Let $K\subseteq G$. Show that if $\chi$ is a monomial character of $G$, then, if $\chi_K \in Irr(K)$, $\chi_K$ is a monomial character of $K$. 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.
 A: 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.$\square$
