central subgroup $C_q = \langle c^{r/q}\rangle $ of order $q$. In this case $G(r,p,q,n)$ is defined as the quotient $$G(r,p,q,n) {\stackrel{\rm def}{=}}G(r,p,n)/C_q.$$ We generally refer to elements of $G(r,p,q,n)$ by the same notation $(\pi,x)$ as for elements $G(r,p,n)$, with the added stipulation that $(\pi,x) =c^{ir/q}\cdot (\pi,x) $ for all $i$. This convention, while slightly abusive, does not present much ambiguity in practice.

We define $|g|$ for $g \in G(r,p,q,n)$ exactly as for $g \in G(r,p,n)$, but the notation $z_i(g)$ is generally no longer well-defined. We also write $$N(r,p,q,n) {\stackrel{\rm def}{=}}\{ g \in G(r,p,q,n) : |g| = 1\}$$ for the normal abelian subgroup of $G(r,p,q,n)$ given by the images of the diagonal matrices in $G(r,p,n)$.

One final piece of notation which we introduce here is the *inverse transpose* or *complex conjugation automorphism* $$\tau \in {\mathrm{Aut}}(G(r,p,q,n)).$$ Explicitly, we define this automorphism by the formula $$\tau(\pi,x) = (\pi,-x),\qquad\text{for }(\pi,x) \in G(r,p,q,n).$$ In words, note that taking the inverse of the transpose of $g \in G(r,p,n)$ has the same effect as replacing all entries of the matrix $g$ by their complex conjugates. If we let $\tau$ denote the automorphism of $G(r,p,n)$ afforded by this operation, then $\tau$ preserves the normal subgroup $C_q$, and so 