For plasma EM waves we have the dispersion relation
$$\omega^2=\omega_p^2+c^2k^2$$
where the plasma frequency $$\omega_p^2=\frac{n_e e^2}{\epsilon_0 m_e}$$
One can show that $v_p v_g=c^2$, i.e., the product of phase and group velocities is the speed of light squared (see edit at bottom).
The critical density is when $$\omega^2=\frac{n_{crit} e^2}{\epsilon_0 m_e}$$
$$\Rightarrow \frac{\omega_p}{\omega}=\sqrt{\frac{n_e}{n_{crit}}}$$
Then subbing into the dispersion relation
$$\omega^2=\frac{n_e}{n_{crit}}\omega^2+c^2k^2$$
$$\Rightarrow \omega^2(1-\frac{n_e}{n_{crit}})=c^2k^2$$
$$\Rightarrow v_p=\frac{\omega}{k}=c \left(1-\frac{n_e}{n_{crit}}\right)^{-1/2}$$
is the phase velocity. The group velocity is $\frac{d\omega}{dk}$, so one might expect $v_g=v_p$ here as $\omega$ seems linear in $k$. But using $v_g=c^2/v_p$, we get
$$v_g=c\left(1-\frac{n_e}{n_{crit}}\right)^{1/2}$$
instead. So why does the group velocity not equal the phase velocity?
Edit
Just to show that $v_p v_g=c^2$
$$\omega=\left(\omega_p^2+c^2 k^2\right)^{1/2}$$
$$\Rightarrow \frac{\omega}{k}=\left(\frac{\omega_p^2}{k^2}+c^2\right)^{1/2}$$
$$\frac{d\omega}{dk}=\frac{1}{2}\left(\omega_p^2+c^2 k^2\right)^{-1/2}2c^2 k$$
$$=c^2\left(\frac{\omega_p^2}{k^2}+c^2\right)^{-1/2}$$
Thus $\frac{\omega}{k}\frac{d\omega}{dk}=c^2$
Best Answer
When the phase velocity is a constant (with respect to wavelength), the group velocity will indeed be equal to it, as you yourself have shown.
What you've got wrong here is the assumption for this case that the product of the phase velocity and group velocity equal the square of the speed of light, which can be true in other cases but not for plasma electromagnetic waves.