Why does an Abelian Symmetry group necessarily imply no degeneracy?

Consider an operator $A$ such that $A^2 = I$ (essentially a representation of $\mathbb{Z}_2$) and a Hamiltonian $H$ such that $[H,A]=0$. Assume that this is the only symmetry. Now, $H$, $A$ have a simultaneous eigenbasis, and $A$ can be divided into a direct sum of irreps, each of dimension 1 (the answer here says that this is the reason for a lack of degeneracy).

However, if the different eigenstates have different eigenvalues of $A$, what prevents $H$ from having two eigenstates with the same eigenvalue? These could then be distinguished by the labels that $A$ provides.

Why does an Abelian Symmetry group necessarily imply no degeneracy?

As an example, consider an operator $A$ such that $A^2 = I$ (essentially a representation of $\mathbb{Z}_2$) and a Hamiltonian $H$ such that $[H,A]=0$. Assume that this is the only symmetry. Now, $H$, $A$ have a simultaneous eigenbasis, and $A$ can be divided into a direct sum of irreps, each of dimension 1 (the answer here says that this is the reason for a lack of degeneracy).

However, if the different eigenstates have different eigenvalues of $A$, what prevents $H$ from having two eigenstates with the same eigenvalue? These could then be distinguished by the labels that $A$ provides.

Why does an Abelian Symmetry group necessarily imply no degeneracy?

Consider an operator $A$ and a Hamiltonian $H$ such that $[H,A]=0$. Assume that this is the only symmetry. Now, $H$, $A$ have a simultaneous eigenbasis, and $A$ can be divided into a direct sum of irreps, each of dimension 1 (the answer here says that this is the reason for a lack of degeneracy).

However, if the different eigenstates have different eigenvalues of $A$, what prevents $H$ from having two eigenstates with the same eigenvalue? These could then be distinguished by the labels that $A$ provides.

Why does an Abelian Symmetry group necessarily imply no degeneracy?

Consider an operator $A$ such that $A^2 = I$ (essentially a representation of $\mathbb{Z}_2$) and a Hamiltonian $H$ such that $[H,A]=0$. Assume that this is the only symmetry. Now, $H$, $A$ have a simultaneous eigenbasis, and $A$ can be divided into a direct sum of irreps, each of dimension 1 (the answer here says that this is the reason for a lack of degeneracy).

However, if the different eigenstates have different eigenvalues of $A$, what prevents $H$ from having two eigenstates with the same eigenvalue? These could then be distinguished by the labels that $A$ provides.