Commutation relation of $J^2$ and $R(\alpha,\beta,\gamma)$

Question

If $R(\alpha,\beta,\gamma)$ is the Rotation operator and $\alpha,\beta,\gamma$ are Euler angles and $J$ is the total angular momentum then how to get to this: $$[J^2,R]~=~0?$$

This is stated in Zettili's book Quantum Mechanics.

Answer 1

See: http://students.washington.edu/tkarin/rotations.pdf

The rotation operator $R$ is generated by the angular momentum operators $J_i$: $R$ can be written as a power series of $J_i$.

Since $[J^2,J_i]=0$ for $i=1,2,3$ the operator $J^2$ will commute with every term in the power series of $R$. By this we have $[R,J^2]=0$.