Let $G$ be finite with $H \leq G$ and $N$ an abelian normal subgroup. Suppose $G=NN_G(H)$ and $H \cap N = \{1\}$. Show that $C_H(N)=1.$

Question

I have encountered the following problem while trying to understand a proof of a result. Here is the gist of it:

Let $G$ be finite solvable group with $H \leq G$ and $N$ a minimal normal subgroup of $G$

Suppose that $G = NN_G(H)$ and $H \cap N = \{1\}$. I need to show that $C_H(N) =1$

The proof argues that $C_H(N) \unlhd N_G(H)$ and $[C_H(N), N] =1$ , which I can comfortably show. I can't seem to get a deduction that $C_H(N)=1$ from these two facts