Exercise 6.11 from Isaac's Character theory of finite groups.

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.

There is a converse to Ito's Theorem that runs as follows.

Proposition Let $N$ be a normal Hall subgroup of $G$ and assume that $\chi(1) \mid |G:N|$ for all $\chi \in Irr(G)$. Then $N$ must be abelian.

Proof Let $\chi \in Irr(G)$. By Clifford's Theorem we have $\chi_N=e\sum_{i=1}^t \vartheta_i$, where $\vartheta_i$'s are the distinct conjugates of a designated irreducible constituent $\vartheta$ of $\chi_N$. Hence $\chi(1)=et\vartheta(1)$ and $\vartheta(1)\mid \chi(1) \mid |G:N|$. But also, $\vartheta(1) \mid |N|$. Since gcd$(|N|,|G:N|)=1$ it follows that all the $\vartheta_i$'s must be linear and hence $N' \subseteq ker(\chi)$. Since $\bigcap_{\chi \in Irr(G)}ker(\chi)=1$, the proposition follows.