# Product of Normal Subgroups

I'm trying to prove something regarding solvable groups, but I got stuck near the end with this problem: Let $K \unlhd H \leq G$ and $N \unlhd G$. I'm trying to prove that $NK \unlhd NH$. I've tried to do this through conjugation-invariance: $$(nh)^{-1}n'k(nh) = h^{-1}n^{-1}n'knh$$ but got stuck. I'd really appreciate a clue :)

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Just continue: \begin{align} (nh)^{-1}n'k(nh) = h^{-1}n^{-1}n'knh & = h^{-1}n^{-1}n'h\cdot h^{-1}kh\cdot h^{-1}nh \\ &\in NKN \end{align} and as $N$ is normal, $KN=NK$ (because $kn=(knk^{-1})\cdot k$).