to clarifying which groups $G(r,p,q,n)$ have GIMs. The following theorem, proved in Section \[n=2-sect\] below, completely solves this problem in the often pathological case $n=2$.

\[n=2-thm\] The projective reflection group $G(r,p,q,2)$ has a GIM if and only if $(r,p,q) = (4,1,2)$ or $G(r,p,q,2) \cong G(r,q,p,2)$.

By Theorem \[thm2\], the condition $G(r,p,q,2) \cong G(r,q,p,2)$ holds if and only if (i) $p$ and $q$ have the same parity or (ii) $\frac{r}{pq}$ is an odd integer.

A few notable differences between complex reflection groups and projective reflection groups complicates the task of determining the existence of GIMs, and in the case $n\neq 2$ our classification is incomplete. For example, the groups $G(r,p,q,n)$ occasionally can have conjugacy class-preserving outer automorphisms (see Proposition \[no-class-prop\] below). The fact that the groups $G(r,p,n)$ never have such automorphisms [@MM Proposition 3.1] was the source of a significant reduction in the proof of [@Ma Theorem 5.1] which is no longer available in many cases of interest. Nevertheless, by carrying out a detailed analysis of the conjugacy classes and automorphisms of $G(r,p,q,n)$, we are able to prove the following theorem.

\[thm3\] Let $G=G(r,p,q,n)$ and assume $n \neq 2$.

1.  If $\gcd(p,n) = 1$ then $G$ has a GIM if $q$ 