Let $G$ be a nilpotent group, $H\subsetneq G$. How to prove $H \subsetneq N$, where $N$ is a normalizer of $H$?

Question

Let $G$ be a nilpotent group, $H\subsetneq G$. How to prove $H \subsetneq N$, where $N$ is a normalizer of $H$?

I tried to use induction. It's easy to show the property works when nilpotency is equal to $1$, but I've got stuck moving further.

Answer 1

Hint: since $G$ is nilpotent it has a central series $\{N_i: 0 \leq i \leq r \}$ and $N_0=1 \subseteq H$ and $N_r=G$ which is not contained in $H$ obviously. So, there must be an index $i$ with $N_i \subseteq H$ and $N_{i+1} \nsubseteq H$. Try to show $N_{i+1} \subseteq N_G(H)$. Then $H \lt N_G(H)$.

Observe that we have a central series here, and centers are always contained in normalizers, so $$N_{i+1}/N_i \subseteq Z(G/N_i) \subseteq N_{G/N_i}(H/N_i) \subseteq N_G(H)/N_i.$$ It follows that $N_{i+1} \subseteq N_G(H)$.