Proof that the conjugacy class association scheme is an association scheme I was looking at the conjugacy class association scheme (where, given some group $G$, each conjugacy class $C_i$ gets a relation $R_i$, where $R_i=\{(x,y)|xy^{-1}\in C_i\}$), and trying to show that it's an association scheme. 
Defining a set $S_{ij}^k(x,y)=\{g\in G|(x,g)\in R_i,(g,y)\in R_j\}$, one needs to show that $\vert S_{ij}^k(x,y)\vert=\vert S_{ij}^k(z,w)\vert$ for all pairs $(x,y),(z,w)\in R_k$.
The approach that I've seen is that, for any $h\in G$, $(xh,yh)\in R_k$, so that if $g$ is such that $(x,g)\in R_i,(g,y)\in R_j$, then $(xh,gh)\in R_i,(gh,yh)\in R_j$. This gives a bijection between $S_{ij}^k(x,y)$ and $S_{ij}^k(xh,yh)$, so for all such pairs in $R_k$, the cardinality of $S_{ij}^k$ is the same. But if the size of the conjugacy class is more than one, there will definitely be some $y'\neq y$ such that $(x,y')\in R_k$, and then there is no $h$ such that $(xh,yh)=(x,y')$. So how can I show that $\vert S_{ij}^k(x,y)\vert=\vert S_{ij}^k(x,y')\vert$?

An alternative approach: Notice that the set $S_{ij}^k(x,y)$ is in bijection with the set $X_c = \{(a,b)\in C_i\times C_j : ab=c\}$, where $c=xy^{-1}\in C_k$. It is easy to see that $|X_c|=|X_{c'}|$ for arbitrary $c,c'\in C_k$. This implies that $S_{ij}^k(x,y)$ is independent of the choice of $(x,y)\in R_k$.
