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  • $\begingroup$ I was about to suggest the counter-example, but I thought it might be facetious and specific to this case, since the Casimir is just the Identity for the spin-1/2 representation. The reverse statement is true, no doubt about it. It's just that many answers seemed to suggest that an Abelian symmetry group precludes the possibility of degeneracy, and I wanted to know if this was true. Could you please elaborate on "accidental degenracies"? $\endgroup$
    – Primo-uomo
    Commented Jul 3, 2017 at 9:52
  • $\begingroup$ Wrong: for instance also $\sigma_x$ commutes with $H$: there is another symmetry against the hypothesis that $A$ is the only symmetry. $\endgroup$
    – Valter Moretti
    Commented Jul 3, 2017 at 14:26
  • $\begingroup$ @ValterMoretti Also $\sigma_y$ commutes with $H$, but neither commutes with $A=\sigma_z$...so i'm not sure how they relate to the statements about irreps of $A$? Is the elaboration in your answer? $\endgroup$
    – Mikael Fremling
    Commented Jul 3, 2017 at 15:48
  • $\begingroup$ I mean $\sigma_y$ is another symmetry since it commutes with $H$ but it is not function of $\sigma_x$ (otherwise it would commute with it). In the hypothesis there must be only one symmetry... $\endgroup$
    – Valter Moretti
    Commented Jul 3, 2017 at 15:50
  • $\begingroup$ I see, so the symmetries need not commute with each other. I missed that point. Thanks for clarifying. $\endgroup$
    – Mikael Fremling
    Commented Jul 3, 2017 at 15:54