Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 19J (Representation Theory)

Over $\mathbb{C}$, the operator representing $r$ has an eigenvector $v$ with eigenvalue $\lambda \ne 0$. The relation $tr{t}^{-1}=r^{-1}$ givesTherefore $\mathrm{span}\{v,tv\}$ is a nonzero $G$-invariant subspace of dimension at most two. If $V$ is irreducible it equals this subspace, so $\dim V\le 2$.

In one dimension the defining relations forcegiving four characters.

Every irreducible two-dimensional representation is, after a choice of basis,where $\lambda \in {\mathbb{C}}^{\times }\setminus \{1,-1\}$. It is irreducible because the two distinct $r$-eigenlines are interchanged by $t$. Conversely, part (a) gives this basis from $v,tv$. If the only eigenvalue of $r$ were $\pm 1$, its eigenspace would be $t$-invariant and contain a $t$-eigenvector, contradicting irreducibility. Parameters $\lambda$ and ${\lambda }^{-1}$ give isomorphic representations, and there are no other identifications.

The quotient $G\to D_{2n}$ adds the relation $r^n=1$. All four one-dimensional characters factor through some such quotient: choose even $n$ when $r$ acts as $-1$. A two-dimensional representation from part (b) factors through $D_{2n}$ exactly when ${\lambda }^n=1$. Hence the irreducible representations that never arise from a finite dihedral quotient are precisely the two-dimensional ones for which $\lambda$ is not a root of unity.

Take $t\mapsto A$ and $r\mapsto B$. Then $A^2=I$ andso these matrices define a representation. The line $\mathbb{C}{e}_1$ is fixed by both matrices and is a one-dimensional subrepresentation.

Any complementary line has a generator $e_2+c{e}_1$. Butwhich cannot lie on the same line. Thus no complement is $B$-invariant, and therefore no $G$-invariant complement exists.