Question: (This is 5.8 in Isaacs CTFG) Let $\chi$ be a monomial character of $G$ and suppose $K\subseteq G$ with $\chi_K\in\text{Irr}(K)$. Show that $\chi_K$ is a monomial character of $K$.
My idea: So, we want to show that $\chi_K=\lambda^K$ where $\lambda$ is a linear character of some subgroup of $K$. So, let $H\subseteq K\subseteq G$ and let $\lambda\in\text{Irr}(H)$. Then, by Frobenius Reciprocity, we have $[\lambda,(\chi_K)_H]=[\lambda^K,\chi_K]=[(\lambda^K)^G,\chi]=[\lambda^G,\chi]$, where the last equality follows by exercise 5.1. I am just not sure if, one, this is correct, and, two, if this will end up leading anywhere. So far, I haven't gotten it to go anywhere useful. I also haven't used the fact that $\chi$ is monomial, but I was hoping that $\chi$ showing up in the final inner product could be used with it being monomial and get us somewhere.
Any help is greatly appreciated! Thank you.
You need the following lemma.
(I urge you to draw a diagram to depict the different groups and characters.)
Lemma Let $H$ and $K$ be subgroups of $G$ and $\varphi$ a character of $H$. Assume that $(\varphi^G)_K \in Irr(K)$. Then $G=HK$.
Proof Put $\varphi^G=\chi$. Observe that since $(\varphi^G)_K$ is irreducible, both $\varphi$ and $\chi$ must be irreducible. Let $\psi \in Irr(H \cap K)$ be an irreducible constituent of $\varphi_{H \cap K}$.
By Frobenius Reciprocity, $\varphi$ must be an irreducible constituent of $\psi^H$, say $\psi^H=a\varphi+\Delta$, with $a$ a positive integer and $\Delta$ a character of $H$ with $[\Delta,\varphi]=0$ or $\Delta=0$. It follows that $\psi^G=(\psi^H)^G=a\varphi^G+\Delta^G=a\chi+\Delta^G$. So again by Frobenius Reciprocity, $[\psi^G,\chi]=[(\psi^K)^G,\chi]=[\psi^K,\chi_K] \geq a.$
On the other hand, $a=[\varphi,\psi^H]=[\varphi_{H \cap K},\psi]$, so $\varphi_{H \cap K}=a\psi+\Gamma$ with $\Gamma$ a character of $H \cap K$ with $[\Gamma,\psi]=0$ or $\Gamma=0$. Hence $(\varphi_{H \cap K})^K=a\psi^K+\Gamma^K$ and it follows that $[(\varphi_{H \cap K})^K, \chi_K]=[a\psi^K+\Gamma^K,\chi_K]=a[\psi^K,\chi_K]+[\Gamma^K,\chi_K] \geq a^2$, by the last formula of the previous paragraph. This means that the irreducible character $\chi_K$ has at least multiplicity $a^2$ as an irreducible constituent of $(\varphi_{H \cap K})^K$. This implies that $$(\varphi_{H \cap K})^K(1)=\varphi(1)|K:H \cap K| \geq a^2\chi(1) \geq \chi(1)=\varphi(1)|G:H|$$ It follows that $|G| \leq \frac{|H| \cdot |K|}{|H \cap K|}=|HK|$. Hence, $G=HK$, since $HK \subseteq G$ as a set.$\square$
Notes (1) This is Problem(5.7) in Isaacs' book. (2) From the last part of the proof it follows that in fact $a=1$.
Corollary 1 Let $\chi$ be a monomial character of $G$ and suppose that $K \leq G$ with $\chi_K \in Irr(K)$. Then $\chi_K$ is monomial.
Proof We can find an $H \leq G$ and a linear $\lambda \in Irr(H)$ with $\lambda^G=\chi$. Since $\chi_K$ is irreducible, the Lemma guarantees that $G=HK$. But then (see Problem(5.2) in Isaacs's CTFG), $\chi_K=(\lambda^G)_K=(\lambda_{H \cap K})^K$, whence $\chi_K$ is monomial.$\square$