 Solution

Because $N$ is an abelian group, every $\phi \in \mathrm{Irr}(N)$ is a linear character, hence a homomorphism to ${\mathbb{C}}^{\times }$. Its inertia group of a character contains $N$, whose inner conjugations act trivially on $N$. Write an arbitrary $g\in G$ as $g=nh$ with $n\in N$ and $h\in H$, using $G=NH$. Conjugation by $n$ does not affect $\phi$, so $g$ stabilizes $\phi$ if and only if $h$ does.

Suppose $h\ne 1$ stabilizes $\phi$. Then $\phi (h^{-1}yh)=\phi (y)$ for every $y\in N$, and its multiplicativity givesPart (iv) makes $y\mapsto [h,y]$ surjective onto $N$, so $\phi (x)=1$ for every $x\in N$. This would be the trivial character, contrary to the hypothesis. Therefore no nonidentity $h\in H$ stabilizes $\phi$, andThe argument uses both abelianness, to make $\phi$ multiplicative, and the fixed-point-free commutator bijection.