What is the relationship between symmetry and degeneracy in quantum mechanics? Let me remind you about the following classical 
examples in quantum mechanics.

Example 1. Bound states in 1-dim potential V(x).
  Let $V(x)$ be a symmetric potential i.e.
  $$V(x) = V(-x)$$
  Let us introduce the parity operator $\hat\Pi$ in the following way:
  $$\hat\Pi f(x) = f(-x).$$
  It is obvious that 
  $$[\hat H,\hat\Pi] = 0.$$
  Therefore, for any eigenfunction of $\hat H$ we have:
  $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$
  i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$

Consider the second example.

Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.
  In that potential we can choose eigenfunction of 
   angular momentum $\hat L^2$ for basis 
  $$|l,m\rangle,$$
  where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.   

My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:


*

*Existence of symmetry $\Rightarrow$ Existence of degeneracy

*Existence of degeneracy $\Rightarrow$ Existence of symmetry


It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true.    
 A: This material seems to be poorly covered in most introductory QM books, so here's the logic:


*

*Suppose there is a group of transformations $G$. Then it acts on the Hilbert space by some set of unitary transformations $\mathcal{O}$.

*The Hilbert space is therefore a representation of the group $G$, and it splits up into subspaces of irreducible representations (irreps). The important thing is that if $|\psi\rangle$ and $|\phi \rangle$ are in the same irrep iff you can get from one to the other by applying operators $\mathcal{O}$.

*If the transformations are symmetries of the Hamiltonian, then the operators $\mathcal{O}$ commute with the Hamiltonian. Then if $|\psi\rangle$ is an energy eigenstate, then $\mathcal{O}|\psi \rangle$ is an energy eigenstate with the same energy.

*Therefore, all states in an irrep have the same energy. So if there are nontrivial irreps of dimension greater than one present, then there will be degenerate states.

*Conversely, if there is any degeneracy at all, we typically think of it as being caused by some symmetry, which may be well hidden. Ideally, there should be no 'accidental' degeneracy.

*If $G$ is an abelian group, then all irreps are one-dimensional, and hence yield no degeneracy.


Below are some examples.


*

*Particle in a 1D symmetric potential. The group is $\mathbb{Z}_2$ and it is generated by the parity operator. The group is abelian, so there's no degeneracy.

*The free particle in 1D. There are two symmetries: translational symmetry and parity symmetry. As a result, the group is not abelian and can have nontrivial irreps. There are irreps of dimension two, and these correspond to the degeneracy of the plane wave states $e^{\pm ikx}$. 

*A particle in 1D with $H = p^3$. There's no degeneracy; the argument for the free particle fails because we don't have parity symmetry, only translational symmetry. This shows that a continuous symmetry (translations) doesn't guarantee degeneracy. It does guarantee a conserved quantity (here, momentum), but that's a different issue.

*Particle in a 3D central potential. The group is $SU(2)$, which is nonabelian. The degenerate sets of states $\{l, m\}_{-l \leq m \leq l}$ are just the irreps of $SU(2)$. 

*Hydrogen atom. There is an additional degeneracy between states with the same $n$ but different $l$ quantum numbers. This comes from a hidden $SO(4)$ symmetry of the Hamiltonian.


In summary, your second point is true (generally, degeneracy implies symmetry), but your first point is false. Continuous symmetries guarantee you get conserved quantities, not degeneracy.
A: knzhou's answer is very well-explained, but it's perhaps worth mentioning that the energy gaps between different symmetry sectors typically decrease with system size, and formally vanish in the thermodynamic limit.  So an infinite-size system can indeed (but doesn't have to) have symmetry-induced degeneracy, even if the symmetry is abelian (regardless of whether the symmetry is discrete or continuous - e.g. the quantum transverse Ising model, which has $\mathbb{Z}_2$ symmetry, has twofold ground-state degeneracy in the thermodynamic limit, and the $X$-$Y$ model, which has $U(1)$ symmetry, has infinite GS degeneracy).  If there is a symmetry-induced degenerate GS manifold in the thermodynamic limit, then the symmetry is typically broken: the physically realistic ground states are not invariant under the symmetry.
Also, even in absence of symmetry, an infinitely large system in a topologically ordered phase can have a finite degeneracy.  This degeneracy is extremely robust because unlike in the symmetry-induced case, no possible perturbation can lift it.
