N-local nilpotency class n implies nilpotency class n

For Lie rings
Suppose $$L$$ is a Lie ring and $$n$$ is a natural number such that $$n \ge 3$$ and the $$n$$-local nilpotency class of $$L$$ is at most $$n$$. Then, $$L$$ is a nilpotent Lie ring and the nilpotency class of $$L$$ is at most $$n$$.

Note that in the case that $$n = 2$$, $$L$$ need not be a Lie ring of nilpotency class two. However, we still must have that $$L$$ is a Lie ring of nilpotency class three.

For groups
 Suppose $$G$$ is a group and $$n$$ is a natural number such that $$n \ge 3$$ and the $$n$$-local nilpotency class of $$G$$ is at most $$n$$. Then, $$G$$ is a nilpotent group and the nilpotency class of $$G$$ is at most $$n$$.

Note that in the case that $$n = 2$$, $$G$$ need not be a group of nilpotency class two. However, we still must have that $$G$$ is a group of nilpotency class three.

Related facts

 * 2-Engel implies class three for Lie rings, 2-Engel implies class three for groups (this is the $$n = 2$$ case)

Facts used

 * 1) uses::Nilpotency class three is 3-local for Lie rings, nilpotency class three is 3-local for groups
 * 2) uses::Relation between local nilpotency class of Lie ring and inner derivation ring, uses::relation between local nilpotency class of group and inner automorphism group,

Proof for Lie rings
Given: A Lie ring $$L$$, $$n \ge 3$$, the $$n$$-local nilpotency class of $$L$$ is at most $$n$$.

To prove: The nilpotency class of $$L$$ is at most $$n$$.

Proof: We use Fact (2) to show that, if we take the quotient of $$L$$ by the $$(n - 3)^{th}$$ member of its Lie ring, i.e., we consider the quotient $$L/Z^{(n-3)}(L)$$, this is a Lie ring whose 3-local nilpotency class is at most 3. Now, we use Fact (1) to show that $$L/Z^{(n-3)}(L)$$ has class at most three. Hence, $$L$$ has class at most $$n$$.

Proof for groups
Given: A group $$G$$, $$n \ge 3$$, the $$n$$-local nilpotency class of $$G$$ is at most $$n$$.

To prove: The nilpotency class of $$G$$ is at most $$n$$.

Proof: We use Fact (2) to show that, if we take the quotient of $$G$$ by the $$(n - 3)^{th}$$ member of its upper central series, i.e., we consider the quotient $$G/Z^{(n-3)}(G)$$, this is a group whose 3-local nilpotency class is at most 3. Now, we use Fact (1) to show that $$G/Z^{(n-3)}(G)$$ has class at most three. Hence, $$G$$ has class at most $$n$$.