Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 19J / b / iii / Solution

PutSince $B=U⋊T$ and $U\cap T=1$, restriction to $U$ gives its regular representation:As $U\cong ({\mathbb{F}}_p,+)$ is abelian, this is the direct sum of all $p$ one-dimensional characters of $U$, each with multiplicity one.

The diagonal subgroup $T$ permutes these one-dimensional weight spaces. It fixes the trivial character, while part (ii) shows that the nontrivial characters split into two orbits, indexed by the nonzero squares and nonsquares. Let $W_0,W_{+},W_{-}$ be the sums of the weight spaces in these three orbits. Thenas $B$-representations, and

Each summand is irreducible. Any nonzero $B$-subrepresentation decomposes into $U$-weight spaces. If it contains one of the one-dimensional weight spaces in an orbit, transitivity of the $T$-action forces it to contain every weight space in that orbit, hence the whole corresponding $W_0$, $W_{+}$, or $W_{-}$. The three summands are pairwise nonisomorphic because their restrictions to $U$ have disjoint character supports. Therefore the induction from the diagonal subgroup of upper-triangular SL2 decomposition has the required three irreducible constituents of dimensions

Solved by gpt-5.6-sol high.