Let $G$ be a finite group and $p\in\pi(G)$. Suppose that 


$\quad$(i) for any non-principal $\chi\in\mathrm{Irr}(G)$, $p\nmid\frac{|G|}{|\mathrm{ker}\chi|\chi(1)}$;


$\quad$(ii) $E$ is the unique minimal normal subgroup $G$;


$\quad$(iii) $N$ is a $p'$-normal subgroup of $G$ and $E\leq N$;


$\quad$(iv) $\varphi\in \mathrm{Irr}{(N)}$ such that $E\nleq\mathrm{ker}\varphi$;


$\quad$(v) $T$ is the inertia group of $\varphi$ in $G$;


$\quad$(vi) $\eta\in\mathrm{Irr}(T)$ is an irreducible constituent of $\varphi^T$.


$\quad$(vii) $\theta=\eta^G\in\mathrm{Irr}(G)$.


QUESTION:


$\quad$ Is $T$ a $p'$-group.



Prove some inertia group $T$ is a $p'$-group
Yes. $\DeclareMathOperator{\Irr}{Irr} \renewcommand{\phi}{\varphi}$ 
Write $\phi^T = \sum\_{\eta} a\_{\eta} \eta$, the sum running over $\eta\in \Irr(T\mid \phi)$. Then $a\_{\eta} = [\phi^T,\eta] = [\phi, \eta\_N]$ and $\eta(1)=a\_{\eta}\phi(1)$. By Clifford theory, $\theta:= \eta^G \in \Irr(G)$ for every $\eta\in \Irr(T\mid \phi)$. Since $E\not\leq \ker(\phi)$, the kernel of every such $\theta$ does not contain the unique minimal subgroup $E$, and so $\ker(\theta)=\{1\}$. Then by (i), 
$$ p \not\mid \frac{\lvert G \rvert }{\theta(1)} 
 = \frac{ \lvert G \rvert }{ \lvert G:T \rvert \eta(1) }
 = \frac{ \lvert T \rvert }{ a\_{\eta}\phi(1) }
 . $$ 
By (iii), $p$ does not divide $\phi(1)$, so it follows that
$$ p \not \mid \frac{\lvert T \rvert }{ a\_{\eta} }
 \quad \text{for all } \eta\in \Irr(T\mid \phi) .$$
Thus $ \lvert P \rvert $ divides $a\_{\eta}$ for all $\eta$, where $P$ is a Sylow $p$-subgroup of $T$. But then $\lvert P \rvert^2$ divides $ \sum\_{\eta} a\_{\eta}^2 = \lvert T:N \rvert$. This is only possible when $P=\{1\}$, so $T$ is indeed a $p'$-group.


