some integer $k$. Multiplication is given by $$(\pi;a,b)(\pi';a',b') = \begin{cases} (\pi\pi';a+a',b+b'),&\text{if }\pi' = 1 \in S_2,\\ (\pi\pi';b+a',a+b'),&\text{if }\pi' \neq 1 \in S_2.\end{cases}$$ We begin with this lemma:

\[npnq\] If $p+p'$ and $q+q'$ are both odd and $\frac{r}{pq}$ is even then $G\not \cong G'$.

Since $pq=p'q'$ we may assume without loss of generality that $p'$ and $q$ are odd and that $p$ and $q'$ are even. By Theorem \[isomo\] we then have that $G\cong G(r,pq,1,2)$ and $G'\cong G(r,1,p'q',2)$, and so it is enough to show that if $p$ and $\frac{r}{p}$ are both even then $G(r,p,1,2) \not\cong G(r,1,p,2)$.

To this end, let $A = \{g^{r/p}:g\in G(r,p,1,2)\}$ and $B = \{g^{r/p}:g\in G(r,1,p,2)\}$. It suffices to show that $|A|=p$ and $|B|=p+1$. It is easy to check that $A$ consists of the distinct elements $(1;\frac{ir}{p},-\frac{ir}{p}) \in G(r,p,1,2)$ for $i \in [p]$. It is likewise a straightforward exercise to show that $B$ consists of the distinct images in $G(r,1,p,2)$ of $(1;0,\frac{ir}{p}) \in G(r,2)$ for $i \in [p]$ together with $(1;\frac{r}{2p},\frac{r}{2p}) \in G(r,2)$.

Our next lemma is similar.

\[1odd\] If exactly one of the four parameters $p$, $p'$, $q$, $q'$ is odd then $G\not \cong G'$.

We may assume that the unique odd parameter is either $q'$ 