$SU(2)$ Symmetry of Hubbard Model

I am confused with the $SU(2)$ spin rotation symmetry of the fermion Hubbard Hamiltonian. If the Hubbard model has $SU(2)$ rotational symmetry, it means that the Hubbard Hamiltonian commutes with the global spin operator in all direction:
begin{equation}
[vec S, H] = 0 ~~,~~ vec S = frac{1}{2}sum_{i}
begin{pmatrix}
c^{dagger}_{i uparrow} & c^{dagger}_{i downarrow}
end{pmatrix}
vec{sigma}
begin{pmatrix}
c_{i uparrow}
c_{i downarrow}
end{pmatrix}
end{equation}
Where $vec sigma$ is the Pauli matrices in vector form. My confusion is that whether $[vec S, H] = 0$ implies $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $. I have proved that they are equal to zero but I thought that my calculation was wrong. The reason why I think my calculation is wrong since if both $S_{x}, S_{y} , S_{z}$ commutes with $H$, it means that they simultaneously share the same eigenstates and $[S_x, S_y] = 0$. However, we know that from elementary QM course:
begin{equation}
[S_{i}, S_{j}] = i epsilon_{ijk} S_{k} ~~,~~ ijk = xyz
end{equation}
In my opinion, $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $ is not true because they cannot satisfy $SU(2)$ commutation relation if all commutes with $H$. May I know is it true that $[vec S, H] = 0$ implies $[S_{x}, H] = [S_{y}, H] = [S_{z}, H] = 0 $?