\phi: & G(r,p,1,2)&\rightarrow &G(r,1,p,2)\\
 & (\pi;i,j) &\mapsto&(\pi;i,j+di)
\end{array}$$ is a well-defined group homomorphism. To show that $\phi$ is an isomorphism it is enough to demonstrate injectivity, so let $g\in G(r,p,1,2)$ such that $\phi(g)=1$. Then $g$ is necessarily of the form $(1;i,j)$ with $$i+j \equiv 0 \modu p)\qquad\text{and}\qquad i\equiv j+di \equiv k\tfrac{r}{p}\modu r)\text{ for some $k \in [p]$},$$ the second congruence following from the assumption that $\phi(1;i,j)=(1;i,j+di)$ represents the identity in $G(r,1,p,2)$. These two congruences imply that $k\frac{r}{p}(2-d)$ is a multiple of $p$. Since $d$ is odd, no number dividing $2-d$ divides either 2 or $d$, and as every odd prime dividing $p$ also divides $d$, it follows that ${\mathrm{GCD}}(2-d,p)=1$. Since $\frac{r}{p}$ is coprime to $p$ by hypothesis, we conclude that $k$ is a multiple of $p$, which implies that $i\equiv  j \equiv 0 \modu r)$ and in turn that $g=1$, as desired.

Gathering together the preceding results yields the following summary theorem.

\[n2-thm\] Assume $pq=p'q'$. Then $G(r,p,q,2)\cong G(r,p',q',2)$ if and only if one of the following mutually exclusive conditions holds:

1.  $p+p'$ and $q+q'$ are both even;

2.  $p+p'$, $q+q'$, and $\frac{r}{pq}$ are all odd integers.

If the first condition holds then $G \cong G'$ by Theorem \[isomo\]. If the 