Finite Gelfand pairs and cracking points of the symmetric groups
Let \(\Gamma\) be a finite group. Consider the wreath product \(G_n := \Gamma^n \rtimes S_n\) and the subgroup \(K_n := \Delta_n \times S_n\subseteq G_n\), where \(S_n\) is the symmetric group and \(\Delta_n\) is the diagonal subgroup of \(\Gamma^n\). For certain values of \(n\) (which depend on the...
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Veröffentlicht in: | arXiv.org 2019-08 |
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Format: | Artikel |
Sprache: | eng |
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Zusammenfassung: | Let \(\Gamma\) be a finite group. Consider the wreath product \(G_n := \Gamma^n \rtimes S_n\) and the subgroup \(K_n := \Delta_n \times S_n\subseteq G_n\), where \(S_n\) is the symmetric group and \(\Delta_n\) is the diagonal subgroup of \(\Gamma^n\). For certain values of \(n\) (which depend on the group \(\Gamma\)), the pair \((G_n, K_n)\) is a Gelfand pair. It is not known for all finite groups which values of \(n\) result in Gelfand pairs. Building off the work of Benson--Ratcliff, we obtain a result which simplifies the computation of multiplicities of irreducible representations in certain tensor product representations, then apply this result to show that for \(\Gamma = S_k, \ k \geq 5\), \((G_n,K_n)\) is a Gelfand pair exactly when \(n = 1,2\). |
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ISSN: | 2331-8422 |