not suspect these conditions are necessary and that the general case also holds.  

Inductive step in brief:
$h \in \zeta_i G \cap H \Rightarrow [g,h] \in \zeta_{i-1}G\ \forall g \in G \Rightarrow\ ({\rm by\ induction})\ [g,h] \in \zeta_{i-1}H\ \forall g \in H \Rightarrow h \in \zeta_iH$.
