The important notion here is known as the *dispersion relation* of the wave. It relates the temporal properties of the wave (i.e., its frequency $\omega$) to the *spatial* properties of the wave (i.e., its wave vector), and this relationship is *different* for different types of wave.  For light in vacuum, the relationship is exactly $\omega=ck$ so that $\omega$ grows linearly with $k$, but for *matter* waves, the relationship is different.  The dispersion relation is computed using the corresponding "wave" equation which represents the equation of motion for the system: it comes from Maxwell's equation in the context of light and from Schrodinger's equation in the context of quantum particles.

Details below the break.

For completeness (and we will use these below):

de Broglie's hypothesis relates the energy of a matter wave to its frequency, i.e., $E=hf = \hbar\omega$, and the momentum to its wavelength, i.e., $p = h/\lambda = \hbar k$.

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Now, the energy of a free particle of mass $m$ is given by just the kinetic energy, $p^2/2m$.  According to the de Broglie *hypothesis*, the momentum of a particle is given by $p=\hbar k = h/\lambda$, i.e, the momentum of a particle is related to the *spatial* characteristics of the corresponding matter wave. Then, using the kinetic energy of a particle, we can derive the dispersion relation for matter waves as
$$
\hbar\omega = E = \frac{p^2}{2m} = \frac{(\hbar k)^2}{2m}\,,
$$
so that
$$
\omega = \frac{\hbar k^2}{2m}\,.
$$

Now, without getting into the details (you should look these up), the *phase* velocity of a monochromatic wave of angular frequency $\omega$ is given by $v_p=\omega/k$, whereas the *group* velocity of a *wave-packet* centered near $k$ is give by $v_g = d\omega/dk$.  Therefore, for a matter wave,
$$
v_p = \frac{\omega}{k} = \frac{\hbar k^2/2m}{k} = \frac{\hbar k}{2m}
=\frac{h}{2m\lambda}\,,
$$
and
$$
v_g = \frac{d\omega}{dk} = \frac{d}{dk}\frac{\hbar k^2}{2m} = \frac{\hbar k}{m}
=\frac{h}{m\lambda}\,.
$$

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We can see that $c$ never appears here.  In the context of light waves, since the dispersion relation is $\omega = c k$, then we can compute the phase and group velocities as
$$
v_{p,\textrm{light}} = \frac{\omega}{k} = \frac{c k}{k} = c\,,
$$
and
$$
v_{g,\textrm{light}} = \frac{d\omega}{dk} = \frac{d}{dk}(c k) = c\,,
$$
and so in that context, the phase and group velocities for light is equal to the constant $c$, as it should.