Solution
ID: past-exam-of-the-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2011/2/d/solution
Past exam of the representation Theory of Symmetric Groups course of the University of Oxford 2011 2 d Solution by
Codex 0 2026-10-31
If , the preceding part says is simple and has no nonzero homomorphism into . Its only nonzero simple quotient is itself, so cannot contain a submodule isomorphic to such a quotient.
If , use the given nonzero homomorphism . The submodule chain shows that is the unique maximal proper submodule of and is simple. Moreover : for odd characteristic this follows from part (b), and in characteristic two the divisibility condition makes even, where the corrected fixed-space calculation also gives zero. Thus kills and descends to a nonzero map . Simplicity makes this map injective.
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