that $G(r,\delta p,q,n)$ and $C_{\delta}$ intersect trivially. For this, it suffices to verify that $$(c^{\frac{r}{\delta q}})^k\in
G(r,\delta p,q,n)\quad\text{iff}\quad \tfrac{rnk}{\delta q}  \equiv 0 \modu \delta p)\quad \text{iff}\quad k\equiv 0  \modu \delta).$$ The first equivalence follows by definition, and the second equivalence follows from the fact that if $\frac{rnk}{\delta q} = \delta p k'$ for some integers $k$, $k'$, then by dividing both sides by $p$ one obtains $\eta k = \delta k'$, which can only hold if $k$ is a multiple of $\delta $ as $\eta$ and $\delta$ are necessarily coprime.

The next pair of results establish Theorem \[thm2\] in the case $n\neq 2$. This generalizes [@Ca1 Theoroem 4.4].

\[isomo\] If $pq=p'q'$, then the groups ${G(r,p,q,n)}$ and $G(r,p',q',n)$ are isomorphic whenever ${\mathrm{GCD}}(p,n) = {\mathrm{GCD}}(p',n)$ and ${\mathrm{GCD}}(q,n) = {\mathrm{GCD}}(q',n)$.

Write $\frac{rn}{pq} = \eta \delta$ as in Proposition \[dec\]. The theorem will follow immediately from Proposition \[dec\] once we show that $G(r,\delta p,q,n)\cong G(r,\delta p',q',n)$. Since $\frac{rn}{q} = \eta \delta p$ and $\frac{rn}{q'} = \eta \delta p'$, it suffices by Proposition \[suff\] to verify that ${\mathrm{GCD}}(\eta\delta p,\delta p') = {\mathrm{GCD}}(\eta\delta p',\delta p),$ which is equivalent to the identity $
{\mathrm{GCD}}(\eta p,p') = {\mathrm{GCD}}(\eta p',p).$ This holds because every prime dividing $\eta$ appears in 