second condition holds then since $pq=p'q'$, exactly one of $p$, $q$ is odd and it follows that $pq$ in fact divides $r$. In this case, we may assume that $p$ and $q'$ are even and that $p'$ and $q$ are odd. Theorem \[isomo\] then implies that $G \cong G(r,pq,1,2)$ and $G'\cong G(r,1,pq,2)$, while Lemma \[coprime\] implies that $G(r,pq,1,2)\cong G(r,1,pq,2)$.

If $p+p' $ and $q+q'$ are both odd but $\frac{r}{pq}$ is even then $G\not\cong G'$ by Lemma \[npnq\]. If $p+p'$ and $q+q'$ have different parities then exactly one of the parameters $p$, $p'$, $q$, $q'$ is odd as $pq=p'q'$, so $G\not\cong G'$ by Lemma \[1odd\].

Combining this theorem with Corollary \[cor3.5\] gives Theorem \[thm2\] in the introduction.

Constructing an isomorphism explicitly {#iso3-sect}
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We present here an alternative proof of Theorem \[isomo\] by constructing an explicit isomorphism between $G = G(r,p,q,n)$ and $G'=G(r,p',q',n)$. It is our hope that this construction may be useful at some point in explaining why groups with $G(r,p,q,n) \cong G(r,q,p,n)$ often tend to have generalized involution models.

Let $r$ and $n$ be any positive integers. Fix positive divisors $p$, $p'$, $q$, $q'$ of $r$ with $pq=p'q'$ dividing $rn$, and for each prime $\pi$ define $$a_\pi,\quad a'_\pi,\quad b_\pi,\quad 