# For $[A,B]=0$, if an eigenfunction of $A$ not an eigenfunction of $B$, does that imply degeneracy of one operator?

Yes, it is true. Let $u$ be the said eigenvector of $A$ with eigenvalue $a$ that is not eigenvector if $B$. Then $ABu=BAu=aBu$. But $Bu\neq bu$ for every $b$. Therefore $Bu$ is an eigenvector of $A$ with eigenvalue $a$ that is lineary independent of $u$. The spectrum of $A$ is therefore degenerate.