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Given an associative ring, $(R, +, \cdot)$, and the bracket operation defined by $[a,b] = ab-ba$ how can I prove $[a,[b,c]] = 0$ for $a,b,c \in R$? I think the ring must be commutative, so $a(bc-cb) = (bc-cb)a$ and thus $=0$, but I do not know how to prove the ring is commutative.

bgschool
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    Commutativity is stronger than that. You only need the image 9f the bracket to commute with all of $R.$ – Thomas Andrews Nov 29 '21 at 03:11
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    The property may hold in noncommutative rings (if the ring is commutative then you already have $[a,b]=0$ for all $a,b\in R$); but I don't think the property holds for all rings, so you must have some extra condition. – Arturo Magidin Nov 29 '21 at 03:12
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    If $R$ is commutative, then $[a,b] = 0$ for all $a$ and $b$, so the identity $[a,[b,c]]=0$ is trivial. But not every associative ring is commutative, and there are associative rings in which the identity does not hold. Did you maybe mean to prove the Jacobi identity instead? – Alex Kruckman Nov 29 '21 at 03:12
  • In a nilpotent group of class $2$, every commutator is central. There is no analogous name for rings with every commutator central, however. – Geoffrey Trang Nov 29 '21 at 03:29
  • [Non-commutative ring such that $[[x,y],z]=0$](https://math.stackexchange.com/q/2162019/29335) – rschwieb Nov 29 '21 at 15:06
  • In $\mathbb Z\langle x,y,z\rangle$, $[x,[y,z]]\neq 0$, so this doesn't hold for all noncommutative rings. On the other hand, these solutions show a nontrivial noncommutative ring CAN have all commutators central. – rschwieb Nov 29 '21 at 15:10

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