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 $\vert S_{ij}^k(x,y)\vert=\vert S_{ij}^k(x,y')\vert$?
 A: Here's what I came up with: suppose $(x,y)$ and $(x,y')$ are in $R_k$. That means there is some $g\in G$ such that $xy^{-1}=gxy'^{-1}g^{-1}$, since they're in the same conjugacy class. Then the function $f:S_{ij}^k(x,y)\rightarrow S_{ij}^k(x,y')$ defined by $f(a)=g^{-1}ay^{-1}gy'$ gets the job done. First, it's injective by basic group multiplication properties. Next, let $a\in S_{ij}^k(x,y)$ (thus $xa^{-1}\in C_i$ and $ay^{-1}\in C_j$) and we have that (using $\sim$ to represent equivalence of conjugacy classes):
$$xf(a)^{-1}=xy'^{-1}g^{-1}ya^{-1}g\sim gxy'^{-1}g^{-1}ya^{-1}=xy^{-1}ya^{-1}=xz^{-1}\in C_i$$
$$f(a)y'^{-1}=g^{-1}ay^{-1}gy'y'^{-1}=g^{-1}ay^{-1}g\sim ay^{-1}\in C_j$$
Thus $f(a)\in S_{ij}^k(x,y')$, as it should be, so for any pairs $(x,y),(x,y')\in R_k$, $\vert S_{ij}^k(x,y)\vert\leq \vert S_{ij}^k(x,y')\vert$, and so applying it to all pairs gives that $\vert S_{ij}^k(x,y)\vert=S_{ij}^k(x,y')\vert$. 
Thus, given an arbitrary pair $(x,y),(x',y')\in R_k$, there is some $z$ such that $x=zx'$, and then from the question it's clear that $\vert S_{ij}^k(x,y)\vert=\vert S_{ij}^k(zx,zy)\vert=\vert S_{ij}^k(x',zy)\vert$. From what I just showed above, $\vert S_{ij}^k(x',zy)\vert=\vert S_{ij}^k(x',y')\vert$, so $\vert S_{ij}^k(x,y)\vert=\vert S_{ij}^k(x',y')\vert$, and thus the conjugacy class scheme is an association scheme.
A: 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 $(x,y)\in R_k$.
