Zero at identity induction restriction for a frobenius complement

Let $h\in H$ and let $x\in G$ satisfy $x^{-1}hx\in H$, with $h\mathrm{\ne }1$. Then $h=x(x^{-1}hx)x^{-1}$ lies in $H\cap xH{x}^{-1}$, and $h\mathrm{\ne }1$; so the complement condition forces $x\in H$. Consequently, for $h\mathrm{\ne }1$, the summation index set $\{x\in G:x^{-1}hx\in H\}$ is contained in $H$.

For $x\in H$ one has $x^{-1}hx\in H$ because $H$ is closed under products and inverses, and then $\theta (x^{-1}hx)=\theta (h)$ because $\theta$ is a class function on $H$.

For $h=1$ one has $x^{-1}1x=1$ for every $x\in G$, so each summand in [F1] is $\theta (1)=0$ and ${\mathrm{Res}}_H^G{\mathrm{Ind}}_H^G\theta (1)=0=\theta (1)$.

If $h\in H\setminus \{1\}$, the elements $x\in G$ with $x^{-1}hx\in H$ are exactly the elements of $H$, by steps 1.1 and 1.2; hence ${\mathrm{Res}}_H^G{\mathrm{Ind}}_H^G\theta (h)=\frac{1}{\mathrm{\mid }H\mathrm{\mid }}{\sum }_{x\in H}\theta (h)=\theta (h)$.

The two cases $h=1$ and $h\mathrm{\ne }1$ cover every element of $H$, so the induced class function restricts to $\theta$, as claimed. ∎