Well, in vacuum the group velocity and phase velocity are identical. The velocity of light in vacuum is not really determined by the properties of the moving charges. It is a fundamental constant of nature: if you like, it is determined by the properties of the vacuum. The important feature of the charges that generate the EM wave is the frequency at which they oscillate, which gives the frequency of the resulting wave.
The distinction between phase and group velocity becomes important when you are in a dispersive medium, meaning a medium in which you have charges that are stuck in one place somehow, e.g. electrons bound to atomic nuclei. (An example of a dispersive medium is a piece of glass.) Now the speed at which light propagates will depend on its frequency, this is what dispersion means.
Basically, dispersion arises because the bound charges have their own frequency that they like to oscillate at. However, the incoming light forces them to oscillate at a different frequency. The way they react to the incoming light (the amplitude of their oscillation) depends on how the light's frequency compares to their own favourite frequency. The bound charges' reaction is really important because the total EM wave that you observe is the sum of the incoming wave and the contribution from the oscillating charges, which re-radiate EM waves at an amplitude depending on their amplitude of oscillation.
Given that different frequencies of light have different speeds (i.e. phase velocities), we can now see why the group velocity is different, in general. The phase velocity is
$$ v_p = \frac{\omega}{k}, $$
and the group velocity is
$$ v_g = \frac{\partial \omega}{\partial k}, $$
which are only the same if $\omega = c k$, with $c$ a constant (the speed of light). This is why $v_p = v_g$ in vacuum. (Here $\omega$ is angular frequency and $k = 2\pi/\lambda$ is wavenumber.) However, if different frequencies have different velocities, then $v_p(\omega)$ is a non-trivial function of $\omega$, which is only true if the relationship between $\omega$ and $k$ (the dispersion relation) is more complicated than just the linear one. In other words, in a dispersive medium, $\omega \neq c k$. But the condition $v_p = v_g$ implies that
$$\frac{\partial \omega}{\partial k} = \frac{\omega}{k} \; \Leftrightarrow \; \omega = c k, $$
which shows that if dispersion is present then $v_p \neq v_g$. To see why group velocity is a useful measure of how a wavepacket moves, which is the usual interpretation, see e.g. Wikipedia.
The microscopic origin of dispersion is non-trivial, so I'll leave it to greater luminaries than I to give a proper explanation. This was really just a comment that got a bit long. Good question though, hopefully someone really clever will go into all the gory details :)
This is a question that was drastically changed and the other answer and the comments to the question and the answer are discordant with the edited question.
I already advised in the comment that you read a simplified article in wikipedia on electromagnetic radiation. Classical electromagnetic radiation cannot be simplified easily by analogies. I will try to explain with the quantum mechanical framework which is the underlying framework for all physics, classical theories emerge from this framework smoothly.
The electromagnetic wave is composed by a huge number of photons. Photons are elementary particles, they have zero mass, are point particles and their energy is characterized by the frequency of the classical wave we observe macroscopically, given by
In addition photons obey the Maxwell Electromagnetic equations in their quantum mechanical form. This means except their spin and energy they are characterized also by the vector potential entering the Maxwell equations.
You must have no problem accepting how a single photon can travel in space, even in vacuum. It is a quantum mechanical particle which will go in a straight line if it finds nothing to interact with. Single photons have been measured by making laser beams very faint so that only one hit happens on the detector at a time. They even have commercial single photon detectors.
The question goes into how zillions of photons can build up the electromagnetic waves that come to us from the stars, for example. They add up in a similar way that water molecules add up and make up streams, except in smaller dimensions and faster times. A stream of photons arrives continuously from those stars with the impetus given when they left so long ago.
If the light is coherent, i.e. the disparate constants describing each photon mathematically have a fixed phase , the individual fields, E and B described macroscopically with a wave equation, can be seen schematically:
This 3D diagram shows a plane linearly polarized wave propagating from left to right. Note that the electric and magnetic fields in such a wave are in-phase with each other, reaching minima and maxima together
The wave is built up by zillions of photons in step so that their spins, and other constants describing them build up the macroscopic wave, even though it is made up of all these elementary particles, all going in a straight line unless they hit something on the way.
Best Answer
Classically, electromagnetic waves are propagating disturbances in the electric and magnetic fields.
Remember, the electric field of a point charge extends to infinity. It does not simply stop somewhere.
When the point charge is briefly accelerated, a disturbance in the field (and the associated magnetic field) propagates with speed c outward away from the point charge.
The disturbance will continue to propagate even after the point charge has stopped accelerating; the disturbance has "a life of its own".
There's a nice applet for visualizing this here:
But, do keep in mind that we must ultimately understand electromagnetic radiation in terms of photons and that requires quantum field theory.
As I wrote above: electromagnetic waves are propagating disturbances in the electric and magnetic fields. Now, how and why does that happen?
Think carefully about what I wrote above: the disturbance has "a life of its own" and think about how that might be.