# Number of conjugacy classes in finite groups

Let $G$ be a finite group. Let $C_1,C_2,\dots,C_k$ be its conjugacy classes. We denote by $C_{j\ '}=\{g^{-1}|\ g\in C_j\}$ the conjugacy class inverse to $C_j$.

Set $$a_{rst} = \frac{1}{|C_t|}|\{(x,y)\in C_r \times C_s|\ xy\in C_t\}|$$

Prove:

1. $a_{rst} \in \mathbb{N}_0 =\{0,1,\dots\}$
2. $a_{rst'} = a_{r' s' t}$

I'm having difficulties proving this problem. Thanks for help !

The claims are relatively straightforward to argue using the group ring $\Bbb Z[G]$. One easily checks that a formal sum in this ring is central iff it commutes with all $g\in G$, iff the coefficients of any two conjugates in $G$ are the same, iff it is a sum of the elements $\sigma_K:=\sum_{x\in K}x$, where $K$ ranges over all the conjugacy classes of $G$. Write $\sigma_i:=\sigma_{K_{\large i}}$. Prove the quotient in (1) is the coefficient of $\sigma_t$ after we expand the product $\sigma_r\sigma_s$ as a sum of $\sigma$s. (The product of central elements is central, and every central element is a sum of $\sigma$s.) Argue $\sigma_r\sigma_s=\sigma_s\sigma_r\Rightarrow a_{rst}=a_{srt}$. Finally, argue an equality equivalent to that in (2), $a_{rst'}=a_{s'r't}$, using $xy=z\Leftrightarrow y^{-1}x^{-1}=z^{-1}$.

After some search I found a more "intuitive" proof as I'm not very familiar with group rings. Feel free to comment & criticize:

Proof of $1.$:
First, if there doesn’t exist $x_0 \in C_r , y_0 \in C_s$ such that $x_0 y_0 \in C_t \Rightarrow a_{rst} = 0 \in \mathbb{N}_0$.

If there exist such a tuple $x_0 , y_0$ , then we can write $$x_0 y_0 = z_0 \Rightarrow gx_0 y_0 g^{−1} = g z_0 g^{−1} \Rightarrow gx_0 g^{−1} gy_0 g^{−1} = gz_0 g^{−1} \text{ with } g \in G.$$ Now we use the definition of a conjugacy class and see that $gx_0 g^{−1} \in C_r \quad gy_0 g^{−1} \in C_s \quad gz_0 g^{−1} \in C_t$.

Therefore if we find such a tuple $x_0 , y_0$ we can construct $|C_t |$ other tuples. Maybe we find more than one choices of $x_0$ and $y_0$ such that $x_0 y_0 \in C_t$ , lets say there are $n$ such tuples altogether $\Rightarrow$ We can construct $n |C_t |$ such tuples and $|\{(x,y)\in C_r \times C_s|\ xy\in C_t\}|=n |C_t |$ .
Hence $a_{rst} = n \in \mathbb{N}_0$

Proof of $2.$:
Assume $x \in C_r , y \in C_s , z \in C_t , x^{−1} \in C_{r'} , y^{−1} \in C_{s'} , z^{−1} \in C_{t'}$ $$z^{−1} = xy ⇒ x^{−1} z^{−1} = y ⇒ x^{−1} = yz ⇒ x^{−1} y^{−1} = yzy^{−1} ∈ C_t$$ Therefore we find for each tuple $(x,y) \in \{(x,y)\in C_r \times C_s|\ xy\in C_{t'}\}$ an other tuple $(x^{−1} , y^{−1}) \in \{(x,y)\in C_{r'} \times C_{s'}|\ xy\in C_{t}\}$.

As $|C_t|=|C_{t'}|$ we finally have $a_{rst'} = a_{r's't}$