[proofplan] The proof is purely algebraic on the common invariant domain $D$. We expand each commutator using the definitions $J_+=J_x+iJ_y$ and $J_-=J_x-iJ_y$, then substitute the angular momentum commutation relations and their antisymmetric consequences. The first two computations show how $J_+$ and $J_-$ commute with $J_z$, and the final computation collects the two nonzero cross terms in $[J_+,J_-]$. [/proofplan]

[step:Expand the commutator of $J_z$ with the raising operator]Fix $\psi \in D$. Since $J_x,J_y,J_z,J_+:D \to D$, all compositions appearing below are defined on $D$. Using the definition of the commutator and the linearity of $J_z$, we compute \begin{align*} [J_z,J_+]\psi=J_z(J_x\psi+iJ_y\psi)-J_+(J_z\psi). \end{align*} By the definition of $J_+$ applied to $J_z\psi \in D$, \begin{align*} J_+(J_z\psi)=J_x(J_z\psi)+iJ_y(J_z\psi). \end{align*} Therefore \begin{align*} [J_z,J_+]\psi=[J_z,J_x]\psi+i[J_z,J_y]\psi. \end{align*} The hypotheses give $[J_z,J_x]\psi=i\hbar J_y\psi$. Also, from $[J_y,J_z]\psi=i\hbar J_x\psi$ and the definition of the commutator, $[J_z,J_y]\psi=-i\hbar J_x\psi$. Substituting these two identities yields \begin{align*} [J_z,J_+]\psi=i\hbar J_y\psi+i(-i\hbar J_x\psi). \end{align*} Since $i(-i)=1$, this becomes \begin{align*} [J_z,J_+]\psi=\hbar J_x\psi+i\hbar J_y\psi. \end{align*} Factoring out $\hbar$ and using $J_+\psi=J_x\psi+iJ_y\psi$, we obtain \begin{align*} [J_z,J_+]\psi=\hbar J_+\psi. \end{align*}[/step]

[guided]Fix $\psi \in D$. The reason the common invariant domain matters is that every operator product in the commutators must be defined. Because $J_x,J_y,J_z,J_+:D \to D$, the vectors $J_+\psi$, $J_z\psi$, $J_x\psi$, and $J_y\psi$ all lie in $D$, so applying the next operator is legitimate. We begin from the definition of the commutator: \begin{align*} [J_z,J_+]\psi=J_z(J_+\psi)-J_+(J_z\psi). \end{align*} Now insert the definition $J_+\psi=J_x\psi+iJ_y\psi$. Linearity of $J_z$ gives \begin{align*} J_z(J_+\psi)=J_z(J_x\psi+iJ_y\psi)=J_z(J_x\psi)+iJ_z(J_y\psi). \end{align*} Since $J_z\psi \in D$, the definition of $J_+$ may also be applied to $J_z\psi$: \begin{align*} J_+(J_z\psi)=J_x(J_z\psi)+iJ_y(J_z\psi). \end{align*} Subtracting these expressions gives \begin{align*} [J_z,J_+]\psi=\bigl(J_z(J_x\psi)-J_x(J_z\psi)\bigr)+i\bigl(J_z(J_y\psi)-J_y(J_z\psi)\bigr). \end{align*} This is exactly \begin{align*} [J_z,J_+]\psi=[J_z,J_x]\psi+i[J_z,J_y]\psi. \end{align*} We now use the angular momentum commutation relations. One relation is already in the required order: \begin{align*} [J_z,J_x]\psi=i\hbar J_y\psi. \end{align*} For the other term, the hypothesis gives $[J_y,J_z]\psi=i\hbar J_x\psi$. Since commutators are antisymmetric by direct expansion, \begin{align*} [J_z,J_y]\psi=-[J_y,J_z]\psi=-i\hbar J_x\psi. \end{align*} Substituting both identities into the expanded commutator gives \begin{align*} [J_z,J_+]\psi=i\hbar J_y\psi+i(-i\hbar J_x\psi). \end{align*} The scalar identity $i(-i)=1$ converts the second term into $\hbar J_x\psi$, so \begin{align*} [J_z,J_+]\psi=\hbar J_x\psi+i\hbar J_y\psi. \end{align*} Factoring out $\hbar$ recovers the raising operator: \begin{align*} [J_z,J_+]\psi=\hbar(J_x\psi+iJ_y\psi)=\hbar J_+\psi. \end{align*}[/guided]

[step:Expand the commutator of $J_z$ with the lowering operator] Again fix $\psi \in D$. Using $J_-\psi=J_x\psi-iJ_y\psi$, the same commutator expansion gives \begin{align*} [J_z,J_-]\psi=[J_z,J_x]\psi-i[J_z,J_y]\psi. \end{align*} Substitute $[J_z,J_x]\psi=i\hbar J_y\psi$ and $[J_z,J_y]\psi=-i\hbar J_x\psi$: \begin{align*} [J_z,J_-]\psi=i\hbar J_y\psi-i(-i\hbar J_x\psi). \end{align*} Since $-i(-i)=-1$, this becomes \begin{align*} [J_z,J_-]\psi=-\hbar J_x\psi+i\hbar J_y\psi. \end{align*} Using $J_-\psi=J_x\psi-iJ_y\psi$, we conclude \begin{align*} [J_z,J_-]\psi=-\hbar J_-\psi. \end{align*} [/step]

[step:Expand the commutator of the two ladder operators] Fix $\psi \in D$. Since $J_+,J_-:D \to D$, the commutator $[J_+,J_-]\psi$ is defined. Expanding from the definitions, \begin{align*} [J_+,J_-]\psi=(J_x+iJ_y)(J_x\psi-iJ_y\psi)-(J_x-iJ_y)(J_x\psi+iJ_y\psi). \end{align*} Using linearity of $J_x$ and $J_y$, the first product equals \begin{align*} J_x(J_x\psi)-iJ_x(J_y\psi)+iJ_y(J_x\psi)+J_y(J_y\psi). \end{align*} The second product equals \begin{align*} J_x(J_x\psi)+iJ_x(J_y\psi)-iJ_y(J_x\psi)+J_y(J_y\psi). \end{align*} Subtracting cancels the two square terms and leaves \begin{align*} [J_+,J_-]\psi=-2iJ_x(J_y\psi)+2iJ_y(J_x\psi). \end{align*} Factoring $-2i$ gives \begin{align*} [J_+,J_-]\psi=-2i\bigl(J_x(J_y\psi)-J_y(J_x\psi)\bigr). \end{align*} The expression in parentheses is $[J_x,J_y]\psi$, so \begin{align*} [J_+,J_-]\psi=-2i[J_x,J_y]\psi. \end{align*} Using $[J_x,J_y]\psi=i\hbar J_z\psi$, we obtain \begin{align*} [J_+,J_-]\psi=-2i(i\hbar J_z\psi)=2\hbar J_z\psi. \end{align*} Thus all three claimed commutation relations hold for the arbitrary vector $\psi \in D$. [/step]