{\stackrel{\rm def}{=}}\( x_{\pi^{-1}(1)}, x_{\pi^{-1}(2)},\dots,x_{\pi^{-1}(n)}\)\qquad\text{for $\pi \in S_n$ and $x \in ({\mathbb{Z}}_r)^n$}.$$ One then checks that if $\pi,\sigma \in S_n$ and $x,y \in ({\mathbb{Z}}_r)^n$ then $$(\pi,x)(\sigma,y) = ( \pi \sigma, \sigma^{-1}(x) + y).$$ We may thus identify $S_n$ and $({\mathbb{Z}}_r)^n$ as the respective subgroups of $G(r,n)$ consisting of all elements $(\pi,x)$ with $x = 0$ and $\pi =1$.

To extract the pair $(\pi,x)$ from an arbitrary element $g \in G(r,n)$, we make the following definition.

Given $g = (\pi,x) \in G(r,n)$ and an integer $i \in [n]$, let $$|g| {\stackrel{\rm def}{=}}\pi \in S_n \qquad\text{and}\qquad z_i(g) {\stackrel{\rm def}{=}}x_i \in {\mathbb{Z}}_r
\qquad\text{and}\qquad
\Delta(g) {\stackrel{\rm def}{=}}\sum_{i=1}^n z_i(g)
.$$ The map $g\mapsto |g|$ affords a homomorphism $G(r,n) \to S_n$, while $g \mapsto \Delta(g)$ affords a homomorphism $ G(r,n) \to {\mathbb{Z}}_r$.

If $p$ is a positive divisor of $r$ then the subgroup $G(r,p,n)$ consists of all elements $g \in G(r,n)$ with $\Delta(g) \in 
p{\mathbb{Z}}_r {\stackrel{\rm def}{=}}\{ p,2p,3p,\dots,rp\} \subset {\mathbb{Z}}_r$. In particular $G(r,1,n)= G(r,n)$ while $G(r,r,n) = \ker( \Delta)$. Throughout, we write $c$ to denote the $n \times n$ scalar matrix $\zeta_r I_n$; this is the central element $$c = (1,e_1+e_2+\dots +e_n) \in G(r,n).$$ If $q$ divides $r$ and $pq$ divides $rn$, then $G(r,p,n)$ contains the cyclic 