Let $G$ be a finite group and $N$ be a normal subgroup $N\lhd G$ such that the factor group $G/N$ is nilpotent. Then there exists a nilpotent subgroup $H$ of $G$ such that $G=HN$.
proof:
I consider an arbitrary nilpotent subgroup $H \leq G$. Then, there exists a number $t$ such that $H^{(t)}=\{e\}$. Since $G/N$ is nilpotent, there exists a number $k$ such that $(G/N)^{(k)}=\{e\}$ implying that $G^k/N = \{e\}$. Also, since $N\lhd G$ and $H\leq G$, $HN$ is a subgroup of $G$. Then, we can write that $|HN|\leq |G|$. If I can show the inequality $|G|<|HN|$, then the proof will be done.
How can I show this equality $|G|<|HN|$?
