When is Noether charge quantized? Noether charge is the generator of the corresponding global symmetry. My question is when a Noether charge is quantized? How does the charge quantization is related to the compactness of the symmetry group?
 A: I am interpreting "quantized" as "discrete",i.e., in proper mathematical terms "pure point spectrum" if referring to the spectrum of the self-adjoint generator representing the charge associated with a continuous dynamical symmetry.
If a Lie group $G$ is compact, then all irreducible representations $G \ni g \mapsto U_g$ are finite dimensional and every unitary (strongly-continuous) representation is a direct orthogonal sum of a countable class finite dimensional representations (assuming as is the standard, that the Hilbert space of the theory is separable).
$$U_g = \oplus_k U^{(k)}_g\:.$$ 
Since, in finite dimensional Hilbert spaces, the spectra of operators is always pure-point spectra (discrete) and  a generator  $Q$ (the conserved charges) of a generic unitary rep is the sum of the corresponding generators  of its finite dim representations,
$$Q= \oplus_k Q^{(k)}\:,$$
 the spectrum of generators are the closure of the union of the spectra of subgenerators. 
$$\sigma(Q)= \overline{\cup_k \sigma(Q^{(k)})}$$
 The final spectrum $\sigma(Q)$ is therefore a pure point-spectrum barring possibly accumulation points of $\cup_k \sigma(Q^{(k)})$ which stay in the continuous part of the spectrum if are not elements of $\cup_k \sigma(Q^{(k)})$ itself. However $\sigma_c(Q)$, if any, has not much relevance here. As a matter of fact, $\sigma_c(Q)$ cannot include any interval $(a,b)$ since the proper eigenvectors of $Q$ form a Hilbert basis and the spectral projector associated to $(a,b)$ should projects onto a closed subspace orthogonal to all eigenvectors.
A typical example is the case of $SU(2)$ representations where the generators (the components of angular momentum) have pure point spectrum since there are no accumulation points of the union of the sets $\sigma_J := \{j = 0,\pm 1/2, \pm 1, \pm 3/2, \ldots \:|\: -J \leq j \leq J\}$ and $J=0,1/2, 1, 3/2, \ldots$. 
Compactness is only a sufficient condition for these features of the spectrum of  generators. Even if $G$ is non-compact some generators may have discrete spectrum: think of $SL(2,\mathbb C)$ and $SL(2,\mathbb R)$ in particular.
Regarding the electric charge, if we assume that this is nothing but the generator of a strongly-continuous unitary representation of $U(1) \ni \theta \mapsto e^{i\theta Q}$, it must be $e^{i 2\pi Q} =I$ so that, passing in spectral representation, we must have $\sigma(Q) \subset \mathbb Z$ independently but in accordance with the reasoning above.
It is worth stressing that this argument does not fix the value of the elementary electric charge since a further parameter is free (depending on the units one adopts) and a rescaling $\theta \to q^{-1}\theta$, $Q \to qQ$ is always possible with the subsequent identification of $[0, 2\pi q^{-1}]$ and $U(1)$. Nevertheless, as soon as that parameter is fixed, we have $\sigma(qQ) \subset q \mathbb Z$ and the spectrum remains discrete.
Finally, the fact that symmetries are represented by (anti) unitary operators up to phases (Wigner-Kadison theorem) here does not matter. Indeed, it is possible to prove that projective continuous representations of $\mathbb R$ (and thus also $U(1)$) are always unitarizable. So we can always reduce to the case treated above regarding electric charge if assuming that it exists as a consequence of a $U(1)$ (continuous projective) symmetry.
