If I have to show that the group velocity of a free particle is twice the phase velocity, is the following argument correct (avoiding to use the wave function and the momentum operator):
For a particle with energy $E$ and momentum $p$ we have the circular frequency
$$\omega = \frac{E}{\hbar}$$
and the wave length
$$\lambda = \frac{2 \pi \hbar}{p}\, .$$As usual for waves we get the phase velocity by the formula $$v_{p} = \frac{\omega \lambda}{2\pi} \, .$$
This gives us $$v_{p} = \frac{\omega \lambda}{2\pi} = \frac{E \cdot 2\pi \hbar }{\hbar p \cdot 2\pi} = \frac{E }{p} = \frac{ p^2 }{2 m p} = \frac{ p}{2 m} = \frac{v}{2}.$$ But the speed of the particle, $v$, is nothing but the group velocity $v_g$ of the corresponding wave function. Therefore $v_g = 2 v_p$.
Best Answer
This is a perfectly correct derivation that uses the correspondence principle nicely: we can identify the group velocity with the classical velocity because a classical particle corresponds to a quantum particle whose wavefunction is a sharply peaked wavepacket, whose velocity is the group velocity.
If you want to do it more formally, you can also start from the usual definitions of the group velocity and phase velocity, $$v_g = \frac{d \omega}{dk}, \quad v_p = \frac{\omega}{k}.$$ The simplest form of the de Broglie relations are $$E = \hbar \omega, \quad p = \hbar k.$$ Your forms are perfectly right too, but a little more complicated because there are $2\pi$'s all over the place. Next, we know that for a free particle, the energy is $$E = \frac12 mv^2 = \frac{p^2}{2m}.$$ Combining this with the de Broglie relations, we have $$\omega = \frac{\hbar k^2}{2m}.$$ Using the definitions of the group and phase velocity $$v_g = \frac{\hbar k}{m}, \quad v_p = \frac{\hbar k}{2m}.$$ Then $v_g = 2 v_p$ as desired. Incidentally, another way to that the classical velocity is $v = v_g$ is to note $$p = \hbar k = m v$$ and comparing with our previous expression gives $v = v_g$.