It turned out they are actually, I solved the problem later on. The hamiltoinian will give you two energystates both with degeneracy 4.
It is also fully possible to solve the states by first finding the spin for two particles and then add the third. I was not looking for the coefficients, so...
I have a hamiltonian:
H = J\left(\frac{1}{2\hbar^2}S_{tot}^2 - \frac{9}{8}\right)
where S_{tot} = S_1 + S_2 + S_3
for 3 spin half particles in a cluster. The magnetic properties are governed entirely by their spin.
I have set up the possible spin combinations from Clebsch-Gordan to be:
From...