Let's say we have a function $f(x_{\mu},t)$ that transforms under the action of an $N$-parameter group $G(a_{\nu})$. Then a projective representation of $G(a_\nu)$ in the $f(x_\mu,t)$ basis would follow
$$ T_G(a_\nu)f(x_\mu,t)=e^{i\theta(a_\nu)}f(x_\mu',t')$$
but $e^{i\theta}$ is also a representation of the $U(1)$ group. So let's say we have a group $G'(a_\nu)$ and $\theta(a_\nu)$ a function of $a_\nu$ that returns a a real number. We have a basis such that
$$ T_{G'}(a_\nu)f(x_\mu,t)=f(x_\mu',t') $$
and
$$ T_{U(1)}(\theta(a_\nu))f(x_\mu,t)=e^{i\theta(a_\nu)}f(x_\mu,t).$$
Then the action of both groups would be
$$ T_{U(1)}(\theta(a_\nu))T_{G'}(a_\nu)f(x_\mu,t)=e^{i\theta(a_\nu)}f(x_\mu',t')=T_G(a_\nu)f(x_\mu,t).$$
I'm thinking of this in the context of the wave-function where under the Galilean transformation
$$ T_{G}(\Delta\vec{x},\Delta t,R(\alpha,\beta, \gamma),v)|\psi(\vec{x},t)|^2=|\psi(\vec{x}',t')|^2$$
which means
$$ T_{G}(\Delta\vec{x},\Delta t,R,v)\psi(x,t)=e^{i\theta(\Delta\vec{x},\Delta t,R,v)}\psi(x',t').$$