Spin is not a relativistic property. The existence of the spin operators can be derived from the assumption that space is rotationally invariant. This is how the argument goes (details omitted because of time and because I only know some of them):
A symmetry is a bijection on the set of unit rays, that preserves probabilities, i.e. S is a symmetry if |\langle\psi'_1|\psi'_2\rangle|^2=|\langle\psi_1|\psi_2\rangle|^2 for all \psi_1,\psi_2\in \mathcal R and all \psi'_1,\psi'_2\in S(\mathcal R). The idea that space is rotationally invariant is incorporated into QM as the assumption that there's a group homomorphism from SO(3) into the group of symmetries.
Now the question is, does this mean that there also exists a group homorphism from SO(3) into the group of unitary linear operators on our Hilbert space H, i.e. is there a representation of the symmetry group on the Hilbert space of state vectors? Wigner showed that the answer is no, and also that it's "almost" yes. We can define a function from the set of continuous curves that start at the identity of SO(3), into the group of unitary linear operators on H. I'll write such operators as U(C), where C is the curve. Note that the other endpoint of C is a rotation matrix. It turns out that these curves can be divided into two equivalence classes, which I'll call 0 and 1, and that U(C) only depends on the endpoint and the equivalence class. So I'll write U([C],R) instead of U(C). It's also possible to prove that U(0,R)=-U(-1,R).
For every R, there's exactly one curve in 0 and one curve in 1 that goes from the identity to R. We can try to use only the 0 curves, hoping that this will enable us to drop the C's completely and just write U(R), but it turns out that we still won't get the simple relationship U(R')U(R)=U(R'R). The U(C) operators satisfy
U(C',R')U(C,R)=U(C'*C,R'R)
where C'*C is the curve that goes through C at twice the normal speed, and then through a translated version of C' from R to R'R at twice the speed. The problem is that C'*C might be in 1 even if C and C' are both in 0. This means that we can write
U(0,R')U(0,R)=U(s,R'R)=(-1)^s U(0,R'R)
where s=[C'*C], and this can't be simplified any further.
This is why we take the symmetry group to be SU(2) instead of SO(3). SO(3) is homeomorphic to a 3-sphere with opposite points identified, and SU(2) is homeomorphic to a 3-sphere. This difference means that there will only be one equivalence class of curves in SU(2) instead of two, and that enables us to write U(r')U(r)=U(r'r), where r and r' are members of SU(2) that correspond to rotation operators, and r and -r correspond to the same rotation operator.
The R's and the r's can be expressed as functions of three parameters (e.g. Euler angles), and therefore both U(R) and U(r) can be expanded in Taylor series:
U(r)=1+\frac i 2\omega_{ij}M_{ij}+\cdots[/itex]<br />
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where second-order terms have been omitted and repeated indices are summed over, and \omega is the first-order term in a series expansion of r. The fact that \omega is symmetric implies that M can be defined as symmetric. The spin operators can now be defined as<br />
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S_i=\epsilon_{ijk}M_{jk}<br />
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where the epsilon is the Levi-Civita symbol. The commutation relations<br />
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[S_i,S_j]=i\epsilon_{ijk}S_k<br />
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can be derived from the fact operators must transform as<br />
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X\rightarrow U(r)^\dagger X U(r)<br />
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in order to make expectation values invariant.<br />
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As you guys can see, this stuff is pretty complicated. I think that's the reason why a proper discussion of spin is usually delayed until an advanced class where no one is going to be bothered by the fact that we consider the restricted Poincaré group instead of just SO(3) or SU(2). SO(3) is a subgroup of the restricted Poincaré group, so when we do the relativistic version of this, we get everything I said above and then some. That includes the concepts concepts "momentum", "energy" and "mass", and we end up with a proper definition of a quantum mechanical "particle".