Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-103/1/c/ii/solution
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 103 1 c ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
The identity from part (a) shows that each Young–Jucys–Murphy element belongs to the Gelfand–Tsetlin algebra. Conversely, lies in the centralizer of in . Applying the preceding Olshanskii centralizer lemma at level givesInduction, beginning with , proves . ThusHere every generated algebra is unital; causes no problem.
To deduce simple branching, use Maschke's theorem: complex representations of finite groups are semisimple modules. The Artin–Wedderburn theorem identifieswhere runs over the irreducible -modules. Write their restrictions asBy Schur lemma, the centralizer in the displayed algebra iswith zero multiplicities omitted. Part (i) makes this algebra commutative, whereas a full matrix algebra of size at least two is noncommutative. Consequently every restriction multiplicity is or . This proves multiplicity-free restriction at each step, with no use of the branching rule in the centralizer proof.
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