= Solution
Let $K$ be the intersection under consideration. The family is nonempty since it contains $G$, and is invariant under conjugation: conjugating an irreducible inducing pair $(H,\varphi)$ gives another such pair. Hence $K$ is a <normal subgroup> of $G$.
For any $\chi\in\operatorname{Irr}(G)$, monomiality gives $\chi=\lambda^G$ with $\lambda$ a <linear character> of some $H$. This $H$ belongs to the family, so $K\leq H$. Because $K$ is normal, $K\leq tHt^{-1}$ for every $t\in G$. In the <induced representation> on
$$
\mathbb C[G]\otimes_{\mathbb C[H]}\mathbb C_\lambda,
$$
each coset line $t\otimes\mathbb C_\lambda$ is therefore invariant under $K$: for $k\in K$,
$$
k(t\otimes v)=t\otimes\lambda(t^{-1}kt)v.
$$
Thus the restriction to $K$ is a direct sum of one-dimensional <group representations>. All representing matrices of $K$ commute, and the <derived subgroup> $K'$ acts as the identity. Equivalently,
$$
K'\leq\ker\chi\qquad\text{for every }\chi\in\operatorname{Irr}(G).
$$
The intersection of the <character kernels> is trivial: the <regular representation> is faithful and is a direct sum of <irreducible representations>. It follows that $K'=1$.
$$
\boxed{K\text{ is abelian}.}
$$
This proves the <abelian intersection of irreducible inducing subgroups> without assuming that every <subgroup> of an M-group is itself monomial.
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