Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 19F (Representation Theory)

Choose a nonzero vector in an irreducible representation. Its finite group orbit spans a finite-dimensional invariant subspace, which must be the whole representation. The representing matrices of an abelian group commute. Choose an eigenspace of one matrix; all other matrices preserve it. Repeatedly choosing an eigenspace for the restrictions of the remaining matrices gives a common eigenvector. Its line is invariant under the whole group. Irreducibility therefore makes the representation that line, proving every irreducible complex representation of a finite abelian group is one-dimensional.

A faithful representation of dimension one embeds $A$ into ${\mathbb{C}}^{\times }$. Every image element is a $|A|$th root of unity by Lagrange's theorem, so the image is a subgroup of the cyclic group of those roots. It is cyclic, and faithfulness makes $A$ cyclic. Conversely a generator of a cyclic group of order $m$ can be sent to $e^{2\pi i/m}$, giving an injective one-dimensional character. Thus such a faithful representation exists exactly when $A$ is a cyclic group.

For $z\in Z$, the eigenspaces of $\rho (z)$ are invariant under $G$, since every representing matrix commutes with $\rho (z)$. Irreducibility forces one such eigenspace to be all of $W$, so $\rho (z)=\chi (z)I$. The multiplication law gives $\chi (z{z}^{\prime })=\chi (z)\chi (z^{\prime })$, making $\chi$ a one-dimensional, hence irreducible, group representation of $Z$. Therefore

Given a one-dimensional character $\chi$ of $Z$, form the nonzero space of functions $f:G\to \mathbb{C}$ satisfying $f(zg)=\chi (z)f(g)$. Right translation $[\rho (h)f](g)=f(gh)$ defines a representation of dimension $[G:Z]$. Centrality gives $\rho (z)f(g)=f(gz)=\chi (z)f(g)$, so $Z$ acts by $\chi$ on the entire space. This representation has an irreducible summand: average any positive Hermitian inner product over the finite group, so invariant subspaces have invariant orthogonal complements, and decompose inductively by dimension. Every irreducible summand has the same central one-dimensional character $\chi$. Thus every irreducible representation of $Z$ occurs by restriction in the requested way.

Let $\chi$ be the central one-dimensional character of a proposed $W$. Since $Z$ is not cyclic, part (a) makes this one-dimensional character nonfaithful. Choose a nonidentity $z\in \ker \chi$. It acts as identity on every tensor power $W^{\otimes n}$ and therefore on every summand. There is a one-dimensional character $\eta$ of $Z$ with $\eta (z)\ne 1$: decompose the finite abelian group into cyclic factors and choose a primitive one-dimensional character on a factor where $z$ has a nonidentity component. By part (b)(ii), some irreducible representation $V$ of $G$ has central one-dimensional character $\eta$ on $Z$. This $V$ cannot occur in any tensor power of $W$, because $z$ acts nontrivially on it. Therefore no such tensor-generating irreducible $W$ exists.