Let $G$ be a nilpotent group of class $\leq n$ and $H\leq G$. There exists a sequence of subgroups $$G=H^1\supset H^2\supset\ldots\supset H^{n+1}=H$$ such that $H^{k+1}\trianglelefteq H^k$ and $H^k/H^{k+1}$ is abelian for all $1\leq k\leq n$.
Since $G$ is a nilpotent group of class $\leq n$, there exists sequence of subgroups of $G$: $$G=G^1\supset G^2\supset\ldots\supset G^{n+1}=\{e\}$$ such that $\left[G,G^k\right]\subset G^{k+1}$ for all $1\leq k\leq n$. Put $H^k:=HG^k$.
I have to show that $HG^{k+1}\trianglelefteq HG^{k}$. It is easy to show that $H^{k+1}$ is normalized by $H$. I want to show that $G^k$ also normalizes $H^{k+1}$, but I am experiencing difficulties. Any suggestions?
Let $y \in G^k$, $x \in G^{k+1}$, $h \in H$. Then $$y^{-1}hxy = h(h^{-1}y^{-1}hy)(y^{-1}xy).$$ We have $h \in H$ and $y^{-1}xy \in G^{k+1}$.
Since $G^k/G^{k+1} \le Z(G/G^{k+1})$ we also have $h^{-1}y^{-1}hy \in G^{k+1}$, so $y^{-1}hxy \in HG^{k+1}$.