Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 4 / 2 / Solution

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: