Past exam of the groups and representations course of the University of Oxford / 2018 / 1

A group is a nonempty set $G$ with an operation $G\times G\to G$, $(g,h)\mapsto gh$, satisfying associativity, an identity element $e$ with $eg=ge=g$, and an inverse element $g^{-1}$ with $g{g}^{-1}=g^{-1}g=e$ for every $g$. These are the group axioms. An Abelian group additionally satisfies $gh=hg$ for every pair. A subgroup is a subset that is itself a group under the restricted operation; equivalently, it is nonempty and closed under $h{k}^{-1}$.

Let $R:G\to \mathrm{GL}(V)$ be a nonzero, finite-dimensional complex irreducible representation of an Abelian group. For each fixed $g$, commutativity givesThus $R(g)$ is an intertwining operator from $R$ to itself. By Schur's lemma over the complex numbers, $R(g)={\lambda }_g{I}_V$. Every line in $V$ is therefore an invariant subspace. Since $R$ is irreducible, this forcesThis is the finite-dimensional convention used throughout the paper. Without it, the literal assertion for arbitrary Abelian groups needs qualification. A concrete counterexample is $G=\mathbb{C}(t{)}^{\times }$ acting by multiplication on the complex vector space $V=\mathbb{C}(t)$ of rational functions. Any nonzero invariant complex vector subspace containing $f$ contains $gf$ for every nonzero rational $g$, hence contains every rational function. This is an infinite-dimensional irreducible representation. The scalar conclusion of Schur's lemma used above depends on finite dimensionality.

The additive cyclic group ${\mathbb{Z}}_n$ is generated by $1$, with $n\cdot 1=0$. Every complex irreducible representation is one-dimensional by a. If $R(1)=z$, the defining relation requires $z^n=1$. Conversely each such root of unity defines a group homomorphism. Hence the complete list isChanging the representative of $g$ changes the exponent by an integer multiple of $2\pi i$, so the formula is well-defined. These representations are pairwise inequivalent because their values on the generator differ. The case $k=0$ is the trivial representation.

Put $\alpha =e^{2\pi i/3}$, so ${\alpha }^3=1$ and $1+\alpha +{\alpha }^2=0$. Since the group is Abelian, every conjugacy class is a singleton. The character table isFor the character inner product, complex conjugation givesFor $j=k$, all three summands are one, giving unit norm. For $k-j\equiv 1(\mathrm{mod}3)$ the sum is $1+\alpha +{\alpha }^2=0$; for $k-j\equiv 2(\mathrm{mod}3)$ it is $1+{\alpha }^2+\alpha =0$. This checks every distinct pair, proving

Use the two complementary projection operatorsThey obey $P_{+}+P_{-}=I_2$, $P_{+}{P}_{-}=0$ and $P_{\pm }^2=P_{\pm }$. The given matrix is $A=P_{+}+\alpha P_{-}$, and consequentlyIt follows that multiplication of these matrices respects addition modulo three:Thus this is a group representation. The eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $\alpha$, respectively. They span invariant lines carrying ${\chi }_0$ and ${\chi }_1$ in the convention of c, soThe other nontrivial irreducible representation, ${\rho }_2$, is absent.

The stated coordinate substitution splits the polynomial basis into two three-cycles:The identity fixes all six basis vectors; neither nonidentity element fixes any. A permutation representation has character equal to the number of fixed basis vectors, henceThe representation multiplicity of each irreducible character is thereforeAn explicit eigenbasis confirms the result. For $k=0,1,2$, putThen $\tilde{R}(1)f_k={\alpha }^k{f}_k$ and $\tilde{R}(1)h_k={\alpha }^k{h}_k$. Thus each three-cycle carries one copy of every irreducible representation. These signs use precisely the source's substitution action; taking an inverse substitution would interchange ${\rho }_1$ and ${\rho }_2$, leaving the total decomposition unchanged.