Dgdrxrt

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 $R_i=xy^-1in C_i$), and trying to show that it's an association scheme.

Defining a set $S_ij^k(x,y)=(x,g)in R_i,(g,y)in R_j$, 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 $hin G$, $(xh,yh)in R_k$, 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 $vert S_ij^k(x,y)vert=vert S_ij^k(x,y')vert$?

Here's what I came up with: suppose $(x,y)$ and $(x,y')$ are in $R_k$. That means there is some $gin G$ such that $xy^-1=gxy'^-1g^-1$, 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 $f(a)=g^-1ay^-1gy'$ gets the job done. First, it's injective by basic group multiplication properties. Next, let $ain S_ij^k(x,y)$ (thus $xa^-1in C_i$ and $ay^-1in C_j$) and we have that (using $sim$ to represent equivalence of conjugacy classes):
$$xf(a)^-1=xy'^-1g^-1ya^-1gsim gxy'^-1g^-1ya^-1=xy^-1ya^-1=xz^-1in C_i$$
$$f(a)y'^-1=g^-1ay^-1gy'y'^-1=g^-1ay^-1gsim ay^-1in C_j$$
Thus $f(a)in S_ij^k(x,y')$, as it should be, so for any pairs $(x,y),(x,y')in R_k$, $vert S_ij^k(x,y)vertleq vert S_ij^k(x,y')vert$, 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.