Conditions for equivalence of $[A,BC] = 0$ and $[AB,C] = 0$

Question

What are the most general conditions for which the commutation relations $[A,BC] = 0$ and $[AB,C] = 0$ equivalent? Alternatively, can it at least be shown to hold if $A$, $B$, and $C$ are all involutant? If $A$, $B$, and $C$ are involutant permutation matrices? I am particularly interested in cases for which none of the commutators $[A,B]$, $[B,C]$, and $[C,A]$ are zero.

Answer 1

For involutant matricies in $M_n$ lets take

$$[A,BC] = ABC - BCA = 0$$

and

$$[AB,C] = ABC - CBA = 0$$

Which holds iff

$$BCA = CBA = ABC$$

If $A,B,C$ are all involutant, this is equivalent to $BC = CB = ABCA$.We also have that $ABCA = BC \iff ABC = BCA$. This can be rewritten as:

$$[A,BC] = 0, [B,C]=0$$

Take for example the pauli matricies. They are involutary but no two pauli matricies commute, therefore it is not the case that for involutary matricies we necessarily have that $[A,BC] = [AB,C]=0$.

What about for involutant matricies coming from a group? We can say more.

Let's take our matricies to lie in the representation of a group $\langle a,b,c \rangle \subseteq G \to M_n$ as involutions $a,b,c \mapsto A,B,C$.

Assume for now that $\langle a \rangle, \langle b \rangle, \langle c \rangle$ are normal. We have that $\langle b \rangle \cap \langle c \rangle = \{1\}$, so $\langle b , c \rangle$ is a commutative subgroup of $G$, since both elements are of order two. Likewise $\langle a, b,c \rangle$ is commutative.

What if $\langle a \rangle$ is not normal? For the following let's just assume that $ \langle b,c \rangle$ is commutative and that $ \langle a \rangle$ is not normal.

We know that $\langle bc \rangle$ is of order two. By orbit-stabilizer, $|G:C_G(bc)|=|Cl(bc)|$, where Cl is the conjugacy class of $bc$. If $a \in C_G(bc)$ then the above commutator identity is satisfied. If $a \notin C_G(bc)$ then $a$ is in $gbcg^{-1}C_g(bc)$ for some $g$, in which case, we use the fact that $gbg^{-1}$ is also involutive, and so $a,gbg^{-1},gcg^{-1}$ satisfy your commutator relation.

When $\langle b,c \rangle$ is commutative we have two cases:

(1) $\langle a \rangle$ is normal in $G$, so your commutator identity is satisfied with $[A,B] = [B,C]=0$.

(2) $a, gbg^{-1}, gcg^{-1}$ satisfies your identity for some $g \in G$.

Now you can use the fact that permutation matricies are representations of $S_n \ .$ You can also take $G = \langle a,b,c \rangle$ so that its easier to find those elements $g$.