= Solution
Use the defining property of the <Frobenius complement>: $H\cap H^x=\{1\}$ whenever $x\notin H$. The <induced character> formula, extended linearly to any <class function>, is
$$
\theta^G(g)=\frac1{|H|}\sum_{x\in G:\ x^{-1}gx\in H}\theta(x^{-1}gx).
$$
For $h\in H\setminus\{1\}$, a summand requires $h\in H\cap xHx^{-1}$. The Frobenius intersection property forces $x\in H$. All $|H|$ remaining terms equal $\theta(h)$ since $\theta$ is a <class function> on $H$, so $\theta^G(h)=\theta(h)$. At the identity,
$$
\theta^G(1)=[G:H]\theta(1)=0=\theta(1).
$$
Together these give
$$
\boxed{(\theta^G)_H=\theta.}
$$
The zero identity value is essential: at nonidentity elements the intersection argument already gives the equality, whereas induction multiplies the identity value by the subgroup index.
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