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Are there some references about the proof of the following fact?

Type $C_n$ Weyl group lies in the centralizer of the longest word $w_0$ in $S_{2n}$.

Thank you very much.

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  • $\begingroup$ So you take the longest element $w_0$ in $S_{2n}$ and want to show that there is a subgroup of the centralizer $C_{S_{2n}}(w_0)$ that is isomorphic to $C_n$? Why do you call it a Weyl group? Is it important or are you ok with any $C_n$ that could be found in the centralizer? $\endgroup$
    – Dirk
    Oct 5, 2017 at 8:53
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    $\begingroup$ The longest word is $(1,2n)(2,2n-1) \ldots (n,n+1)$ of cycle type $(2^n)$. So its centralizer is $C_2 \wr S_n$, which is the Weyl group of type $B$ or $C$. Reference: any textbook on Lie theory / Coxeter groups, plus e.g. James and Kerber, The representation theory of the symmetric groups, Chapter 4. $\endgroup$ Oct 5, 2017 at 10:22
  • $\begingroup$ Small linguistic question: does your expression "contains in" (in the header and question) actually mean "lies in"? (Also, a tag 'reference-request' is useful here.) $\endgroup$ Oct 5, 2017 at 14:02
  • $\begingroup$ @JimHumphreys, thank you very much. Yes, I need to use "lies in". I will add a tag "reference-request". $\endgroup$ Oct 5, 2017 at 15:18

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Although the comment of Mark Wildon certainly addresses the question, here is another (maybe more root system-y) perspective. Conjugation by $w_0$ in $S_{2n}$ corresponds to the non-identity involutive automorphism of the Dynkin diagram of Type $A_{2n-1}$ (i.e., reflect the diagram across its vertical axis of symmetry). The centralizer of $w_0$ is exactly the fixed point subgroup of this automorphism. And in general, for any diagram automorphism, the fixed point subgroup is naturally isomorphic to the Weyl group of the "folded" diagram: in this case the folded diagram is the Type $B_{n}$ diagram. See Stembridge's write-up on folding (http://www.math.lsa.umich.edu/~jrs/papers/folding.pdf), in particular, Claim 3 there.

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  • $\begingroup$ (Sorry, when I answered it, I didn't notice this question was so old!) $\endgroup$ Nov 28, 2019 at 15:12
  • $\begingroup$ (I also see this is discussed in more detail in a follow-up question of Jim Humphreys mathoverflow.net/questions/282806/…) $\endgroup$ Nov 28, 2019 at 15:17

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