Solvable groups in group theory If $N \unlhd G$, and $M,K \leq G$ such that $M \unlhd K$, then does it imply that $MN \unlhd KN$? If yes, how?

First of all, $N$ normalizes $MN$. In fact, if $n_{1}, n_{2} \in N$, and $m \in M$, then
$$
(m n_{1})^{n_{2}} = n_{2}^{-1} m n_{1} n_{2}  = m (m^{-1} n_{2}^{-1} m) n_{1} n_{2} = m (n_{2}^{-1})^{m} n_{1} n_{2}\in MN,
$$
as $N$ is normal in $G$.
And then $K$ also normalizes $MN$. If $k \in K$, then
$$
(m n_{1})^{k} = m^{k} n_{1}^{k} \in M N,
$$
as $N$ is normal in $G$, and $M$ is normal in $K$.
Therefore $\langle K, N \rangle = KN$ normalizes $MN$.
I am writing $x^{y} = y^{-1} x y$.
