$U(n)$ and $SU(n)$ Symmetry I am a first year post graduate student and studying symmetries in quantum mechanics. I am not getting the point why we need $U(n)$ and $SU(n)$ symmetry in general and how does groups come into role? 
 A: Other answers already explained on some aspects of the unitary groups as used in quantum mechanics and QFT. However, a thing that made me curious as student was that one sometimes workes with $SU(2)$ instead of $SO(3)$ when dealing with rotations in 3-space. For example, I learned that a spin 1/2 particle is a representation of the former. Without doubt, $SO(3)$ is the rotation group in 3 dimensions, so what's the thing about $SU(2)$?
There are various ways to answer that question, but I want to focus on a mathematical nice one. Begin with noting that when describing a quantum system by a Hilbert space, we have quite a redundancy, since states are equivalent if they are proportional. This fact has a crucial effect on the treatment of symmetries. Instead of just ordinary linear representations $$\phi:G\rightarrow GL(\mathcal{H})$$
we are allowed to use the bigger class of projective representations, which you can think of as representations up to proportionallity. This is just a consequence of the redundant description of states. 
Now, how to work with this? Luckily enough, we already know how to describe projective representations by means of ordinary ones. However, you have to dive deeper into group theory and work with so called central extensions of your symmetry group $G$. It was shown, that the projective representations of $G$ correspond to ordinary representations of central extensions of $G$. 
But again, we are lucky and don't have to deal with central extensions in our case, since our interest is in a simple finite-dimensional Lie group. According to Bargmann's theorem in the case of semi-simple Lie groups (which $SO(3)$ belongs to) the projective representations of $G$ are in one-to-one correspondence to the ordinary representations of a single central extension, namely the universal covering group of $G$. In our case of $G=SO(3)$ this universal covering happens to be $SU(2)$. So the reason we use $SU(2)$ instead of $SO(3)$ is that the ordinary representations of $SU(2)$ correspond to projective representations of $SO(3)$! Hurray! Another prominent example is the Lorentz Group $SO(1;3)$ with its universal covering $SL(2;\mathbb{C})$, which is important for the characterization of particle types in QFT. 
For a more consequent description of all that (with actual definitions of what I was talking about) check for example the Wikipedia entry for projective representations.
Just a fun fact at the end: The projective representations of $SO(3)$ or $SO(1;3)$ that can be "lifted" to ordinary representations of $SO(3)$ or $SO(1;3)$ respectivelly are what we know as Bosons. The ones that have no corresponding ordinary representation are what we know as Fermions. Note that here I really mean "corresponding to ordinary reps of $G$" and not "of the universal cover of $G$".
I hope this helps you to find your way through qunatum theory.
Cheers!
A: I think that an important aspect that could help you understand why we need unitary groups in quantum mechanics is the following. (if there is anything wrong let me know).
Consider that at a very informal level a simmetry is the characteristic of an object to conserve properties under a transformation.
In QM the predictive aspect of the theory is the probability of finding a system in a certain physical state. So to find symmetries you should search for transformations that leaves the probability unchanged, and so operators U such that
$$P(\psi_i \rightarrow \psi_j)=P(\phi_i \rightarrow \phi_j)$$ where $P(\psi_i \rightarrow \psi_j)$ stands for the porbability of finding state $\psi_i$ in $\psi_j$ after a measure and  $$\phi_i=U\psi_i $$$$\phi_j=U\psi_j$$ 
Wigner theorem state that these transformations are given by unitary (or antiunitary) operators, that's one possible reason to study U(n) and SU(n) in QM among many others.
When you have studied quantum mechanics you should have already seen this in cases of transformations like rotations, translation or the time evolution operator.
I hope this is useful
Trying to answer why we need groups, i think the main reason is that groups help us to study the system. As an example, suppose the hamiltonian H is invariant under the transformations of a group G, then given a U in G we have $$H=U^{-1}HU \rightarrow HU=UH$$ so consider the eigenvalue problem
$$H\psi=E\psi$$
$$H(U\psi)=UH\psi=U(E\psi)=E(U\psi) \rightarrow H(U\psi)=E(U\psi)$$ 
you can see that U transform the state $\psi$ in linear combinations of states with same energy E, so we can study degeneracy of the system with the help of groups.
I am studying group theory on the book Group theory in a nutshell (by A. Zee) and it is all focused on the physical importance of group theory 
A: Groups play a very crucial role in physics, and in particular within particle physics. To ask why we need them is rather complicated to answer, as one could say that gives us a way to construct theories which match the experimental evidence. But if I had to give a purely theoretical explanation, first of all I would say that groups are almost natural, if you think about it humans group together certain objects within a category. Of course the mathematical definition of group is much more technical, but surely you have noticed how we tend to create groups out of everything.
Going more into physics, I would say that groups come into role due to symmetries. Symmetries play a central role in theoretical physics, in particular with Noether's theorem, which states that to every symmetry there is a conserved current. As Heisenberg said himself, "We will have to abandon the philosophy of Democritus and the concept of elementary particles. We should accept instead the concept of elementary symmetries". This explains why group theory is so important and comes into play: it gives us the setup to play around with the symmetries of the theory. And just as an example for you to see how important symmetries are, if one allows for a $U(1)$ symmetry in QFT, one obtains the gauge field corresponding to the photon. Therefore, we introduce the photon to ensure the invariance (symmetry) of the Lagrangian under $U(1)$. In conclusion, symmetries are present in our theories, and if you wish to study such symmetries, you need to use group theory.
As for the difference between introducing $U(n)$ and $SU(n)$, recall for a start that $U(n)=U(1)\times SU(n)$ where $\times$ indicates the group product. Therefore, you can see that they are related. But again, the presence of $U(n)$ or $SU(n)$ will depend on the symmetry of your theory. Taking the example of the photon, it appears when we introduce a $U(1)$ symmetry, while the gluon, for example, appears with a $SU(3)$ symmetry. If you wish to ask why such groups and not other ones, the answer here is surely that such groups give rise to correct predictions. Much work has been done in trying to explore symmetries such as $SU(5)$, and although it might be theoretically plausible, it never gave predictions (so far), so we don't consider it to correspond to nature in our current understanding of quantum field theory.
A: They are internal degrees, if you could connect to an external thing, it would not be internal degrees. They are particles, there are no external degrees. Of course, it is not a clue to select a group SU(n), all the n´s are possible. Theorethical physics has no limits,  not this way, anyway. in spite SU(n) simmetries are crucial in particle physics, it is neccesary a more general principle if we wish deduce all the physics. EFT are the best aproximattion till the moment. The SU(n) are not  a priori, that is the main con about it. If you read in physics text $ U(1) \cdot SU(2)....SU(n) $ , think that $U(1) $ is mathematically SU(1)
