[proofplan]
The proof is a direct tensor-product computation. First we check that the algebraic tensor domain is invariant under the operators $J_i$, so that the commutators $[J_i,J_j]$ are well-defined on $D$. Then we compute $J_iJ_j$ and $J_jJ_i$ on an elementary tensor, where the mixed tensor-factor terms cancel because the $A$-operators act only on $H_A$ and the $B$-operators act only on $H_B$. The two remaining commutators are exactly the angular momentum commutators for $A$ and $B$, and they combine into the required commutator for $J$.
[/proofplan]
[step:Verify that the tensor product domain is invariant under each $J_i$]
Let $i \in \{1,2,3\}$. Since $D_A$ is invariant under $A_i$ and $D_B$ is invariant under $B_i$, for every elementary tensor $u \otimes v \in D_A \odot D_B$ we have $A_i u \in D_A$ and $B_i v \in D_B$. Hence
\begin{align*}
J_i(u \otimes v) = A_i u \otimes v + u \otimes B_i v \in D_A \odot D_B.
\end{align*}
By linearity, $J_i(D) \subset D$. Therefore $J_iJ_j$ and $J_jJ_i$ are well-defined on $D$ for all $i,j \in \{1,2,3\}$.
[/step]
[step:Expand the commutator on an elementary tensor]
Fix $i,j \in \{1,2,3\}$ and let $u \in D_A$, $v \in D_B$. Using the definition of $J_j$ and then the definition of $J_i$, we obtain
\begin{align*}
J_iJ_j(u \otimes v) = A_iA_j u \otimes v + A_j u \otimes B_i v + A_i u \otimes B_j v + u \otimes B_iB_j v.
\end{align*}
Similarly,
\begin{align*}
J_jJ_i(u \otimes v) = A_jA_i u \otimes v + A_i u \otimes B_j v + A_j u \otimes B_i v + u \otimes B_jB_i v.
\end{align*}
Subtracting the second identity from the first gives
\begin{align*}
[J_i,J_j](u \otimes v) = [A_i,A_j]u \otimes v + u \otimes [B_i,B_j]v.
\end{align*}
[guided]
We compute on an elementary tensor because every element of the algebraic [tensor product](/page/Tensor%20Product) $D_A \odot D_B$ is a finite linear combination of such tensors, and all operators involved are linear. Fix $i,j \in \{1,2,3\}$, $u \in D_A$, and $v \in D_B$. First,
\begin{align*}
J_j(u \otimes v) = A_j u \otimes v + u \otimes B_j v.
\end{align*}
Since the domain is invariant under the $A$-operators and $B$-operators, the two tensors on the right still lie in $D_A \odot D_B$, so we may apply $J_i$ to each term. This gives
\begin{align*}
J_iJ_j(u \otimes v) = A_iA_j u \otimes v + A_j u \otimes B_i v + A_i u \otimes B_j v + u \otimes B_iB_j v.
\end{align*}
Interchanging $i$ and $j$ gives
\begin{align*}
J_jJ_i(u \otimes v) = A_jA_i u \otimes v + A_i u \otimes B_j v + A_j u \otimes B_i v + u \otimes B_jB_i v.
\end{align*}
Now subtract. The mixed terms cancel exactly: $A_j u \otimes B_i v$ appears once with positive sign and once with negative sign, and $A_i u \otimes B_j v$ also appears once with positive sign and once with negative sign. These mixed terms are the algebraic expression of the fact that operators acting on different tensor factors commute on the algebraic tensor product. The remaining terms are
\begin{align*}
[J_i,J_j](u \otimes v) = (A_iA_j u - A_jA_i u) \otimes v + u \otimes (B_iB_j v - B_jB_i v).
\end{align*}
By the definition of the commutator, this is
\begin{align*}
[J_i,J_j](u \otimes v) = [A_i,A_j]u \otimes v + u \otimes [B_i,B_j]v.
\end{align*}
[/guided]
[/step]
[step:Insert the angular momentum commutation relations for the two factors]
By the hypotheses on the angular momentum triples $A_1,A_2,A_3$ and $B_1,B_2,B_3$, we have
\begin{align*}
[A_i,A_j]u = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} A_k u
\end{align*}
and
\begin{align*}
[B_i,B_j]v = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} B_k v.
\end{align*}
Substituting these identities into the formula from the preceding step gives
\begin{align*}
[J_i,J_j](u \otimes v) = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} A_k u \otimes v + i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} u \otimes B_k v.
\end{align*}
Combining the two finite sums,
\begin{align*}
[J_i,J_j](u \otimes v) = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk}(A_k u \otimes v + u \otimes B_k v).
\end{align*}
By the definition of $J_k$ on elementary tensors,
\begin{align*}
[J_i,J_j](u \otimes v) = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} J_k(u \otimes v).
\end{align*}
[/step]
[step:Extend the identity from elementary tensors to the whole algebraic tensor product]
Let $\xi \in D$. By definition of the algebraic tensor product, there exist a positive integer $m$, vectors $u_1,\dots,u_m \in D_A$, and vectors $v_1,\dots,v_m \in D_B$ such that
\begin{align*}
\xi = \sum_{\ell=1}^{m} u_\ell \otimes v_\ell.
\end{align*}
The operators $J_i$, $J_j$, and $J_k$ are linear on $D$, and therefore $[J_i,J_j]$ is linear on $D$. Applying the elementary-tensor identity term by term gives
\begin{align*}
[J_i,J_j]\xi = i\hbar \sum_{k=1}^{3}\varepsilon_{ijk} J_k \xi.
\end{align*}
Since $i,j \in \{1,2,3\}$ were arbitrary, the triple $(J_1,J_2,J_3)$ satisfies the angular momentum commutation relations on $D_A \odot D_B$.
[/step]
