EDIT:

After a lot of research I have cleared up some of the doubts concerning the problem I mentioned above. I was helped a lot by this PDF (link: http://sharif.edu/~kavehvash/Group_Phase_Velocity.pdf) which explains what happens to the phase and group velocities according to the dispersion relation. In particular, two cases are distinguished:

1. The dispersion relation is linear: this causes group velocity and phase velocity to be equal. This happens because in this condition it should be $n=n_0$ and from the formula of $\beta_1$ we get $v_g=v_p$.

2. The dispersion relation is non-linear: in this case, the two velocities are always different from each other (unless you have special restrictive conditions). This is due to the fact that $n=n(\omega)$ and each component moves at a different speed. From the PDF: "A very important consequence of this is that our initial wave package broadens out with time because the partial waves forming it gradually move out of phase with each other" and this because, if $v_p=v_p(\omega)$, then $v_g=v_g(\omega)$. The only alternative for which $v_g$ does not depend on $\omega$ and does not broaden is that it is $0$, i.e. there is a standing wave.

If what I have written is correct I am left with the last doubt: is it possible to have $v_g\neq v_p$ without obtaining a time broadening as shown in the following two links?

Link 1: https://en.wikipedia.org/wiki/File:Wave_packet_propagation_(phase_faster_than_group,_nondispersive).gif

Link2 (Figure 1): https://www.rp-photonics.com/group_velocity.html

If yes, how?