 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 $(H,\phi )$ gives another such pair. Hence $K$ is a normal subgroup of $G$.

For any $\chi \in \mathrm{Irr}(G)$, monomiality gives $\chi ={\lambda }^G$ with $\lambda$ a linear character of some $H$. This $H$ belongs to the family, so $K\le H$. Because $K$ is normal, $K\le tH{t}^{-1}$ for every $t\in G$. In the induced representation oneach coset line $t\otimes {\mathbb{C}}_{\lambda }$ is therefore invariant under $K$: for $k\in K$,Thus the restriction to $K$ is a direct sum of one-dimensional group representations. All representing matrices of $K$ commute, and the derived subgroup $K^{\prime }$ acts as the identity. Equivalently,The intersection of the character kernels is trivial: the regular representation is faithful and is a direct sum of irreducible representations. It follows that $K^{\prime }=1$.

This proves the abelian intersection of irreducible inducing subgroups without assuming that every subgroup of an M-group is itself monomial.