G(r,1,p,2)$.

If $p$ is odd then $G(r,p,1,2) \cong G(r,1,p,2)$ by Theorem \[isomo\], so assume that $\frac{r}{p}$ is odd. Let $p'$ be the largest power of 2 dividing $p$ (and hence also $r$), and let $q=1$ and $q' = p/p'$. With respect to these choices of $p$, $p'$, $q$, $q'$, the special primes are precisely the odd primes dividing $p$. Write $\frac{2r}{p} = \frac{rn}{pq} = \eta \delta$ as in Proposition \[dec\], so that $\eta$ is a product of non-special primes and $\delta$ is a product of special primes, and we have $$G(r,p,1,2) \cong G(r,\delta p,1,2) \times {\mathbb{Z}}_\delta$$ and $$G(r,1,p,2) \cong G(r,\delta,p,2) \times {\mathbb{Z}}_\delta
\cong G(r,1,\delta p,2) \times {\mathbb{Z}}_\delta,$$ the second congruence on the right following from Theorem \[isomo\] as $\delta$ is odd. Because $\frac{r}{p}$ is also odd, $\eta$ is even and $\frac{\eta}{2} = \frac{r}{\delta p}$ is odd; thus $\frac{r}{\delta p}$ is a product of odd primes not dividing $p$, and so is coprime to both $p$ and $\delta$ and in particular to $\delta p$.

Since $G(r,p,1,2)\cong G(r,1,p,2)$ if $G(r,\delta p,1,2) \cong G(r,1,\delta p,2)$, the preceding argument shows that we may assume without loss of generality that $\frac{r}{p}$ and $p$ are coprime. One checks that for $d = r/p'$ the map $$\begin{array}{cccc}
 