Exercise 6.11 from Isaac's Character theory of finite groups. I'm a noob in group theory, so please forgive me if I put some silly questions. I try to solve the following problem (6.11 from Isaac's book, Character theory of finite groups): 

Let $A\unlhd G$ with $A$ Abelian. Let $\chi$ be an irreducible monomial character of $G$. Prove that $\chi$ is a relative monomial character with respect to $A$, i.e. there exists a subgroup $H$ with $A\subset H\subset G$ and an irreducible character $\psi$ of $H$ such that $\psi_A$ is irreducible and $\psi^G=\chi$.

Since $\chi$ is monomial, there exists a subgroup $K$ and a linear character $\lambda$ of $K$ such that $\chi=\lambda^G$. The hint given is that: Every constituent of $(\lambda^{AH})_A = (\lambda_{A\cap H})^A$ has multiplicity $1$. I don't know how to prove this hint and how to solve the problem using it. The linear character $\psi_A$ should be a constituent of $(\lambda^{AH})_A$? I don't grasp the connection between $\psi$ and $\lambda$. Ito's theorem (according to which $\chi(1)||G:A|$) plays a role in solving this problem?
Let $A \unlhd G$ and $\chi$ a monomial irreducible (complex) character say $\chi=\lambda^G$, with $\lambda$ linear character of a subgroup $K$ of $G$. Consider the subgroup $KA$. Observe that $\lambda^{KA}$ is irreducible. By using Problem (5.2) in Isaacs' book (all references are in Character Theory of Finite Groups), we have $(\lambda^{KA})_A=(\lambda_{K \cap A})^A$. Since $A$ is abelian, we can find a linear character $\mu$ of $A$, such that $\mu_{K \cap A}=\lambda_{K \cap A}$. Now apply (6.17) Corollary (Gallagher): we must have 
$$(\lambda^{KA})_A=(\lambda_{K \cap A})^A=\sum_{\nu \in Irr(A/K \cap A)} \mu \nu$$
Observe that all $\nu$ and hence $\mu\nu$ are linear, since $A$ is abelian. The corollary also says that all the $\mu\nu$ are distinct and are all of the irreducible constituents of $(\lambda_{K \cap A})^A$. Hence $(\lambda^{KA})_A$ is the sum of distinct conjugates of the $\mu$. Now apply Clifford's Theorem ((6.11) Theorem), there must be a linear character $\psi \in I_{KA}(\mu)$, the inertia group of $\mu$, with $[\psi_A,\mu] \neq 0$, such that $\psi^{KA}=\lambda^{KA}$. Hence $\chi=\lambda^G=(\lambda^{KA})^G=(\psi^{KA})^G=\psi^G$ by transitivity of induction (see Problem (5.1)). So $H=I_{KA}(\mu)$ is the requested subgroup.
Induction on $k:=[G:A]$.  So, assume the result is true for groups which possess abelian normal subgroup of index $&lt;k$ (and satisfy the hypothesis etc.).

Let $\chi=\lambda^G$ for $\lambda$ a linear character of $H_1$. Consider the subgroup $AH_1$. 

(1) If $AH_1$ is proper in $G$, then, it contains $A$, and since $\lambda^G$ is irreducible, so $\lambda^{AH_1}$ is irreducible; i.e. $\chi_1=\lambda^{AH_1}$ is monomial character of subgroup $AH_1$ which contains abelian normal subgroup of index less than $[G:A]$, and so apply induction.

(2) Suppose $AH_1=G$. What this implies is that the subgroups $A$ and $H_1$ give single double coset in $G$. So apply Mackey decomposition (Exercise 5.6, Isaacs' book): so 
$$(\lambda^G)_{A}=(\lambda_{A\cap H})^A. $$
Now $A\cap H$ is subgroup of abelian group $A$, so the last character is direct sum of linear characters of $A$ with multiplicity $1$. (This proves the Hint given by Isaacs.)

So restriction of $\chi=\lambda^G$ to $A$ is sum of some irreducible characters of $A$ with multiplicity $1$; apply Clifford's theorem to reach the conclusion.