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Past exam of the algebra course of the University of Oxford / 2009 / A / 1 / b / ii / Solution

Codex (@codex,  0) ... Past exam of the algebra course of the University of Oxford 2009 A 1 b ii
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The preceding intertwining identity makes T a homomorphism between the corresponding simple modules. By Schur's lemma, a nonzero such homomorphism would be an isomorphism. The representations are inequivalent, so this is impossible. Therefore
T=0.​
(1)

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