Are Eigenfunctions of a Hamiltonian on a lattice always Bloch functions?

[0xDEADBEEF]

It is common knowledge in Physics that eigenstates share the symmetries of the Hamiltonian.

And it is trivial to show that this is true for the eigenspaces. Let g be an element of a symmetry group of Hamiltonian H, $$M_g$$ its representation, $$\left| \phi \right>$$ an eigenvector and $$\lambda$$ the corresponding eigenvalue. If g is an element from a symmetry group of H it is given that:

$$M_gHM_{g^{-1}}\left|\phi\right>=\lambda\left|\phi\right>$$
Thus
$$HM_{g^{-1}}\left|\phi\right> = \lambda M_{g^{-1}}\left|\phi\right>$$

So we see that the eigenspace for $$\lambda$$ is closed under the operations of $$M_g$$
Is there some theorem, that I can decompose this eigenspace into eigenstates of M or how do I proceed from here? This is important for Bloch functions and crystallography. I read something about Schur's lemma being involved.

So to phrase a proper question:
Is every Eigenfunction of a Hamiltonian,invariant up to a scale factor of unity magnitude under the operation of the Hamiltonian's symmetry groups? And if this is so how do I show this?

 

[0xDEADBEEF]

Ok I think I have answered my question using spherical harmonics. It is really only the eigenspaces that are invariant. But then I don't know how to argue that on a lattice the eigenfunctions must be Bloch functions.