Commutator between Angular Momentum J and Cross Product I'm trying to show that a cross product $C$ of two vectorial operators (let's say $A$ and $B$)  it's a vector by it's own, which means, I want to show 
$$\left [J_i,C_j  \right ] = i\hbar \epsilon_{ijk}C_k.$$
Attempt of solution:
$A$ and $B$ are vectorial operators, so
$$\left [J_i,A_j  \right ] = i\hbar \epsilon_{ijk}A_k,$$
$$\left [J_i,B_j  \right ] = i\hbar \epsilon_{ijk}B_k.$$
$C$ is the cross product of $A$ and $B$, so
$$C_i =  \epsilon_{ijk}A_jB_k$$
$$C_j =  \epsilon_{jik}A_iB_k$$
$$C_k =  \epsilon_{kij}A_iB_j.$$
Building the commutator of $J_i$ and $C_j$, I found this:
$$\left [ J_i,C_j \right ] = \left [ J_i,\epsilon_{ijk}A_iB_k \right ] = \epsilon_{ijk}(\left [ J_i,A_i\right ]B_k + A_i\left [ J_i,B_k\right ]).$$
The term with $\left [ J_i,A_i\right ]$ is zero, so we have
$$\left [ J_i,C_j \right ] = \epsilon_{ijk}(A_i\left [ J_i,B_k\right ]) = \epsilon_{ijk} A_i (\epsilon_{ikj}B_j) = i\hbar\epsilon_{ikj}\epsilon_{ijk} A_iB_j = i\hbar \epsilon_{ikj}C_k = -i\hbar \epsilon_{ijk}C_k.$$
I was not supposed to find this minus signal, but I cannot find the first mistake in my calculations.
 A: There is the general statement about commutator between momenta operators, $J_i$ with scalar, vector and tensor operators. In your case the answer is
$$[J_i,C_j]\sim\epsilon_{ijk}C_k,$$
where you substitute $C_k=\epsilon_{kab}A_aB_b$, so
$$\epsilon_{ijk}\epsilon_{kab}A_aB_b=\epsilon_{ijk}\epsilon_{abk}A_aB_b=(\delta_{ia}\delta_{jb}-\delta_{ib}\delta_{ja})A_aB_b=A_iB_j-A_jB_i.$$
I am not sure what do you want to show, but let me scetch the derivation of general  fact, $[J_i,C_j]=\epsilon_{ijk}C_K$, where $C_j$ is an arbitrary vector operator. Consider coordinate system $r'$ and coordinate system $r$, which differs in rotation, $R_{\phi}$. Two coordinate systems relates to each other by the relation
$$r'=R_{\phi}r,$$
where $R_{\phi}$ is rotation matrix (and it will be written explicitly below). Then consider an arbitrary vector operator $\hat{C}$. Its components should transform by the similar relation,
$$\hat{C'}=e^{-i\phi(n\cdot J)}\hat{C}e^{+i\phi(n\cdot J)},$$
which comes from the fact that operator $J$ describes rotations. For small angle $\delta\phi$, we can write down
$$\hat{C'}\approx (1-i\delta\phi(J\cdot n))\hat{C}(1+i\delta\phi(J\cdot n))\approx\hat{C}-i\delta \phi [(n\cdot J),\hat{C}]\approx\hat{C}-\delta\phi[n\times\hat{C}],$$
where the last term comes from the Euler relation for rotation by small angle and this term can be rewritten in tensor denotations as
$$n_i[J_i,\hat{C}_j]=-ie_{kil}n_i\hat{C}_l=i\epsilon_{ikl}n_i\hat{С}_l,$$
where you should understand unit vector $n_i$ as a direction of rotation. The last expression gives exactly the desired general statement for an arbitrary vector operator. Hope that this will help.
A: Reading the comments of jacob1729, Artem Alexandrov and Cosmas Zachos, I found out the so many mistakes of my calculations. Here comes the complete solution in case anyone needs this proof in the future:
$C_j$ can be writen as $C_j = \epsilon_{jqr}A_qB_r$, so we have
$$[J_i,C_j] = [J_i,\epsilon_{jqr}A_qB_r] = \epsilon_{jqr} [J_i,A_qB_r] = \epsilon_{jqr} ([J_i,A_q,]B_r + A_q[J_i,B_r])$$
Using the comutation relations for A and B
$$\left [J_i,A_q  \right ] = i\hbar \epsilon_{iqp}A_p$$
$$\left [J_i,B_r  \right ] = i\hbar \epsilon_{irs}B_s$$
we can found 
$$[J_i,C_j] = \epsilon_{jqr} (i\hbar \epsilon_{iqp}A_pB_r + A_qi\hbar \epsilon_{irs}B_s) = i\hbar\epsilon_{qjr}\epsilon_{qip}A_pB_r - i\hbar\epsilon_{rjq}\epsilon_{ris}A_qB_s$$
Remembering the levi-civita "multiplication" identity, we obtain
$$[J_i,C_j] = i\hbar(\delta_{ji}\delta_{rp}-\delta_{jp}\delta_{ri})A_pB_r - i\hbar(\delta_{ji}\delta_{qs}-\delta_{js}\delta_{qi})A_qB_s = i\hbar(A_rB_r - A_jB_i-A_sB_s+A_iB_j)$$
Using the Einstein Sum Notation we can see the equality between $A_rB_r$ and $A_sB_s$, and because of this, these terms annihilate each other.
In the end of the day, we have this:
$$[J_i,C_j] =  i\hbar(A_iB_j- A_jB_i) = i\hbar\epsilon_{ijk}\epsilon_{klm}A_lB_m = i\hbar\epsilon_{ijk}C_k$$
