# Multiplication in $Z(\mathbb{C}S_n)$ [duplicate]

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

• You mean $3x_{(1,1,1)}+6x_{(3)}$? – user43326 May 16 at 16:59
• 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) – Qfwfq May 16 at 17:04
• This is the same as mathoverflow.net/questions/62088. – Richard Stanley May 16 at 17:08
• @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/> – Filip92 May 16 at 19:57
• @Richard Stanley - true, thanks for pointing that out! – 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 $$C_{i,j,t} = \frac{k_ik_j}{|G|} \sum_{\chi \in Irr(G)} \frac{\chi(g_i)\chi(g_j)\overline{\chi(g_t)}}{\chi(1)}$$