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Let S_n be the group of permutations of n elements. Consider map S_n -> S_mn of block permutations, and an irreducible representation of S_mn (over complex numbers), corresponding to Young diagram Y. Naturally, it decomposes into direct sum of irreducible representations for S_n. Is it possible to give a formula for the decomposition?

Block permutations: Consider S_n as matrices in GL(n), embed GL(n) into GL(mn) as block matrices with scalar blocks of size m * m, then you got the embedding S_n -> S_mn

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  • $\begingroup$ The keyword you're looking for is "plethysm." See, e.g., Chapter 7 Appendix 2 of Stanley's "Enumerative Combinatorics, Vol. 2." But, for most plethysms, we do not have any kind of nice formula. So likely the answer is "it is not (at the moment) possible to give a formula for the decomposition." $\endgroup$ Jan 25 at 2:10
  • $\begingroup$ Thank you, Sam, checking it out $\endgroup$ Jan 25 at 2:33
  • $\begingroup$ It is not what I am looking for. The symmetric group changes as well $\endgroup$ Jan 25 at 2:43
  • $\begingroup$ Dear Sam, may be you can figure it out better than I will. The "block permutations" are actually permutations of the tensor product of n-dimensional by m-dimensional vector spaces, and it is easy to decompose the respective tensor powers as S_n representation. My doubts are related to the "rest" of the Young diagram for S_mn. Is it "negligible" in any reasonable way? $\endgroup$ Jan 25 at 2:57
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    $\begingroup$ No, no @SamHopkins. Let S_n act on V_n, S_m on V_m. Then S_n x S_m act on V_n \otimes V_m, which has dimension n*m. I am not talking about the usual product of Schur functions. $\endgroup$ Jan 25 at 17:26

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