= Solution
Suppose first that $N$ is the <Frobenius kernel>. An element $g\notin N$ fixes a point in the Frobenius action. If $n\in C_N(g)$, then $n$ maps that point to another point fixed by $g$. Since $g$ has only one fixed point and $N$ acts regularly, $n=1$. Thus
$$
C_N(g)=1\qquad(g\notin N).
$$
No nonidentity <conjugacy class> of $N$ can be fixed by such a $g$. If the class of $n\ne1$ were fixed, $gng^{-1}=vnv^{-1}$ for some $v\in N$, so $v^{-1}g\notin N$ would centralize $n$, a contradiction. Conjugation by $g$ consequently fixes just the identity class. By <Brauer's permutation lemma> it fixes just one <irreducible character>, necessarily $1_N$.
For every nonprincipal $\varphi\in\operatorname{Irr}(N)$, its <inertia group of a character> is therefore $I_G(\varphi)=N$. Since $N$ is normal, the restriction of the <induced character> is the sum of its conjugates, counted over $G/N$:
$$
(\varphi^G)_N=\sum_{gN\in G/N}\varphi^g.
$$
<Frobenius reciprocity> and <character orthogonality> give
$$
\langle\varphi^G,\varphi^G\rangle_G
=\langle\varphi,(\varphi^G)_N\rangle_N
=|I_G(\varphi):N|=1.
$$
An actual <character> of norm one is irreducible. \b[Every nonprincipal <irreducible character> of $N$ therefore induces irreducibly to $G$.]
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