Group Delay for Hilbert Transformer and Resulting Dispersion An ideal Hilbert Transformer shifts the phase for all positive frequencies by $-\pi/2$ and all negative frequencies by $+\pi/2$ while maintaining constant magnitude everywhere. Group Delay is the negative derivative of phase with respect to frequency, so apart from the discontinuity at frequency $\omega=0$ the group delay would be 0. 

Linear phase filters do not cause distortion to a waveform that has many frequency components as each of these components would remain aligned in time at the output of the filter as they were at the input.

Yet the Hilbert Transformer is clearly dispersive meaning different frequency components of a waveform would experience different delays in order for the phase shift to maintain  a $\pi/2$ relationship between the input and output of the network at all frequencies.

The phase of the Hilbert Transformer is linear, so in this case we observe linear phase but still the system is dispersive. How is this explained as this appears to be an exception that I haven’t really seen in definitions of Group Delay or linear phase?

Further imagine a case that is often used in practice with two phase tracking networks used to split a signal into two over a wide frequency with a quadrature relationship. We could imagine a condition with the first network as a linear phase filter and the second network having the same linear phase with a $\pi/2$ offset at all frequencies of interest (generalized linear phase). Here the derivative of the phase with respect to frequency is not zero but constant yet we would have the same dispersive results in order to maintain the constant $\pi/2$ offset.
A real-valued system that doesn't distort the shape of the input signal must have the following input-output relation:

$$y(t)=Ax(t-t_0)\tag{1}$$

with arbitrary real-valued constants $A&gt;0$ and $t_0$.

In the frequency domain, Eq. $(1)$ corresponds to

$$Y(\omega)=Ae^{-j\omega t_0}X(\omega)\tag{2}$$

Consequently, the corresponding system is an LTI system with frequency response

$$H(\omega)=Ae^{-j\omega t_0}\tag{3}$$

i.e., the magnitude response must be constant and the phase response must be linear:

$$\phi(\omega)=-\omega t_0\tag{4}$$

Here, linear is meant in the narrow sense of the word, i.e., it must not have an additive constant term (hence, it must not be affine). Note that for real-valued systems, the phase must be  an odd function of frequency, so the sign of a possible additive constant in the phase would need to change depending on the sign of the frequency. Consequently, if there's such a constant phase term (as is the case for a Hilbert transformer) then the phase has a discontinuity at $\omega=0$, which implies that the group delay is undefined / infinite at $\omega=0$.

The phase response of an ideal Hilbert transformer with a possibly non-zero delay $t_0$ does not satisfy $(4)$ because it is given by

$$\phi(\omega)=-\frac{\pi}{2}\textrm{sgn}(\omega)-\omega t_0\tag{5}$$

As mentioned above, the phase of an ideal Hilbert transform has a discontinuity at $\omega=0$.