= Solution
An <M-group> is a finite group for which every <irreducible character> is induced from a <linear character> of some subgroup. Let $L\triangleleft K$ have both $L$ and $K/L$ abelian, and fix an irreducible $K$-module $V$ with character $\chi$. Since $L$ is abelian, $V_L$ has a linear constituent. Choose, among all pairs $(A,\varphi)$ with $L\subseteq A\subseteq K$ and a <linear character> $\varphi$ occurring in $\chi_A$, one for which $|A|$ is maximal.
Every subgroup containing $L$ is normal in $K$: its image in the abelian quotient $K/L$ is normal, and it is the full preimage of that image. Thus $A\triangleleft K$, so the <inertia group of a character> $I_K(\varphi)$ is defined and contains $A$.
Suppose $x\in I_K(\varphi)\setminus A$. The nonzero $\varphi$-<isotypic component> $V_\varphi$ is preserved by $x$, and every $a\in A$ acts there as the scalar $\varphi(a)$ because $\varphi(1)=1$. The finite-order operator representing $x$ is diagonalizable over $\mathbb C$; choose an eigenvector $0\ne v\in V_\varphi$. Then $\mathbb Cv$ is invariant under both $A$ and $x$, hence under $D=\langle A,x\rangle$. This line affords a <linear character> $\psi$ of $D$ with $\psi_A=\varphi$. Since it is a subrepresentation of $V_D$, it is a constituent of $\chi_D$. But $D$ strictly contains $A$ and still contains $L$, contradicting maximality.
Therefore $I_K(\varphi)=A$. Apply the <Clifford correspondence> with <normal subgroup> $A$. Its inertia group is $A$ itself, and $\operatorname{Irr}(A\mid\varphi)=\{\varphi\}$, so the unique irreducible $K$-character lying over $\varphi$ is $\varphi^K$. In particular,
$$
\boxed{\chi=\varphi^K,\qquad\varphi(1)=1.}
$$
Since $\chi$ was arbitrary, every <irreducible character> is a <monomial character>, proving that <finite metabelian groups are monomial>:
$$
\boxed{\text{Every finite metabelian group is an M-group}.}
$$
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