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:


*

*What does $K_\lambda$ mean in the above equation?

*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 $K_2K_3$? 

 A: 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.
