of groups $G(r,p,q,n)$ defined as follows.

For positive integers $r,p,n$ with $p$ dividing $r$, let $G(r,p,n)$ denote the subgroup of $G(r,n)$ consisting of the matrices whose nonzero entries, multiplied together, form an $(r/p)$th root of unity. Apart from thirty-four exceptions, the irreducible finite complex reflection groups are all groups $G(r,p,n)$ of this kind. The projective reflection group $G(r,p,q,n)$ is defined as the quotient $$G(r,p,q,n) {\stackrel{\rm def}{=}}G(r,p,n) / C_q$$ where $C_q$ is the cyclic subgroup of scalar $n\times n$ matrices of order $q$. Note that for this quotient to be well-defined we must have $C_q \subset G(r,p,n)$, which occurs precisely when $q$ divides $r$ and $pq$ divides $rn$. Observe also that $G(r,n) = G(r,1,n)$ and $G(r,p,n) = G(r,p,1,n)$.

There is an interesting notion of duality for projective reflection groups; by definition, the projective reflection group *dual* to $G=G(r,p,q,n)$ is $G^* {\stackrel{\rm def}{=}}G(r,q,p,n)$. The starting point of the present collaboration is now the following theorem which reformulates the main result of [@Ma].

\[thm1\] The complex reflection group $G=G(r,p,1,n)$ has a GIM if and only if $G \cong G^*$; i.e., if and only if $G(r,p,1,n) \cong G(r,1,p,n)$.

Explicitly, $G$ has a GIM if and only if (i) $n\neq 2$ and $\gcd(p,n) = 1$ 