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Let $L◃G$ and both $L,G/L$ be abelian. For an irreducible $G$-module, choose a subgroup $A\supseteq L$ of maximal order whose restriction contains a linear character $\phi$. Every subgroup containing $L$ is normal. If an element $x\notin A$ stabilizes $\phi$, its action on the nonzero $\phi$-isotypic component has an eigenvector; that line is invariant under $⟨A,x⟩$ and affords a larger linear constituent, contradicting maximality. Thus $I_G(\phi )=A$, and the Clifford correspondence gives $\chi ={\phi }^G$.

Every complex irreducible representation of a finite p-group is induced from a one-dimensional representation of a subgroup. Induct on the group order. Nonfaithful representations reduce to a smaller quotient. For a faithful representation of a nonabelian group, choose an abelian normal subgroup not contained in the centre. Restriction cannot be isotypic, since it would make that subgroup scalar and faithfulness would then force it central. The Clifford correspondence induces the representation from a proper subgroup; induction there and transitivity of induction complete the proof. Abelian groups have only one-dimensional complex irreducible representations.

Let $V$ be an irreducible $G$-module with character $\chi \in \mathrm{Irr}(G\mid \theta )$, and let $W=V_{\theta }\ne 0$ be its $\theta$-isotypic component. It is invariant under the inertia group of a character $T$. The sum of its $G$-translates is a nonzero $G$-submodule, so irreducibility makes it all of $V$. Translates associated with different cosets $gT$ have distinct $N$-types and hence form a direct sum:To see that $W$ is irreducible over $T$, let $0\ne W_0\subseteq W$ be a $T$-submodule. Then ${\sum }_{g\in \mathcal{R}}g{W}_0$ is a nonzero $G$-submodule of $V$, hence equals $V$. Its $\theta$-isotypic part is exactly $W_0$, whereas the corresponding part of $V$ is $W$. Thus $W_0=W$.

Let $\xi$ be the irreducible character of $W$ as a $T$-module. It lies over $\theta$, and the natural map of induced representationsis an isomorphism: it identifies each summand with the corresponding $gW$, and these summands are direct and exhaust $V$. Consequently $\chi ={\xi }^G$, proving surjectivity of induction on the stated sets.

For injectivity, the construction in (a) shows that the $\theta$-isotypic component of ${\mathrm{Ind}}_T^{G}W$ is exactly $1\otimes W$, with its original $T$-action. Isomorphic induced $G$-modules therefore have isomorphic intrinsic $\theta$-components as $T$-modules. Since complex representations with the same character are isomorphic, ${\xi }_1^G={\xi }_2^G$ forces ${\xi }_1={\xi }_2$. Hence the Clifford correspondence is the bijectionwhose inverse takes the character of the $\theta$-isotypic component.

An M-group is a finite group for which every irreducible character is induced from a linear character of some subgroup. Let $L◃K$ 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 $(A,\phi )$ with $L\subseteq A\subseteq K$ and a linear character $\phi$ 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 $A◃K$, so the inertia group of a character $I_K(\phi )$ is defined and contains $A$.

Suppose $x\in I_K(\phi )\setminus A$. The nonzero $\phi$-isotypic component $V_{\phi }$ is preserved by $x$, and every $a\in A$ acts there as the scalar $\phi (a)$ because $\phi (1)=1$. The finite-order operator representing $x$ is diagonalizable over $\mathbb{C}$; choose an eigenvector $0\ne v\in V_{\phi }$. Then $\mathbb{C}v$ is invariant under both $A$ and $x$, hence under $D=⟨A,x⟩$. This line affords a linear character $\psi$ of $D$ with ${\psi }_A=\phi$. 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 $I_K(\phi )=A$. Apply the Clifford correspondence with normal subgroup $A$. Its inertia group is $A$ itself, and $\mathrm{Irr}(A\mid \phi )=\{\phi \}$, so the unique irreducible $K$-character lying over $\phi$ is ${\phi }^K$. In particular,Since $\chi$ was arbitrary, every irreducible character is a monomial character, proving that finite metabelian groups are monomial: