= Solution
The <phase velocity> is the speed of a constant-phase surface, $v_{\rm ph}=\omega/k$. The <group velocity> is the envelope speed of a narrow packet, $v_{\rm gr}=d\omega/dk$, requiring a local <dispersion relation> rather than only a single monochromatic wave. In an isotropic transparent medium with $k=n(\omega)\omega/c$,
$$
\boxed{v_{\rm ph}=\frac c{n(\omega)},\qquad
v_{\rm gr}=\frac c{n(\omega)+\omega\,dn/d\omega}.}
$$
For a vacuum wave, $n\equiv1$ over the packet bandwidth, so both speeds are $c$ and $\boxed{v_{\rm ph}v_{\rm gr}=c^2}$. No <superluminal signalling> is implied by a phase speed in a dispersive material.
Strictly, the value $n(\omega_0)=1$ at one frequency alone does not prove the product identity. For example $n(\omega)=1+q(\omega-\omega_0)/\omega_0$ gives $v_{\rm ph}(\omega_0)=c$ but $v_{\rm gr}(\omega_0)=c/(1+q)$. The vacuum interpretation, or an equivalent nondispersive assumption, is necessary.
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