 Solution

Insert the two observed spins into the configuration sum for the transfer matrix for a classical spin chain trace. Since the diagonal matrix ${\sigma }^z$ has entry $s$ on state $s$, multiplication by it at positions $j$ and $j+r$ produces the factor ${\sigma }_j{\sigma }_{j+r}$. The trace representation is consequentlyThere are exactly $N$ transfer factors: $(j-1)+r+(N+1-j-r)=N$. The distance between the two insertions is $r$ bonds, not $r-1$. Cyclicity of the trace combines the outside factors and gives the equivalent expressionwhich makes independence of the starting site manifest. No assumption that $T$ commutes with ${\sigma }^z$ is used; in general it does not. For $r=0$ the two insertions multiply to the identity and $g(0)=1$, a useful normalization check.