In an excercise regarding two particles with the same angular momentum $j_1=j_2$, I was given an operator $P$ (I don't think it is relevent here what $P$ actually is).
I was asked to show that $P$ commutes with $J^z$ and $J^\pm$ in the space spanned by the simultanuous eigenstates of the commuting operators ${J_1^2, J_2^2, J_1^z, J_2^z}$, denoted by $|j_1,j_1,m_1,m_2\rangle$. Then it is claimed without explanation, as if it was immediate, that every eigenstate $|j_1,j_1,j,m\rangle$ is also an eigenstate of $P$.
I fail to understand why this is true.
The eigenstates $|j_1,j_1,j,m\rangle$ are simultaneous eigenstates of the commuting operators ${J_1^2, J_2^2, J^2, J^z}$. The claim that every eigenstate $|j_1,j_1,j,m\rangle$ is also an eigenstate of $P$, is equivalent to the claim that we can add $P$ to the set of commuting operators ${J_1^2, J_2^2, J^2, J^z}$, so $P$ commutes with every single one of them (commuting isn't transitive).
In the exercise I showed that $P$ commutes with $J^z$, and now I'm trying to understand why commuting with $J^\pm$ and $J^z$ means commuting with $J_1^2, J_2^2,$ and $ J^2$.
