Just to add another look to the answer of Paul. As you know any thermodynamic potential has its own variables. As the function of its variables it achieves a minimum in equilibrium. Entropy is one of such potentials that can be used on the equal basis with the others. The only difference is that it achieves minimum rather than maximum. It appeared historically this way, that entropy has been defined as it is. Put minus in front of it, and it will be as all the others.
OK, its own variables are the internal energy, E, the volume, V, and the number of particles N. The first of these, the internal energy, is very inconvenient in use. One typically cannot measure this parameter independently. It is not often evident, how to calculate it into measurable parameters to compare with experiment. For this reason the entropy is rarely used.
In fact the philosophy behind is that you first define the set of variables that is adequate to the problem, and then work with the corresponding potential. The set (V, T, N) corresponds to the free energy. Use that!
Second. You write:
[...] and more disorder means less symmetry in the system. (Or is this statement wrong?)
You are definitely wrong. Below I just give you a counter-example that will clarify the situation:
a)The symmetry group of a crystalline solid is a discrete group with (i) translations over a discrete set of vectors, rotation over discrete set of angles, and reflection in a discrete set of planes. Eventually the symmetry might be even more complex, but this does not influence my example.
b) Introducing a disorder one may transform this crystal into amorphous solid. This has the symmetry group with continuous translations and rotations and the infinite set of the mirror planes. This group is continuous, the so-called, Euclidean motion group. Any symmetry group of any crystal is its subgroup. Thus, by increasing the disorder we increased the symmetry.
c) By further increasing the disorder (say, by increasing the temperature) we may bring our solid into the liquid state. Here the symmetry is still higher, since it allows all movements with a constant volume. The Euclidean motion group is the subgroup of this one.
One may also give a number of such examples within the solid state.
Though it is typical that during phase transitions a high-temperature phase has a higher symmetry, there are also opposite examples. They are too specific and I will not give them here. There are also examples for transitions between solid states with the symmetry groups that have no group-subgroup relations between one-another. In this case one cannot decide which symmetry is higher.
To summarize all this one should say that there is no solid rule about the relation between disorder and symmetry, though increasing disorder is often followed by the symmetry increase.
The low-entropy initial state of the universe is an open problem without a satisfactory answer. Your question is the first time I've heard the suggestion that the initial state should have been a crystal; you remind me that the quark-gluon plasma, which was the state of the universe while it was too hot for nucleons to be stable, has been shown to be a minimum-entropy fluid.
Sean Carroll wrote a nice book on the subject a couple of years ago, which I think was an extension of this paper.
Best Answer
Yes, one can observe a decrease in entropy of an isolated system.
The statistics of these observation are quantified by the fluctuation theorem. The logic of it is based on what you (the OP) suggest: since statistical mechanics is about statistical laws, one would expect there to be fluctuations.
The first paragraph of wikipedia states this well, so I'll just copy it with some emphasis added:
The paper in the first note (pdf) there gives access to a PRL with an experimental observation of a decrease in entropy in an isolated system. Of course, these observations were done in very small systems; as the systems get larger, the chance of observing such fluctuations decreases.