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

We use complex representations. By Maschke's theorem, restriction to any finite subgroup is a semisimple module, so it has a canonical isotypic decomposition. In particular, the $\theta$-isotypic component $W_{\theta }$ is the sum of all copies of the irreducible $N$-module with character $\theta$. The central character idempotentprojects onto this component. These projections preserve every $N$-submodule, which consequently decomposes as the direct sum of its intersections with the isotypic components.

Let $W$ afford $\xi \in \mathrm{Irr}(T\mid \theta )$. The component $W_{\theta }$ is nonzero. Since $T=I_G(\theta )$ fixes $\theta$, it preserves $W_{\theta }$. Irreducibility of $W$ as a $T$-module therefore gives $W=W_{\theta }$: its restriction to $N$ consists entirely of copies of $\theta$.

Form the induced representationwhere $\mathcal{R}$ represents the left cosets $G/T$. Because $N◃G$, for $n\in N$ we haveThus $g\otimes W$ is $N$-isotypic of type ${\theta }^g$, where ${\theta }^g(n)=\theta (g^{-1}ng)$. These types are distinct for distinct cosets $gT$, precisely by the definition of the inertia group of a character $T$.

Let $Y$ be a nonzero $G$-submodule of $V$. Its isotypic decomposition as an $N$-module shows that it meets some $g\otimes W$ nontrivially. Acting by $g^{-1}$ gives $Y\cap (1\otimes W)\ne 0$. This intersection is a $T$-submodule of the irreducible $W$, hence equals $1\otimes W$. The $G$-translates of that component span $V$, so $Y=V$. The induced representation is therefore irreducible, and its restriction contains $\theta$. Its induced character satisfiesThis proves the irreducibility assertion directly from isotypic components; no form of the correspondence being proved has been assumed.

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: