= Solution
Let $V$ afford the given <group representation> of $H$. Its <induced representation> is the left $G$-module
$$
\boxed{\operatorname{Ind}_H^G V=\mathbb C[G]\otimes_{\mathbb C[H]}V}.
$$
Choose representatives $t$ of the left cosets $G/H$; then it is the direct sum of the spaces $t\otimes V$. If $gt=t'h$ with $h\in H$, the action sends $t\otimes v$ to $t'\otimes\Psi(h)v$. This is independent of the representatives up to equivalence, and its dimension is $[G:H]\dim V$.
Only cosets fixed by $g$ contribute to its trace. Their contribution is $\psi(t^{-1}gt)$, giving the <induced character> formula
$$
\boxed{\psi^G(g)=\sum_{\substack{tH\in G/H\\t^{-1}gt\in H}}\psi(t^{-1}gt)
=\frac1{|H|}\sum_{\substack{x\in G\\x^{-1}gx\in H}}\psi(x^{-1}gx)}.
$$
For an arbitrary <class function> $f$ on $H$, the same averaging formula defines its induced <class function> on $G$. Equivalently, extend $f$ by zero outside $H$ and average $|H|^{-1}\sum_x f(x^{-1}gx)$. This definition is linear and does not require $f$ to be an actual <character>.
For a complex <character> $\psi$ of a <finite group> $G$ afforded by a <representation> $\rho$, its <determinant character> is
$$
\boxed{(\det\psi)(g)=\det\rho(g)}.
$$
Multiplicativity of the determinant makes this a <linear character>; it does not mean the determinant of the scalar value $\psi(g)$. The definition depends only on the <character> because <characters> determine the equivalence class of a finite-dimensional complex <representation>. In particular the <determinant character> is trivial on the <derived subgroup> $G'$.
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