Let $G$ be a finite group and let $\text{Irr}(G)$ be the set of irreducible complex characters of $G$. A character $\chi\in\text{Irr}(G)$ is monomial if there exists a subgroup $H\leq G$ and a linear character $\lambda\in\text{Lin}(H)$ such that $\chi=\lambda^G$. Let $\text{Irr}_m(G)$ be the set of irreducible monomial characters of $G$. Then a group $G$ is monomial if $\text{Irr}(G)=\text{Irr}_m(G)$. I'm interested in solvable groups even if probably this is not necessary.
My question is: given two finite solvable groups $G$ and $H$ is it true that $$\text{Irr}_m(G\times H)=\{\varphi\times \psi\mid \varphi\in\text{Irr}_m(G),\psi\in\text{Irr}_m(H)\}$$
Using GAP I checked some (really small) cases and this always works.
What is this useful for? I'm interested in this question because a positive answer will give an easy way to construct a family of solvable groups $(G_n)_{n\in\mathbb{N}}$ such that
$$\lim\limits_{n\to \infty} \frac{|\text{Irr}_m(G_n)|}{|\text{Irr}(G_n)|}=0$$ To construct such a family fix a solvable nonmonomial group $G$ (e.g. $\text{SL}_2(3)$) and define $G_n$ to be the direct product of $n$-copies of $ G$.
Update: The above question has a negative answer, in general, by work of van der Waall "Direct products and monomial characters". However, a positive answer is given when at least one of the factors, say $G$, satisfies the following property: either $G=1$ or every maximal subgroup $M$ of $G$ has a normal subgroup $N\unlhd M$ with only abelian Sylow subgroups (for every prime) and such that $M/N$ is nilpotent. In particular, the above family can be constructed considering $G=\text{SL}_2(3)$.
