First recall that for a linear wave of the form
\[
y = A \cos(kx - wt) = A\cos[k(x - (w/k)t],
\]

the phase velocity and group velocities are defined as
\[
c_p = \frac{w}{k}, \qquad c_g = \dd{w}{k}.
\]

If we assume that the energy and momentums are of the form
\[
E = \frac{h}{2\pi} w, \qquad p = \frac{h}{2\pi} k,
\]

then we can write the velocities in the form
\[
c_p = \frac{E}{p}, \qquad c_g = \dd{E}{p}.
\]

As a warm-up, consider non-relativistic motion, with $p = m_0 v$ and $E = \frac{1}{2} m_0 v^2$. The key is that the energy is related to the momentum by $E = \frac{1}{2} p^2/m_0$. Thus we can simply check that the velocities are
\[
c_p = \frac{E}{p} = \frac{v}{2}, \qquad c_g = \dd{E}{p} = \frac{p}{m_0} = v,
\]

and the phase velocity is half the group velocity.

Consider now relativistic motion, with $p = m(v) v$ and $E = m(v) c^2$ where $m(v)$ is the relative mass of the particle as a function of its velocity. The phase velocity is now
\[
c_p = \frac{E}{p} = \frac{c^2}{v}.
\]

The group velocity is harder to compute, however, since the mass is no longer constant, and us a function of the velocity, which is linked to the momentum. Considering the energy as a function of the velocity, then by the chain rule,
\[
c_g = \dd{E}{p} = \dd{v}{p}\dd{E}{v} = \frac{1}{\frac{m}{(1-v^2)^{3/2}}} \frac{mv}{(1-v^2)^{3/2}} = v.
\]

Thus again in this case, the group velocity is the particle velocity.

*Open question:* This seems to be a very deep fact. Why is it important that when taking into account relativistic effects, that $c_p = c^2/v$ instead of $c_p = v/2$?