2-Engel implies class three for Lie rings

Statement
Suppose $$L$$ is a 2-Engel Lie ring. Then, $$L$$ is a nilpotent Lie ring of nilpotency class at most three.

Note that for the proof, we use the cyclic symmetry formulation of the 2-Engel condition: $$[a,[b,c]] = [b,[c,a]] = [c,[a,b]]$$ for all $$a,b,c$$ in the Lie ring.

Similar facts

 * Nilpotency class three is 3-local for Lie rings
 * 2-Engel and 3-torsion-free implies class two for Lie rings
 * 3-Engel and (2,5)-torsion-free implies class six for Lie rings
 * 4-Engel and (2,3,5)-torsion-free implies nilpotent for Lie rings

Analogue in groups

 * 2-Engel implies class three for groups

Applications

 * n-local nilpotency class n implies nilpotency class n plus 1

Facts used

 * 1) uses::2-Engel implies third member of lower central series is in 3-torsion for Lie rings: This say that $$3[a,[b,c]] = 0$$ for all $$a,b,c$$ in the ring.

Direct proof
Given: Lie ring $$L$$ such that $$[a,[b,c]] = [b,[c,a]] = [c,[a,b]]$$ for all $$a,b,c \in L$$. This is the cyclic symmetry formulation of the 2-Engel condition.

To prove: $$[w,[x,[y,z]]] = 0$$ for all $$w,x,y,z \in L$$.

Proof: We assume $$w,x,y,z$$ are fixed for the proof.

Proof via 3-local class
This is a somewhat shorter version of the above proof. The idea is to first use the 2-Engel condition to show that the 3-local class is at most three. Then, we use that nilpotency class three is 3-local for Lie rings.