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I am trying to multiply two generators of center $Z(\mathbb{C}[S_n])$ of ring algebra of symmetric group of $n$ elements. We know that these generators are given by sums of conjugacy classes in $S_n,$ which are given by various partitions. E.g. for $S_3$ one gets $$Z(\mathbb{C}[S_3])=\mathbb{C}\langle (1)(2)(3), (12)(3)+(23)(1)+(13)(2), (123)+(132) \rangle.$$ Now the question is: Giving two partitions $p_1$ and $p_2$ of $n$ how to get the product of the associated generators $x_{p_1} \cdot x_{p_2}$ of $Z(\mathbb{C}[S_n])?$
E.g, in the given example $S_3$ we get $$x_{(2,1)}\cdot x_{(2,1)}=3x_{(1,1,1)}+3x_{(3)}.$$

By saying ''to get'' I mean an algorithm that takes two partitions and spits out the partitions that appear in the product, together with the corresponding coefficients.


marked as duplicate by Community May 16 at 21:18

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  • $\begingroup$ You mean $3x_{(1,1,1)}+6x_{(3)}$? $\endgroup$ – user43326 May 16 at 16:59
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    $\begingroup$ I just made a quick google search: it's not the solution (you are looking for the structure constants of $Z[\mathbb{C}S_n]$ with respect to that set of algebra generators) but maybe it has some info: core.ac.uk/download/pdf/82470256.pdf - Look e.g. on page 727 (page 3 of the pdf doc) $\endgroup$ – Qfwfq May 16 at 17:04
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    $\begingroup$ This is the same as mathoverflow.net/questions/62088. $\endgroup$ – Richard Stanley May 16 at 17:08
  • $\begingroup$ @user43326 No - multiplying you get 9 terms, 3 of them go to neutral = $x_{(111)}$, and 6 others are 3 times $x_{(3)}=(123)+(132).$ <br/> $\endgroup$ – Filip92 May 16 at 19:57
  • $\begingroup$ @Richard Stanley - true, thanks for pointing that out! $\endgroup$ – Filip92 May 16 at 20:01

This is just a small comment. Let $G$ be a finite group and let $K_1$, $\ldots$, $K_n$ be the class sums that form a basis of $Z(\mathbb{C}[G])$. For $K_i$ let $k_i$ be the size and $g_i \in G$ a representative of the corresponding conjugacy class.

Then in general $$K_i K_j = \sum_{t} C_{i,j,t} K_t$$ where $C_{i,j,t}$ are non-negative integers and $$C_{i,j,t} = \frac{k_ik_j}{|G|} \sum_{\chi \in Irr(G)} \frac{\chi(g_i)\chi(g_j)\overline{\chi(g_t)}}{\chi(1)}$$

This is very well known, see for example exercise 3.9 in the character theory book of Isaacs. Anyway, this shows that we have a formula for the coefficients in terms of character values. It might be useless in practice for computations though.


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