with different multiplicities. Since $pq = p'q'$, a prime is special if and only if it also appears in $q$ and $q'$ with different multiplicities. We now have the following proposition.

\[dec\] Assume that $${\mathrm{GCD}}(p,n) = {\mathrm{GCD}}(p',n) \qquad\text{and}\qquad
{\mathrm{GCD}}(q,n) = {\mathrm{GCD}}(q',n),$$ and write $\frac{rn}{pq} = \eta \delta$ where $\eta$ (respectively, $\delta$) is a positive integer equal to a product of non-special (respectively, special) primes. Then $ G(r,\delta p,q,n)$ is well-defined and $${G(r,p,q,n)}\cong G(r,\delta p,q,n)\times {\mathbb{Z}}_{\delta}.$$

Since ${\mathrm{GCD}}(q,n) = {\mathrm{GCD}}(q',n)$, the multiplicity of any special prime in $n$ is not greater than the corresponding multiplicity in $q$. As $n$ divides $\eta \delta q = \frac{rn}{p}$, it follows that $n$ divides $\eta q = \frac{rn}{\delta p}$. Thus $\delta p$ divides $r$, and since $\delta pq$ likewise divides $rn$ as $\frac{rn}{\delta pq} = \eta$, we conclude that $G(r,\delta p,q,n)$ is well-defined.

A symmetric argument using the assumption that ${\mathrm{GCD}}(p,n) = {\mathrm{GCD}}(p',n)$ shows that $\delta q$ likewise divides $r$. Therefore $c^{\frac{r}{\delta q}}$ is a well-defined element of order $\delta$ in $G$; let $C_\delta \cong {\mathbb{Z}}_\delta$ be the cyclic subgroup it generates. Both $G(r,\delta p,q,n)$ and $C_\delta$ are normal subgroups of ${G(r,p,q,n)}$, so to complete the proof of the proposition, we have only to show 