= Solution
We prove closure of $N$ through <character kernels>, rather than assuming that a conjugacy-invariant set is a subgroup. Let $\eta\in\operatorname{Irr}(H)\setminus\{1_H\}$, write $d=\eta(1)$, and define the <generalised character>
$$
\Psi_\eta=(\eta-d1_H)^G+d1_G.
$$
Here $1_H,1_G$ are the <trivial characters>. Since $(\eta-d1_H)(1)=0$, part (i) gives $(\Psi_\eta)_H=\eta$ and $\Psi_\eta(1)=d$. Put $\vartheta=\eta-d1_H$. By <Frobenius reciprocity> and <character orthogonality>,
$$
\langle\vartheta^G,\vartheta^G\rangle_G=\langle\vartheta,(\vartheta^G)_H\rangle_H=\langle\vartheta,\vartheta\rangle_H=1+d^2,\qquad \langle\vartheta^G,1_G\rangle_G=-d.
$$
Consequently
$$
\langle\Psi_\eta,\Psi_\eta\rangle_G=(1+d^2)-2d^2+d^2=1.
$$
To justify the resulting irreducibility, write any <generalised character> as an integer combination $\sum_\chi a_\chi\chi$ of <irreducible characters>. Its squared norm is $\sum a_\chi^2$, so norm one makes it $\chi$ or $-\chi$ for one irreducible $\chi$. Positive degree rules out the negative sign. Thus $\Psi_\eta$ is an <irreducible character>, an instance of the fact that <positive-degree norm-one virtual characters are irreducible>.
If $x\in N\setminus\{1\}$, no conjugate of $x$ lies in $H$, so the induction formula gives $\vartheta^G(x)=0$. At $1$ it is also zero. Hence $\Psi_\eta(x)=d$ for all $x\in N$. For a finite-group representation we may choose an invariant positive Hermitian form, making every representing matrix unitary. Its trace equals its dimension only when all its unit-modulus eigenvalues are $1$. Thus the <kernel of a character> is $\{g:\Psi_\eta(g)=\Psi_\eta(1)\}$, and $N\subseteq\ker\Psi_\eta$.
Conversely, if $g\notin N$, then $g$ is conjugate to some $h\in H\setminus\{1\}$. The character of the <regular representation> of $H$ is
$$
\rho_H=\sum_{\eta\in\operatorname{Irr}(H)}\eta(1)\eta,\qquad \rho_H(1)=|H|,\qquad \rho_H(h)=0.
$$
If every nontrivial $\eta$ satisfied $\eta(h)=\eta(1)$, the sum at $h$ would instead equal $\sum\eta(1)^2=|H|$, a contradiction. Choose an $\eta$ for which these values differ. Since $\Psi_\eta$ restricts to $\eta$, it has $\Psi_\eta(g)=\eta(h)\ne d$, so $g\notin\ker\Psi_\eta$. Therefore
$$
\boxed{N=\bigcap_{\eta\in\operatorname{Irr}(H)\setminus\{1_H\}}\ker\Psi_\eta\triangleleft G.}
$$
This proves the <Frobenius kernel theorem>'s subgroup assertion. The definition also gives $N\cap H=\{1\}$. Using the permitted cardinality $|N|=[G:H]$, normality ensures $NH$ is a subgroup and
$$
|NH|=\frac{|N||H|}{|N\cap H|}=|G|,\qquad \boxed{NH=G\quad\text{and}\quad G=N\rtimes H.}
$$
Thus $N$ is the <Frobenius kernel>, and $H$ supplies the complementary factor.
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