# How to calculate product of two conjugacy classes of a finite group?

I learnt from here that the product of two conjugacy classes can be calculated as

$$K_{\mu}K_{\nu} = \sum_{\lambda}C_{\mu,\nu}^{\lambda}K_{\lambda}$$

But I can make no sense of this (I don't have a strong mathematical background so I just make no sense what is explained in the link above).

My questions are:

1. What does $K_\lambda$ mean in the above equation?
2. For P(3) group, there are three conjugacy classes: $K_1 =$ {$E$}, $K_2 =$ {$A, B, C$}, $K_3=$ {$D, F$}. Then how to calculate $K_2K_3$?
• You should read the details of the given answer in order to understand it. To begin with, the calculation takes place in the group algebra and $K_\mu$ is not a set but a certain sum in the group algebra. It is far from clear from you question that this is what you are actually interested in. Oct 25, 2017 at 18:35
• @MoisheCohen the thing that I want to make sense is how the product is calculated, and if there is an explicit example that would be extremely helpful for me to make sense of the given answer. So I really want to see for the simple example of P3 group how the product of class sums is calculated. Oct 25, 2017 at 23:24

I know how to do the calculation now. I was just scared by the equation in the first place.

For the P(3) group, I want to calculate the product of class sums of $K_{2}$ and $K_3$. For performing this calculation:

$$K_2K_3 = (A+B+C)(D+F)$$

$$K_2K_3 = AD +BD+CD+AF+BF+CF$$

Then from the multiplication table, we have:

$$K_2K_3=2(A+B+C) = 2K_2$$

Then I can use this simple result to make sense of the equation:

$$K_{\mu}K_{\nu}=\sum_{\lambda}C_{\mu,\nu}^{\lambda}K_{\lambda}$$

Take P(3) group for instance:

For $\mu = 1$, we essentially have

$$K_{\mu} \left(\begin{matrix} K_1\\ K_2\\ K_3\\ \end{matrix}\right) = \left(\begin{matrix} C_{1}^{1} & C_{1}^2 & C_1^3 \\ C_2^1 & C_2^2 & C_2^3\\ C_3^1 & C_3^2 & C_3^3\\ \end{matrix}\right) \left(\begin{matrix} K_1 \\K_2\\k_3\\ \end{matrix}\right)$$

In the matrix above, $C_{\nu}^{\lambda}$ serves as entries.

From this observation, I can rewrite the first equation as:

$$K_{\mu}K_{\nu}=\sum_{\lambda}(M_{\mu})_{\nu, \lambda}K_{\lambda}$$

So, at this point, althought I don't know how to calculate those coefficients, I make sense of what is calculated.