# Why do we consider spin degeneracy in graphene quantum hall effect and not in the conventional one?

When dealing with quantum hall effect in graphene we say that each landau level (with $$n\neq 0$$) has 4 times the degeneracy of a simple landau level derived for an electron in a magnetic field because of spin and valley degeneracy so that each of them contributes to conductivity with a factor $$\frac{4e^2}{h}$$. Why in conventional quantum hall effect (in other 2D materials like ) we neglect spin degeneracy and say that hall conductivity is just $$\frac{e^2}{h}n$$? shouldn't it be $$\frac{2e^2}{h}n$$?

The integer $$n$$ is given by the number of completely filled Landau levels and is more commonly denoted and referred to as the filling factor $$\nu$$. Ignoring Zeeman splitting, the filling factor should be $$\nu=0,2,4,.\ldots$$ for a conventional system with parabolic dispersion (i.e. a 2DEG in GaAs quantum well), where the factor of two arises from the spin degeneracy of the Landau levels.
On the other hand, in graphene (again ignoring Zeeman splitting), the filling factor sequence for the conduction band is given by $$\nu=2,6,10,\ldots=2,2+4,2+2\cdot 4,\ldots$$ because the Landau levels are fourfold degenerate and the zeroth Landau level is already half filled when the Fermi level is at the Dirac point (see this question).
If the broadening of the Landau levels due to temperature and disorder is much smaller than the Zeeman splitting $$\propto gB$$ where $$g$$ is the g-factor and $$B$$ the magnitude of the magnetic field, the observed quantum Hall plateaus should split into two. For the conventional system, the sequence then becomes $$\nu=0,1,2,\ldots$$, while in graphene, if the Zeeman splitting is still much smaller than the original energy difference between the Landau levels, it should be $$\nu=1,2,4,6,\ldots$$.