Let $\zeta_{i}G$ be the upper central series of G. Why does the definition of central series imply that $H\zeta_{i}G$ is normal in $H\zeta_{i+1}G$.?
I am trying to show that a finite group $G$ being nilpotent implies that every subgroup of $G$ is subnormal. The proof I have found states this as the crucial step, and I do not understand the reasoning.
Hint: use induction on $i$, start of the induction: you need to show that $H\zeta(G) \unlhd H\zeta_2(G)$. Remember that by definition $[G,\zeta_2(G)] \subseteq \zeta(G)$. Now take an $h \in H$ and $kz \in H\zeta(G)$, then $h^{-1}kzh=(h^{-1}kh)z \in H\zeta(G)$, since $z$ is central and commutes with any element. So $H$ normalizes $H\zeta(G)$. Let $x \in \zeta_2(G)$, then $x^{-1}kzx=x^{-1}kxz=x^{-1}kxk^{-1}kz=[x,k^{-1}]kz=k[x,k^{-1}]z \in H\zeta(G)$, since $[x,k^{-1}]\in \zeta(G)$. Hence, $\zeta_2(G)$ normalizes $H\zeta(G)$ too. And then certainly $H\zeta_2(G)$ normalizes $H\zeta(G)$. Can you finish?