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Consider a set of operators $\{A_i\}_{i=1}^N$ and their commutation relations defined by $[A_i,A_j]=\Omega_{ij}$. For the description of my problem, let me introduce the following notation: $[i,j]\equiv [A_i, A_j]$ and $[ij,kl]\equiv [A_iA_j,A_kA_l]$.

Let's find out what is the commutator $[ij,kl]$. Since [i,j] = ij-ji, we can write: $$ \begin{equation} [ij,kl] = ijkl - klij \end{equation} $$ I start by transforming $ijkl$: $$ \begin{align} ijkl & = [i,j]kl + jikl \\ & = [i,j]kl + j[i,k]l + jkil \\ & = [i,j]kl + j[i,k]l + jk[i,l] + jkli \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + kjli \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + klji \\ & = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + kl[j,i] + klij. \end{align} $$ So, we get $$ [ij,kl] = [i,j]kl + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + kl[j,i]. $$ On the other hand, if we start with $klij$: $$ \begin{align} klij & = [k,l]ij + lkij \\ & = [k,l]ij + l[k,i]j + likj \\ & = [k,l]ij + l[k,i]j + li[k,j] + lijk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + iljk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ijlk \\ & = [k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ij[l,k] + ijkl. \end{align} $$ So, in this case, we get $$ [ij,kl] = -([k,l]ij + l[k,i]j + li[k,j] + [l,i]jk + i[l,j]k + ij[l,k]) $$ Now if we set all pairwise commutators to zero, except [i,j], then we get a contradiction, since the first version will give non-zero for the commutator, but the second will give zero. For example, consider the commutator $$ [x_0p_0, p_1p_1], ~\textrm{with} ~ [p_i,p_j]=[x_i,x_j]=0 ~ \textrm{and} ~ [x_i,p_j] = i\delta_{ij} $$ Can someone point my mistake as i guess, there should not be any contradiction here.

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In your first worked-out example, you first comuuted $i$ to the right (past the $j$ to the right of it), and then commuted $j$ to the right (past the $i$ which is now at the rightmost position.

In your third displayed equation, and under the assumption that $k$ and $l$ commute with everything, you'll find $$[ij,kl] = \underbrace{[i,j]kl}_a + j[i,k]l + jk[i,l] + [j,k]li + k[j,l]i + \underbrace{kl[j,i]}_b\\=\left[[i,j],kl\right]=0\,.$$ Note that only terms $a$ and $b$ survive, and they cancel each other.

In your procedure, you perform six commutations ($i$ past $j$, $k$, $l$ and $j$ past $k$, $l$, $i$). But you really only need four of them ($i$ and $j$ past $k$ and $l$). In your second computation, you omit the unneccessary $ij$ swith and reswitch (but do it for $kl$), so you dont't notice the problem.

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In the first case you get

$$[ij,kl]=[i,j]kl-kl[i,j]=\left[[i,j],kl\right]=\left[j,[kl,i]\right]+\left[i,[j,kl]\right]$$

using the Jacobi identity. So you see that assuming only $[i,j]$ to be nonzero also reduces the first result to zero.

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