Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 4 / 19G (Representation Theory)

Work over the complex numbers. By Maschke's theorem, restriction to the normal subgroup $A$ is a direct sum of its isotypic components. For an irreducible $A$-type $W$, denote its isotypic component in $V$ by $V_W$. Normality makes $G$ permute these components, by conjugating the $A$-types. Any union of orbits spans a $G$-invariant subspace, so irreducibility of $V$ means that all occurring components form a single orbit.

If there is only one component, restriction is isotypic, as claimed. Otherwise take one component $V_W$ and its stabilizer $H=\{g:g{V}_W=V_W\}$, a proper subgroup. This is an irreducible representation of $H$: if $0\ne U⊊V_W$ were $H$-invariant, the direct sum of its translates over coset representatives for $G/H$ would be a nonzero proper $G$-invariant subspace of $V$. It is direct because the translates lie in distinct isotypic components, and proper because $\dim U<\dim V_W$.

The natural map from the induced representation sends $g\otimes v$ to $gv$. It identifies the direct-sum coset decomposition with all isotypic components of $V$, so is bijective and $G$-equivariant. ConsequentlyThis proves the dichotomy directly, including the irreducibility of the representation being induced.

Induct on the order of the p-group. Every complex irreducible representation of an abelian group is one-dimensional: commuting group operators preserve a common eigenspace, and irreducibility makes that eigenspace the whole space. This handles the abelian cases.

If $V$ is not faithful, it is an irreducible representation of a smaller quotient $G/K$, where $K$ is its kernel. Induction expresses it as induced from a one-dimensional representation of a subgroup of $G/K$. Pulling back the subgroup and inflating the character gives the same induction statement for $G$, because the coset spaces of the quotient subgroup and its inverse image agree.

Now suppose $V$ is faithful and $G$ is nonabelian. Use the allowed abelian normal subgroup $A$ not contained in the centre. If restriction to $A$ were isotypic, its irreducible type would be one-dimensional, so every $a\in A$ would act as a scalar. It would therefore commute with every representing matrix of $G$. Faithfulness would imply $ga=ag$ in $G$ for every $g$, contradicting the choice of $A$. By part (a), $V={\mathrm{Ind}}_H^{G}W$ for an irreducible representation of a proper subgroup $H$. Induction on $H$ gives $W={\mathrm{Ind}}_K^H\chi$ with $\chi$ one-dimensional. Transitivity of induction yieldsThis proves the monomiality of irreducible representations of finite p-groups.