 Solution

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.