Say I have $\left|1,\psi\right>$ which is a vector in one representation of $\mathcal L$, where $\mathcal L$ is a Lie algebra and $\left|2,\phi\right>$ which is a vector in another representation, and A,B $\in$ $\mathcal L$. I want to show:
$$[A,B](\left|1,\psi\right> \otimes \left|2,\phi\right>)=([A,B]\left|1,\psi\right>)\otimes\left|2,\phi\right>+\left|1,\psi\right>\otimes([A,B]\left|2,\phi\right>)$$
I know:
(1) $(v_1+v_2)\otimes w=v_1\otimes w + v_2 \otimes w$
(2) $v \otimes(w_1+w_2)=v\otimes w_1 + v \otimes w_2$
(3) $\alpha(v\otimes w)=(\alpha v)\otimes w=v \otimes (\alpha w)$
(4) $(A\otimes B)(C \otimes D)=AC\otimes BD$
However I'm not sure how any of these help, since (3) only applies if $\alpha$ is a scalar, which [A,B] is not (I don't think), and (1) and (2) only help if you're directly using the tensor product. (4) also doesn't seem to apply because [A,B] does not equal (A $\otimes$ B).
I'm not sure how to initially expand the equation. My instinct says I'd do:
$[A,B](\left|1,\psi\right> \otimes \left|2,\phi\right>)= [A,B]\left|1,\psi\right> \otimes [A,B]\left|2,\phi\right>$
but this just looks wrong to me. To be clear this is part of a problem set in groups and symmetries. I'm not looking for an entire step by step answer, but a hint in the right direction would be greatly appreciated. A hint given in the question is that one intermediate step should have 8 terms.
If $\mathcal{L}$ is a Lie algebra and $V$ and $W$ are two representations of $\mathcal{L}$, then the tensor product representation on $V\otimes W$ is defined by $$X(v\otimes w)=(Xv)\otimes w+v\otimes(Xw)$$ for each $X\in\mathcal{L}$. In particular, letting $X=[A,B]$, the equation you wish to prove is true by definition (the right-hand side is the definition of the left-hand side!).